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CTJxDgKJxL1pyelJYoyhf | physics | simple-harmonic-motion | simple-harmonic-motion | A particle executes simple harmonic motion and is located at x = a, b and c at times t<sub>0</sub>, 2t<sub>0</sub> and 3t<sub>0</sub> respectively. The freqquency of the oscillation is : | [{"identifier": "A", "content": "$${1 \\over {2\\,\\pi \\,{t_0}}}{\\cos ^{ - 1}}\\left( {{{a + c} \\over {2b}}} \\right)$$"}, {"identifier": "B", "content": "$${1 \\over {2\\,\\pi \\,{t_0}}}{\\cos ^{ - 1}}\\left( {{{a + b} \\over {2c}}} \\right)$$"}, {"identifier": "C", "content": "$${1 \\over {2\\,\\pi \\,{t_0}}}{\\co... | ["A"] | null | In general equation of simple harmonic motion, y = A sin $$\omega $$t
<br><br>$$\therefore\,\,\,$$ a = A sin $$\omega $$t<sub>0</sub>
<br><br>$$\,\,\,\,\,\,$$ b = A sin 2$$\omega $$t<sub>0</sub>
<br><br>$$\,\,\,\,\,\,\,$$c = A sin 3$$\omega $$t<sub>0</sub>
<br><br>a + c = A[sin $$\omega $$t<sub>0</sub> + sin 3$$\omega ... | mcq | jee-main-2018-online-16th-april-morning-slot | 12,669 |
SsY8ijRyGV2Y3ETBy1GQ3 | physics | simple-harmonic-motion | simple-harmonic-motion | Two simple harmonic motions, as shown below, are at right angles. They are combined to form Lissajous figures.
<br/>x(t) = A sin (at + $$\delta $$)
<br/>y(t) = B sin (bt)
<br/><br/>Identify the <b>correct</b> match below. | [{"identifier": "A", "content": "<b>Parameters</b> A $$ \\ne $$ B, a = b; $$\\delta $$ = 0;\n<br><b>Curve</b> Parabola"}, {"identifier": "B", "content": "<b>Parameters</b> A = B, a = b; $$\\delta $$ = $${\\raise0.5ex\\hbox{$\\scriptstyle \\pi $}\n\\kern-0.1em/\\kern-0.15em\n\\lower... | ["C"] | null | <p>The given simple harmonic motions to form Lissajous figures are $$x(t) = A\sin (at + \delta )$$ and $$y(t) = B\sin (bt)$$.</p>
<p>For parabola, conditions should be</p>
<p>A = B or A $$\ne$$ B, a = 2b, $$\delta$$ = $$\pi$$/2</p>
<p>For line, conditions should be</p>
<p>A = B, a = b, $$\delta$$ = $$\pi$$</p>
<p>For c... | mcq | jee-main-2018-online-15th-april-evening-slot | 12,670 |
h9jlYSnorP6W5PIDbR1TC | physics | simple-harmonic-motion | simple-harmonic-motion | A particle is executing simple harmonic motion (SHM) of amplitude A, along the x-axis, about x = 0. When its potential Energy (PE) equals kinetic energy (KE), the position of the particle will be : | [{"identifier": "A", "content": "$${A \\over 2}$$"}, {"identifier": "B", "content": "$${A \\over {2\\sqrt 2 }}$$"}, {"identifier": "C", "content": "$${A \\over {\\sqrt 2 }}$$"}, {"identifier": "D", "content": "A"}] | ["C"] | null | Total energy of particle = $${1 \over 2}k{A^2}$$
<br><br>Potential energy (v) = $${1 \over 2}$$ kx<sup>2</sup>
<br><br>Kinetic energy (K) = $${1 \over 2}$$ kA<sup>2</sup> $$-$$ $${1 \over 2}$$kx<sup>2</sup>
<br><br>According to the question,
<br><br>Potential energy = Kinetic energy
<br><br>$$ \therefore $$  ... | mcq | jee-main-2019-online-9th-january-evening-slot | 12,671 |
v2ptj6ziRkQiRpQdcytDD | physics | simple-harmonic-motion | simple-harmonic-motion | A particle undergoing simple harmonic motion has time dependent displacement given by x(t) = Asin$${{\pi t} \over {90}}$$. The ratio of kinetic to potential energy of this particle at t = 210 s will be: | [{"identifier": "A", "content": "$${1 \\over 9}$$"}, {"identifier": "B", "content": "3"}, {"identifier": "C", "content": "2"}, {"identifier": "D", "content": "1"}] | ["B"] | null | K = $${1 \over 2}$$m$${\omega ^2}$$A<sup>2</sup>cos<sup>2</sup>$$\omega $$t
<br><br>U = $${1 \over 2}m{\omega ^2}$$ A<sup>2</sup> sin<sup>2</sup> $$\omega $$t
<br><br>$${k \over U}$$ = cot<sup>2</sup> $$\omega $$t = cot<sup>2</sup> $${\pi \over {90}}$$(210) = $${1 \over 3}$$
<br><br>Hence ratio is 3 (most appropriate) | mcq | jee-main-2019-online-11th-january-morning-slot | 12,673 |
cm46ByyrxvFyOpZp5rXol | physics | simple-harmonic-motion | simple-harmonic-motion | A particle executes simple harmonic motion with an amplitude of 5 cm. When the particle is at 4 cm from the mean position, the magnitude of its velocity in SI units is equal to that of its acceleration. Then, its periodic time in seconds is - | [{"identifier": "A", "content": "$${{4\\pi } \\over 3}$$"}, {"identifier": "B", "content": "$${3 \\over 8}\\pi $$"}, {"identifier": "C", "content": "$${7 \\over 3}\\pi $$"}, {"identifier": "D", "content": "$${{8\\pi } \\over 3}$$"}] | ["D"] | null | $$v = \omega \sqrt {{A^2} - {x^2}} \,\,$$ . . .(1)
<br><br>$$a = - {\omega ^2}x$$ . . .(2)
<br><br>$$\left| v \right| = \left| a \right|$$ &nb... | mcq | jee-main-2019-online-10th-january-evening-slot | 12,674 |
vtHIpHcwHiT8EDHhpTjgy2xukexsohwd | physics | simple-harmonic-motion | simple-harmonic-motion | The displacement time graph of a particle
executing S.H.M is given in figure :<br/>(sketch is
schematic and not to scale)
<img src="data:image/png;base64,UklGRrYQAABXRUJQVlA4IKoQAADQZwCdASr0Ad4APm0ylkkkIqGhIbC6OIANiWlu/HyYw+tQ0v0g/m/5GeBH9k/Jb9yfXf8Q+Ufrv5Mf2D/te7t/EdHN8l90r7H/D/63/pf7h+3Hv5/g/5v+434c+2vkr9gX03/a/4//Rv... | [{"identifier": "A", "content": "(B), (C) and (D)"}, {"identifier": "B", "content": "(A), (B) and (C)"}, {"identifier": "C", "content": "(A) and (D)\n"}, {"identifier": "D", "content": "(A), (B) and (D)"}] | ["B"] | null | (A) F = ma and a = $$-\omega^{2} x$$
<br><br>At $$\frac{3T}{4} $$ displacement zero (x = 0), so a = 0
<br><br>$$ \therefore $$ F = 0
<br><br> (B) at t = T,
Particle is at extreme.
<br><br> displacement (x) = A $$ \Rightarrow $$ x maximum, <br><br>So acceleration is maximum.
<br><br>(C) V = $$\omega \sqrt{A^{2}-x^{2}}... | mcq | jee-main-2020-online-2nd-september-evening-slot | 12,676 |
b287D16YoPT48A5Ui71klrnt483 | physics | simple-harmonic-motion | simple-harmonic-motion | When a particle executes SHM, the nature of graphical representation of velocity as a function of displacement is : | [{"identifier": "A", "content": "circular"}, {"identifier": "B", "content": "straight line"}, {"identifier": "C", "content": "parabolic"}, {"identifier": "D", "content": "elliptical"}] | ["D"] | null | Since, the particle is executing SHM.<br/><br/>Therefore, displacement equation of wave will be<br/><br/>$$y = A\sin \omega t$$<br/><br/>$$ \Rightarrow y/A = \sin \omega t$$<br/><br/>and wave velocity equation will be<br/><br/>$${v_y} = {{dy} \over {dt}} = A\omega \cos \omega t$$<br/><br/>$$ \Rightarrow {v_y}/A\omega ... | mcq | jee-main-2021-online-24th-february-evening-slot | 12,677 |
5nvIqbcRtSrVqlqPO91klt2xeyn | physics | simple-harmonic-motion | simple-harmonic-motion | Y = A sin($$\omega$$t + $$\phi$$<sub>0</sub>) is the time-displacement equation of a SHM. At t = 0 the displacement of the particle is $$Y = {A \over 2}$$ and it is moving along negative x-direction. Then the initial phase angle $$\phi$$<sub>0</sub> will be: | [{"identifier": "A", "content": "$${{5\\pi } \\over 6}$$"}, {"identifier": "B", "content": "$${{\\pi } \\over 3}$$"}, {"identifier": "C", "content": "$${{2\\pi } \\over 3}$$"}, {"identifier": "D", "content": "$${{\\pi } \\over 6}$$"}] | ["A"] | null | <picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265522/exam_images/nnnxfrvnyztmazv46ql1.webp"><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264534/exam_images/cnhwyhqzkhibk3zceapx.webp"><source media="(max-wid... | mcq | jee-main-2021-online-25th-february-evening-slot | 12,678 |
z3Q9KUwuSIlVdPQJO01klune0zd | physics | simple-harmonic-motion | simple-harmonic-motion | A particle executes S.H.M., the graph of velocity as a function of displacement is : | [{"identifier": "A", "content": "a parabola"}, {"identifier": "B", "content": "a helix"}, {"identifier": "C", "content": "an ellipse"}, {"identifier": "D", "content": "a circle"}] | ["C"] | null | For a body performing SHM, relation between velocity and displacement<br><br>$$v = \omega \sqrt {{A^2} - {x^2}} $$<br><br>now, square both side<br><br>$${v^2} = {w^2}({A^2} - {x^2})$$<br><br>$$ \Rightarrow {v^2} = {w^2}{A^2} - {\omega ^2}{x^2}$$<br><br>$${v^2} + {\omega ^2}{x^2} = {\omega ^2}{A^2}$$<br><br>divide whole... | mcq | jee-main-2021-online-26th-february-evening-slot | 12,680 |
fz2CdSfh9kQeBYn46B1klunlynu | physics | simple-harmonic-motion | simple-harmonic-motion | A particle executes S.H.M. with amplitude 'a', and time period 'T'. The displacement of the particle when its speed is half of maximum speed is $${{\sqrt x a} \over 2}$$. The value of x is __________. | [] | null | 3 | For a particle executes S.H.M.<br><br>$$V = \omega \sqrt {{a^2} - {x^2}} $$<br><br>Given, $$V = {{{V_{\max }}} \over 2} = {{A\omega } \over 2}$$<br><br>$${{{A^2}{\omega ^2}} \over 4} = {\omega ^2}{a^2} - {\omega ^2}{x^2}$$<br><br>$$x = {{\sqrt 3 } \over 2}a$$ | integer | jee-main-2021-online-26th-february-evening-slot | 12,681 |
9I5Qk79S5teMX13cDS1kmj25mzo | physics | simple-harmonic-motion | simple-harmonic-motion | For what value of displacement the kinetic energy and potential energy of a simple harmonic oscillation become equal ? | [{"identifier": "A", "content": "x = $${A \\over 2}$$"}, {"identifier": "B", "content": "x = $$\\pm$$ A"}, {"identifier": "C", "content": "x = $$\\pm$$ $${A \\over {\\sqrt 2 }}$$"}, {"identifier": "D", "content": "x = 0"}] | ["C"] | null | KE = PE<br><br>$${1 \over 2}k({A^2} - {X^2}) = {1 \over 2}K{X^2}$$<br><br>$$ \Rightarrow $$ $${A^2} - {X^2} = {X^2}$$<br><br>$$ \Rightarrow $$ $$2{X^2} = {A^2}$$<br><br>$$ \Rightarrow $$ $${X^2} = {{{A^2}} \over {\sqrt 2 }}$$<br><br>$$ \Rightarrow $$ $$X = \pm {A \over {\sqrt 2 }}$$ | mcq | jee-main-2021-online-17th-march-morning-shift | 12,682 |
n2pzyKmh8p6bT9jrLG1kmkrnuqk | physics | simple-harmonic-motion | simple-harmonic-motion | A particle performs simple harmonic motion with a period of 2 second. The time taken by the particle to cover a displacement equal to half of its amplitude from the mean position is $${1 \over a}$$s. The value of 'a' to the nearest integer is _________. | [] | null | 6 | Time period (T) = 2 sec.<br><br>X = A sin ($$\omega$$t + $$\phi$$) ($$\phi$$ = 0 at M.P.)<br><br>$$ \Rightarrow $$ $${A \over 2} = A\sin {{2\pi } \over T}t$$<br><br>$$ \Rightarrow $$ $${{2\pi } \over 2}t = {\pi \over 6}$$<br><br>$$ \Rightarrow $$ $$t = {1 \over 6}$$<br><br>$$ \therefore $$ a = 6 | integer | jee-main-2021-online-18th-march-morning-shift | 12,683 |
AAufQa38qwmRel8etd1kmlhlvfe | physics | simple-harmonic-motion | simple-harmonic-motion | The time period of a simple pendulum is given by $$T = 2\pi \sqrt {{l \over g}} $$. The measured value of the length of pendulum is 10 cm known to a 1mm accuracy. The time for 200 oscillations of the pendulum is found to be 100 second using a clock of 1s resolution. The percentage accuracy in the determination of 'g' u... | [{"identifier": "A", "content": "2%"}, {"identifier": "B", "content": "3%"}, {"identifier": "C", "content": "5%"}, {"identifier": "D", "content": "4%"}] | ["B"] | null | $$T = 2\pi \sqrt {{l \over g}} $$<br><br>$${T^2} = 2\pi \left( {{l \over g}} \right)$$<br><br>$$g = 2\pi {l \over {{T^2}}}$$<br><br>$${{\Delta g} \over g} = {{\Delta l} \over l} + {{2\Delta T} \over T}$$<br><br>$${{\Delta g} \over g} = {{1 \times {{10}^{ - 3}}} \over {1 \times {{10}^{ - 2}}}} + {{2 \times 1} \over {100... | mcq | jee-main-2021-online-18th-march-morning-shift | 12,684 |
FsHhtR64PT7qW01aGE1kmlw7pej | physics | simple-harmonic-motion | simple-harmonic-motion | The function of time representing a simple harmonic motion with a period of $${\pi \over \omega }$$ is : | [{"identifier": "A", "content": "cos($$\\omega$$t) + cos(2$$\\omega$$t) + cos(3$$\\omega$$t)"}, {"identifier": "B", "content": "sin<sup>2</sup>($$\\omega$$t)"}, {"identifier": "C", "content": "sin($$\\omega$$t) + cos($$\\omega$$t)"}, {"identifier": "D", "content": "3cos$$\\left( {{\\pi \\over 4} - 2\\omega t} \\right)... | ["D"] | null | General equation of SHM<br><br>x = A sin($$\omega$$'t $$\pm$$ $$\phi$$)<br><br>We know, $$\omega$$ = $${{2\pi } \over T}$$<br><br>Given, $$T = {\pi \over \omega }$$<br><br>$$ \therefore $$ $$\omega$$' = $${{2\pi } \over {{\pi \over \omega }}}$$ = 2$$\omega$$<br><br>$$ \therefore $$ Equation becomes,<br><br>x = a sin(... | mcq | jee-main-2021-online-18th-march-evening-shift | 12,685 |
1krqc9uhv | physics | simple-harmonic-motion | simple-harmonic-motion | A particle is making simple harmonic motion along the X-axis. If at a distances x<sub>1</sub> and x<sub>2</sub> from the mean position the velocities of the particle are v<sub>1</sub> and v<sub>2</sub> respectively. The time period of its oscillation is given as : | [{"identifier": "A", "content": "$$T = 2\\pi \\sqrt {{{x_2^2 + x_1^2} \\over {v_1^2 - v_2^2}}} $$"}, {"identifier": "B", "content": "$$T = 2\\pi \\sqrt {{{x_2^2 + x_1^2} \\over {v_1^2 + v_2^2}}} $$"}, {"identifier": "C", "content": "$$T = 2\\pi \\sqrt {{{x_2^2 - x_1^2} \\over {v_1^2 + v_2^2}}} $$"}, {"identifier": "D",... | ["D"] | null | $${v^2} = {\omega ^2}({A^2} - {x^2})$$<br><br>$${A^2} = x_1^2 + {{v_1^2} \over {{\omega ^2}}} = x_2^2 + {{v_2^2} \over {{\omega ^2}}}$$<br><br>$${\omega ^2} = {{v_2^2 - v_1^2} \over {x_1^2 - x_2^2}}$$<br><br>$$T = 2\pi \sqrt {{{x_2^2 - x_1^2} \over {v_1^2 - v_2^2}}} $$ | mcq | jee-main-2021-online-20th-july-evening-shift | 12,686 |
1krunrhe8 | physics | simple-harmonic-motion | simple-harmonic-motion | A pendulum bob has a speed of 3 m/s at its lowest position. The pendulum is 50 cm long. The speed of bob, when the length makes an angle of 60$$^\circ$$ to the vertical will be (g = 10 m/s<sup>2</sup>) ____________ m/s. | [] | null | 2 | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264426/exam_images/zyhhmbabjuax5wywnga7.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 25th July Morning Shift Physics - Simple Harmonic Motion Question 68 English Explanation"> <br><br... | integer | jee-main-2021-online-25th-july-morning-shift | 12,687 |
1krywgvzm | physics | simple-harmonic-motion | simple-harmonic-motion | A particle starts executing simple harmonic motion (SHM) of amplitude 'a' and total energy E. At any instant, its kinetic energy is $${{3E} \over 4}$$ then its displacement 'y' is given by : | [{"identifier": "A", "content": "y = a"}, {"identifier": "B", "content": "$$y = {a \\over {\\sqrt 2 }}$$"}, {"identifier": "C", "content": "$$y = {{a\\sqrt 3 } \\over 2}$$"}, {"identifier": "D", "content": "$$y = {a \\over 2}$$"}] | ["D"] | null | $$E = {1 \over 2}K{a^2}$$<br><br>$${{3E} \over 4} = {1 \over 2}K({a^2} - {y^2})$$<br><br>$$ \Rightarrow $$ $${3 \over 4} \times {1 \over 2}K{a^2} = {1 \over 2}K({a^2} - {y^2})$$<br><br>$$ \Rightarrow $$ $${y^2} = {a^2} - {{3{a^2}} \over 4}$$<br><br>$$ \Rightarrow $$ $$y = {a \over 2}$$ | mcq | jee-main-2021-online-27th-july-morning-shift | 12,689 |
1ks0katmf | physics | simple-harmonic-motion | simple-harmonic-motion | An object of mass 0.5 kg is executing simple harmonic motion. It amplitude is 5 cm and time period (T) is 0.2 s. What will be the potential energy of the object at an instant $$t = {T \over 4}s$$ starting from mean position. Assume that the initial phase of the oscillation is zero. | [{"identifier": "A", "content": "0.62 J"}, {"identifier": "B", "content": "6.2 $$\\times$$ 10<sup>$$-$$3</sup> J"}, {"identifier": "C", "content": "1.2 $$\\times$$ 10<sup>3</sup> J"}, {"identifier": "D", "content": "6.2 $$\\times$$ 10<sup>3</sup> J"}] | ["A"] | null | $$T = 2\pi \sqrt {{m \over k}} $$<br><br>$$0.2 = 2\pi \sqrt {{{0.5} \over k}} $$<br><br>k = 50$$\pi$$<sup>2</sup><br><br>$$ \approx $$ 500<br><br>x = A sin ($$\omega$$t + $$\phi$$)<br><br>= 5 cm sin $$\left( {{{\omega T} \over 4} + 0} \right)$$<br><br>= 5 cm sin $$\left( {{\pi \over 2}} \right)$$<br><br>= 5 cm<br><br>... | mcq | jee-main-2021-online-27th-july-evening-shift | 12,690 |
1ktbvioxk | physics | simple-harmonic-motion | simple-harmonic-motion | Two simple harmonic motions are represented by the equations<br/><br/> $${x_1} = 5\sin \left( {2\pi t + {\pi \over 4}} \right)$$ and $${x_2} = 5\sqrt 2 (\sin 2\pi t + \cos 2\pi t)$$. The amplitude of second motion is ................ times the amplitude in first motion. | [] | null | 2 | $${x_2} = 5\sqrt 2 \left( {{1 \over {\sqrt 2 }}\sin 2\pi t + {1 \over {\sqrt 2 }}\cos 2\pi t} \right)\sqrt 2 $$<br><br>$$ = 10\sin \left( {2\pi t + {\pi \over 4}} \right)$$<br><br>$$\therefore$$ $${{{A_2}} \over {{A_1}}} = {{10} \over 5} = 2$$ | integer | jee-main-2021-online-26th-august-evening-shift | 12,692 |
1kte6oj87 | physics | simple-harmonic-motion | simple-harmonic-motion | The variation of displacement with time of a particle executing free simple harmonic motion is shown in the figure.<br/><br/><img src="data:image/png;base64,UklGRnwHAABXRUJQVlA4IHAHAACwNACdASpwAXQAPm02l0gkIyKhI5M6sIANiWlu4MCUZ2dd/6R/zjth/u/9K6j3y97S+aTzAHeG/Cfxn+d+fH94/kXkfwAvSv+U/mf85/zfkq9viAD6Zf3z+gfzb3C/RP7Txon+4/j... | [{"identifier": "A", "content": "<img src=\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734263926/exam_images/efwc50k67zw95pv0ylen.webp\" style=\"max-width: 100%;height: auto;display: block;margin: 0 auto;\" loading=\"lazy\" alt=\"JEE Main 2021 (Online) 27th August Morning Shift Physics - Simple Harmonic Motion... | ["D"] | null | Potential energy is maximum at maximum distance from mean. | mcq | jee-main-2021-online-27th-august-morning-shift | 12,693 |
1ktjpuusx | physics | simple-harmonic-motion | simple-harmonic-motion | For a body executing S.H.M. :<br/><br/>(1) Potential energy is always equal to its K.E.<br/><br/>(2) Average potential and kinetic energy over any given time interval are always equal.<br/><br/>(3) Sum of the kinetic and potential energy at any point of time is constant.<br/><br/>(4) Average K.E. in one time period is ... | [{"identifier": "A", "content": "(3) and (4)"}, {"identifier": "B", "content": "only (3)"}, {"identifier": "C", "content": "(2) and (3)"}, {"identifier": "D", "content": "only (2)"}] | ["A"] | null | In S.H.M. total mechanical energy remains constant and also <k.e.> = <p.e.> = $${{1 \over 4}}$$KA<sup>2</sup> (for 1 time period)</p.e.></k.e.> | mcq | jee-main-2021-online-31st-august-evening-shift | 12,695 |
1l54v46ak | physics | simple-harmonic-motion | simple-harmonic-motion | <p>The motion of a simple pendulum executing S.H.M. is represented by the following equation.</p>
<p>$$y = A\sin (\pi t + \phi )$$, where time is measured in second. The length of pendulum is</p> | [{"identifier": "A", "content": "97.23 cm"}, {"identifier": "B", "content": "25.3 cm"}, {"identifier": "C", "content": "99.4 cm"}, {"identifier": "D", "content": "406.1 cm"}] | ["C"] | null | <p>$$\omega = \pi = \sqrt {{g \over l}} $$</p>
<p>So, $$l = {g \over {{\pi ^2}}}$$</p>
<p>$$ \simeq 99.4$$ cm (Nearest value)</p> | mcq | jee-main-2022-online-29th-june-evening-shift | 12,696 |
1l568jzrd | physics | simple-harmonic-motion | simple-harmonic-motion | <p>Motion of a particle in x-y plane is described by a set of following equations $$x = 4\sin \left( {{\pi \over 2} - \omega t} \right)\,m$$ and $$y = 4\sin (\omega t)\,m$$. The path of the particle will be :</p> | [{"identifier": "A", "content": "circular"}, {"identifier": "B", "content": "helical"}, {"identifier": "C", "content": "parabolic"}, {"identifier": "D", "content": "elliptical"}] | ["A"] | null | <p>$$x = 4\sin \left( {{\pi \over 2} - \omega t} \right)$$</p>
<p>$$ = 4\cos (\omega t)$$</p>
<p>$$y = 4\sin (\omega t)$$</p>
<p>$$ \Rightarrow {x^2} + {y^2} = {4^2}$$</p>
<p>$$\Rightarrow$$ The particle is moving in a circular motion with radius of 4 m.</p> | mcq | jee-main-2022-online-28th-june-morning-shift | 12,697 |
1l56a398u | physics | simple-harmonic-motion | simple-harmonic-motion | <p>A pendulum is suspended by a string of length 250 cm. The mass of the bob of the pendulum is 200 g. The bob is pulled aside until the string is at 60$$^\circ$$ with vertical as shown in the figure. After releasing the bob, the maximum velocity attained by the bob will be ____________ ms<sup>$$-$$1</sup>. (if g = 10 ... | [] | null | 5 | <p>$${1 \over 2}m{v^2} = mgl(1 - \cos \theta )$$</p>
<p>$$ \Rightarrow v = \sqrt {2gl(1 - \cos \theta )} $$</p>
<p>$$ = \sqrt {2 \times 10 \times 2.5 \times {1 \over 2}} $$</p>
<p>$$ = 5$$ m/s</p> | integer | jee-main-2022-online-28th-june-morning-shift | 12,698 |
1l6gmqiah | physics | simple-harmonic-motion | simple-harmonic-motion | <p>When a particle executes Simple Hormonic Motion, the nature of graph of velocity as a function of displacement will be :</p> | [{"identifier": "A", "content": "Circular"}, {"identifier": "B", "content": "Elliptical"}, {"identifier": "C", "content": "Sinusoidal"}, {"identifier": "D", "content": "Straight line"}] | ["B"] | null | <p>Let $$x = A\sin \omega t$$</p>
<p>$$ \Rightarrow v = A\omega \cos \omega t$$</p>
<p>$$ \Rightarrow v = \, \pm \,\omega \sqrt {{A^2} - {x^2}} $$</p>
<p>$$ \Rightarrow {{{v^2}} \over {{\omega ^2}}} + {x^2} = {A^2}$$</p>
<p>$$\Rightarrow$$ Ellipse</p> | mcq | jee-main-2022-online-26th-july-morning-shift | 12,701 |
1ldohglfg | physics | simple-harmonic-motion | simple-harmonic-motion | <p>The amplitude of a particle executing SHM is $$3 \mathrm{~cm}$$. The displacement at which its kinetic energy will be $$25 \%$$ more than the potential energy is: __________ $$\mathrm{cm}$$</p> | [] | null | 2 | $A=3 \mathrm{~cm}$
<br/><br/>$$
\begin{aligned}
& K=1.25 U \\\\
& \Rightarrow K+\frac{K}{1.25}=K_{\max } \\\\
& \Rightarrow \frac{9}{5} K=K_{\max } \\\\
& \Rightarrow \frac{9}{5} \frac{1}{2} m v^{2}=\frac{1}{2} m v_{\max }^{2} \\\\
& \Rightarrow \frac{9}{5}\left[\omega \sqrt{A^{2}-x^{2}}\right]^{2}=\omega^{2} A^{2} ... | integer | jee-main-2023-online-1st-february-morning-shift | 12,702 |
1ldpm3li8 | physics | simple-harmonic-motion | simple-harmonic-motion | <p>The maximum potential energy of a block executing simple harmonic motion is $$25 \mathrm{~J}$$. A is amplitude of oscillation. At $$\mathrm{A / 2}$$, the kinetic energy of the block is</p> | [{"identifier": "A", "content": "9.75 J"}, {"identifier": "B", "content": "37.5 J"}, {"identifier": "C", "content": "18.75 J"}, {"identifier": "D", "content": "12.5 J"}] | ["C"] | null | $\mathrm{U}_{\max }=\frac{1}{2} \mathrm{~m} \omega^{2} \mathrm{~A}^{2}=25 \mathrm{~J}$
<br/><br/>$\mathrm{KE}$ at $\frac{\mathrm{A}}{2}=\frac{1}{2} m v_{1}^{2}=\frac{1}{2} m \omega^{2}\left(A^{2}-\frac{A^{2}}{4}\right)$
<br/><br/>$=\frac{1}{2} \mathrm{~m} \omega^{2} \frac{3 \mathrm{~A}^{2}}{4}=\frac{3}{4}\left(\frac{... | mcq | jee-main-2023-online-31st-january-morning-shift | 12,703 |
ldqw7fqd | physics | simple-harmonic-motion | simple-harmonic-motion | The velocity of a particle executing SHM varies with displacement $(x)$ as $4 v^{2}=50-x^{2}$. The time period of oscillations is $\frac{x}{7} s$. The value of $x$ is ___________. $\left(\right.$ Take $\left.\pi=\frac{22}{7}\right)$ | [] | null | 88 | <p>$$4{v^2} = 50 - {x^2}$$</p>
<p>or $$v = {1 \over 2}\sqrt {50 - {x^2}} $$</p>
<p>Comparing the above equation with $$v = \omega \sqrt {{A^2} - {x^2}} $$</p>
<p>$$ \Rightarrow \omega = {1 \over 2}$$</p>
<p>& $$A = \sqrt {50} $$</p>
<p>so $${{2\pi } \over T} = {1 \over 2}$$</p>
<p>$$ \Rightarrow T = 4\pi \sec $$</p>
<... | integer | jee-main-2023-online-30th-january-evening-shift | 12,704 |
1ldr2x650 | physics | simple-harmonic-motion | simple-harmonic-motion | <p>The general displacement of a simple harmonic oscillator is $$x = A\sin \omega t$$. Let T be its time period. The slope of its potential energy (U) - time (t) curve will be maximum when $$t = {T \over \beta }$$. The value of $$\beta$$ is ______________.</p> | [] | null | 8 | <p>$$U = {1 \over 2}m{\omega ^2}{A^2}{\sin ^2}\omega t$$</p>
<p>So, $${{dU} \over {dt}} = {{m{\omega ^3}{A^2}} \over 2}\sin 2\omega t$$</p>
<p>This value will be maximum when $$\sin 2\omega t = 1$$</p>
<p>or $$2\omega t = {\pi \over 2}$$</p>
<p>$$2 \times {{2\pi } \over T}t = {\pi \over 2}$$</p>
<p>$$ \Rightarrow t =... | integer | jee-main-2023-online-30th-january-morning-shift | 12,705 |
1ldsbkkq8 | physics | simple-harmonic-motion | simple-harmonic-motion | <p>A particle of mass 250 g executes a simple harmonic motion under a periodic force $$\mathrm{F}=(-25~x)\mathrm{N}$$. The particle attains a maximum speed of 4 m/s during its oscillation. The amplitude of the motion is ___________ cm.</p> | [] | null | 40 | <p>$$F = - 25x$$</p>
<p>$$.250{{{d^2}x} \over {d{t^2}}} = - 25x$$</p>
<p>$${{{d^2}x} \over {d{t^2}}} = - 100x$$</p>
<p>$$ \Rightarrow \omega = 10$$ rad/sec</p>
<p>& $$\omega A = {v_{\max }}$$</p>
<p>$$10\,A = 4$$</p>
<p>$$ \Rightarrow A = 0.4$$ m</p>
<p>$$ = 40$$ cm</p> | integer | jee-main-2023-online-29th-january-evening-shift | 12,706 |
1ldtxfpwr | physics | simple-harmonic-motion | simple-harmonic-motion | <p>A particle executes simple harmonic motion between $$x=-A$$ and $$x=+A$$. If time taken by particle to go from $$x=0$$ to $$\frac{A}{2}$$ is 2 s; then time taken by particle in going from $$x=\frac{A}{2}$$ to A is</p> | [{"identifier": "A", "content": "4 s"}, {"identifier": "B", "content": "1.5 s"}, {"identifier": "C", "content": "3 s"}, {"identifier": "D", "content": "2 s"}] | ["A"] | null | $x=A \sin (\omega t)$
<br/><br/>
$$
\begin{aligned}
& x=\frac{A}{2}=A \sin (\omega t) \\\\
& \frac{1}{2}=\sin (\omega t) \\\\
& t=\left(\frac{\pi}{6 \omega}\right)=2 \\\\
& \frac{\pi}{\omega}=12 \sec
\end{aligned}
$$
<br/><br/>
$x=A=A \sin (\omega t)$
<br/><br/>
$\omega t=\left(\frac{\pi}{2}\right)$
<br/><br/>
$t=\left... | mcq | jee-main-2023-online-25th-january-evening-shift | 12,707 |
1ldws9wn6 | physics | simple-harmonic-motion | simple-harmonic-motion | <p>A mass m attached to free end of a spring executes SHM with a period of 1s. If the mass is increased by 3 kg the period of the oscillation increases by one second, the value of mass m is ___________ kg.</p> | [] | null | 1 | $\because 2 \pi \sqrt{\frac{m}{k}}=1$
<br/><br/>
Finally
<br/><br/>
$$
2 \pi \sqrt{\frac{m+3}{k}}=1+1=2
$$
<br/><br/>
Equation $\frac{(1)}{(2)}$ gives
<br/><br/>
$\sqrt{\frac{m}{m+3}}=\frac{1}{2}$
<br/><br/>
$\therefore m=1 \mathrm{~kg}$ | integer | jee-main-2023-online-24th-january-evening-shift | 12,708 |
lgny52r8 | physics | simple-harmonic-motion | simple-harmonic-motion | In a linear Simple Harmonic Motion (SHM)
<br/><br/>
(A) Restoring force is directly proportional to the displacement.
<br/><br/>
(B) The acceleration and displacement are opposite in direction.
<br/><br/>
(C) The velocity is maximum at mean position.
<br/><br/>
(D) The acceleration is minimum at extreme points.
<br/><... | [{"identifier": "A", "content": "${\\text {(A), (B) and (D) only }}$"}, {"identifier": "B", "content": "(C) and (D) only"}, {"identifier": "C", "content": "(A), (B) and (C) Only"}, {"identifier": "D", "content": "(A), (C) and (D) only"}] | ["C"] | null | The correct options are:
<br/><br/>
(A) Restoring force is directly proportional to the displacement. - True (this is a defining characteristic of SHM)
<br/><br/>
(B) The acceleration and displacement are opposite in direction. - True (the acceleration is proportional to the displacement but in the opposite direction)
... | mcq | jee-main-2023-online-15th-april-morning-shift | 12,709 |
1lgp0iu9i | physics | simple-harmonic-motion | simple-harmonic-motion | <p>A particle executes SHM of amplitude A. The distance from the mean position when its's kinetic energy becomes equal to its potential energy is :</p> | [{"identifier": "A", "content": "$$\\frac{1}{\\sqrt{2}} A$$"}, {"identifier": "B", "content": "$$\\frac{1}{2} A$$"}, {"identifier": "C", "content": "$$2 \\mathrm{~A}$$"}, {"identifier": "D", "content": "$$\\sqrt{2 A}$$"}] | ["A"] | null | The total energy of a particle executing simple harmonic motion (SHM) is given by:
<br/><br/>
$$E = \frac{1}{2}m\omega^2A^2$$
<br/><br/>
where $$m$$ is the mass of the particle, $$\omega$$ is the angular frequency of the SHM, and $$A$$ is the amplitude of the motion.
<br/><br/>
At any point during SHM, the kinetic ener... | mcq | jee-main-2023-online-13th-april-evening-shift | 12,710 |
1lgq19504 | physics | simple-harmonic-motion | simple-harmonic-motion | <p>Which graph represents the difference between total energy and potential energy of a particle executing SHM vs it's distance from mean position ?</p> | [{"identifier": "A", "content": "<img src=\"https://app-content.cdn.examgoal.net/fly/@width/image/1lgqyf1yu/b8e018ec-168b-4ebc-800c-3b2143aaffb7/0df41560-e07c-11ed-85e7-0108400886a8/file-1lgqyf1yv.png?format=png\" data-orsrc=\"https://app-content.cdn.examgoal.net/image/1lgqyf1yu/b8e018ec-168b-4ebc-800c-3b2143aaffb7/0df... | ["A"] | null | $$
\begin{aligned}
& \text {T.E. }- \text { P.E. }=\text { K.E. } \\\\
& \text {K.E. }=\frac{1}{2} m \omega^2\left(A^2-x^2\right)
\end{aligned}
$$<br/><br/>
Which is the equation of downward parabola. | mcq | jee-main-2023-online-13th-april-morning-shift | 12,711 |
1lgq3fgwd | physics | simple-harmonic-motion | simple-harmonic-motion | <p>At a given point of time the value of displacement of a simple harmonic oscillator is given as $$\mathrm{y}=\mathrm{A} \cos \left(30^{\circ}\right)$$.
If amplitude is $$40 \mathrm{~cm}$$ and kinetic energy at that time is $$200 \mathrm{~J}$$, the value of force constant is $$1.0 \times 10^{x} ~\mathrm{Nm}^{-1}$$. T... | [] | null | 4 | Given the general equation for displacement in a simple harmonic oscillator:
<br/><br/>
$$x = A \sin(\omega t + \phi)$$
<br/><br/>
At the given time, we have:
<br/><br/>
$$\omega t + \phi = 30^\circ$$
<br/><br/>
Given the amplitude $$A = 40 \,\text{cm}$$ and the displacement $$x = 40 \times \frac{\sqrt{3}}{2} \,\text{c... | integer | jee-main-2023-online-13th-april-morning-shift | 12,712 |
1lgxxnshu | physics | simple-harmonic-motion | simple-harmonic-motion | <p>A particle executes S.H.M. of amplitude A along x-axis. At t = 0, the position of the particle is $$x=\frac{A}{2}$$ and it moves along positive x-axis. The displacement of particle in time t is $$x = A\sin (wt + \delta )$$, then the value of $$\delta$$ will be</p> | [{"identifier": "A", "content": "$$\\frac{\\pi}{2}$$"}, {"identifier": "B", "content": "$$\\frac{\\pi}{3}$$"}, {"identifier": "C", "content": "$$\\frac{\\pi}{4}$$"}, {"identifier": "D", "content": "$$\\frac{\\pi}{6}$$"}] | ["D"] | null | <p>The initial condition states that the particle is at position $x=\frac{A}{2}$ at $t=0$. </p>
<p>If we substitute these initial conditions into the equation for the displacement of the particle:</p>
<p>$x = A \sin(wt + \delta)$</p>
<p>We have:</p>
<p>$\frac{A}{2} = A \sin(\delta)$</p>
<p>Dividing both sides by $A$ gi... | mcq | jee-main-2023-online-10th-april-morning-shift | 12,715 |
1lgyqagtl | physics | simple-harmonic-motion | simple-harmonic-motion | <p>For particle P revolving round the centre O with radius of circular path $$\mathrm{r}$$ and angular velocity $$\omega$$, as shown in below figure, the projection of OP on the $$x$$-axis at time $$t$$ is</p>
<p><img src="data:image/png;base64,UklGRnAQAABXRUJQVlA4IGQQAADQ4wCdASoAAwQCP4G82GW2LqynIdBpQsAwCWlu4WhBG/Pt9Kb... | [{"identifier": "A", "content": "$$x(t)=\\operatorname{cos}\\left(\\omega t-\\frac{\\pi}{6} \\omega\\right)$$"}, {"identifier": "B", "content": "$$x(t)=\\operatorname{cos}(\\omega t)$$"}, {"identifier": "C", "content": "$$x(t)=r \\cos \\left(\\omega t+\\frac{\\pi}{6}\\right)$$"}, {"identifier": "D", "content": "$$x(t)=... | ["C"] | null | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1ljho68xa/3cc2825c-b3c6-4fa4-9ea5-e035d4e39cf1/960bace0-16c5-11ee-84dd-7526dde12945/file-6y3zli1ljho68xb.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1ljho68xa/3cc2825c-b3c6-4fa4-9ea5-e035d4e39cf1/960bace0-16c5-11ee-84... | mcq | jee-main-2023-online-8th-april-evening-shift | 12,716 |
lsbl5lz1 | physics | simple-harmonic-motion | simple-harmonic-motion | A simple pendulum of length $1 \mathrm{~m}$ has a wooden bob of mass $1 \mathrm{~kg}$. It is struck by a bullet of mass $10^{-2} \mathrm{~kg}$ moving with a speed of $2 \times 10^2 \mathrm{~ms}^{-1}$. The bullet gets embedded into the bob. The height to which the bob rises before swinging back is. (use $\mathrm{g}=10 \... | [{"identifier": "A", "content": "$0.20 \\mathrm{~m}$"}, {"identifier": "B", "content": "$0.40 \\mathrm{~m}$"}, {"identifier": "C", "content": "$0.30 \\mathrm{~m}$"}, {"identifier": "D", "content": "$0.35 \\mathrm{~m}$"}] | ["A"] | null | <p>The initial momentum of the system (bullet + bob) is the momentum of the bullet because the bob is initially at rest. The momentum of the bullet is given by its mass times its velocity :</p>
<p>$$ p_{\text{initial}} = m_{\text{bullet}} \times v_{\text{bullet}} $$</p>
<p>After the collision, the bullet and the bob ... | mcq | jee-main-2024-online-1st-february-morning-shift | 12,719 |
jaoe38c1lsc4fpbd | physics | simple-harmonic-motion | simple-harmonic-motion | <p>A particle executes simple harmonic motion with an amplitude of $$4 \mathrm{~cm}$$. At the mean position, velocity of the particle is $$10 \mathrm{~cm} / \mathrm{s}$$. The distance of the particle from the mean position when its speed becomes $$5 \mathrm{~cm} / \mathrm{s}$$ is $$\sqrt{\alpha} \mathrm{~cm}$$, where $... | [] | null | 12 | <p>$$\begin{aligned}
& \mathrm{V}_{\text {at mean position }}=\mathrm{A} \omega \Rightarrow 10=4 \omega \\
& \quad \omega=\frac{5}{2} \\
& \mathrm{~V}=\omega \sqrt{\mathrm{A}^2-\mathrm{x}^2} \\
& 5=\frac{5}{2} \sqrt{4^2-\mathrm{x}^2} \Rightarrow \mathrm{x}^2=16-4 \\
& \mathrm{x}=\sqrt{12} \mathrm{~cm}
\end{aligned}$$</... | integer | jee-main-2024-online-27th-january-morning-shift | 12,720 |
jaoe38c1lscoytpx | physics | simple-harmonic-motion | simple-harmonic-motion | <p>A ball suspended by a thread swings in a vertical plane so that its magnitude of acceleration in the extreme position and lowest position are equal. The angle $$(\theta)$$ of thread deflection in the extreme position will be :</p> | [{"identifier": "A", "content": "$$\\tan ^{-1}\\left(\\frac{1}{2}\\right)$$\n"}, {"identifier": "B", "content": "$$2 \\tan ^{-1}\\left(\\frac{1}{2}\\right)$$\n"}, {"identifier": "C", "content": "$$2 \\tan ^{-1}\\left(\\frac{1}{\\sqrt{5}}\\right)$$\n"}, {"identifier": "D", "content": "$$\\tan ^{-1}(\\sqrt{2})$$"}] | ["B"] | null | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lt1ithfq/d002178a-4d82-4151-9be6-57cd9a588dfe/44dae260-d3de-11ee-b732-196aad4e0551/file-1lt1ithfr.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lt1ithfq/d002178a-4d82-4151-9be6-57cd9a588dfe/44dae260-d3de-11ee-b732-196aad4e0551... | mcq | jee-main-2024-online-27th-january-evening-shift | 12,721 |
jaoe38c1lsf2apoc | physics | simple-harmonic-motion | simple-harmonic-motion | <p>When the displacement of a simple harmonic oscillator is one third of its amplitude, the ratio of total energy to the kinetic energy is $$\frac{x}{8}$$, where $$x=$$ _________.</p> | [] | null | 9 | <p>$$\begin{aligned}
& \text { Let total energy }=\mathrm{E}=\frac{1}{2} \mathrm{KA}^2 \\
& \mathrm{U}=\frac{1}{2} \mathrm{~K}\left(\frac{\mathrm{A}}{3}\right)^2=\frac{\mathrm{KA}^2}{2 \times 9}=\frac{\mathrm{E}}{9} \\
& \mathrm{KE}=\mathrm{E}-\frac{\mathrm{E}}{9}=\frac{8 \mathrm{E}}{9} \\
& \text { Ratio } \frac{\text... | integer | jee-main-2024-online-29th-january-morning-shift | 12,723 |
jaoe38c1lsfm9nix | physics | simple-harmonic-motion | simple-harmonic-motion | <p>A simple harmonic oscillator has an amplitude $$A$$ and time period $$6 \pi$$ second. Assuming the oscillation starts from its mean position, the time required by it to travel from $$x=$$ A to $$x=\frac{\sqrt{3}}{2}$$ A will be $$\frac{\pi}{x} \mathrm{~s}$$, where $$x=$$ _________.</p> | [] | null | 2 | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsr7fmgr/887c7701-605e-4b32-bf15-965c4595219e/e87f24b0-ce31-11ee-9412-cd4f9c6f2c40/file-6y3zli1lsr7fmgs.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1lsr7fmgr/887c7701-605e-4b32-bf15-965c4595219e/e87f24b0-ce31-11ee... | integer | jee-main-2024-online-29th-january-evening-shift | 12,724 |
luz2u43a | physics | simple-harmonic-motion | simple-harmonic-motion | <p>The position, velocity and acceleration of a particle executing simple harmonic motion are found to have magnitudes of $$4 \mathrm{~m}, 2 \mathrm{~ms}^{-1}$$ and $$16 \mathrm{~ms}^{-2}$$ at a certain instant. The amplitude of the motion is $$\sqrt{x}, \mathrm{~m}$$ where $$x$$ is _________.</p> | [] | null | 17 | <p>Let's begin by understanding the equations related to simple harmonic motion (SHM). For a particle executing SHM, the position $$x$$, velocity $$v$$, and acceleration $$a$$ are given by the following equations:</p>
<p>1. Position: $$x = A \cos(\omega t + \phi)$$</p>
<p>2. Velocity: $$v = -A \omega \sin(\omega t + ... | integer | jee-main-2024-online-9th-april-morning-shift | 12,726 |
lv3veg1w | physics | simple-harmonic-motion | simple-harmonic-motion | <p>An object of mass $$0.2 \mathrm{~kg}$$ executes simple harmonic motion along $$x$$ axis with frequency of $$\left(\frac{25}{\pi}\right) \mathrm{Hz}$$. At the position $$x=0.04 \mathrm{~m}$$ the object has kinetic energy $$0.5 \mathrm{~J}$$ and potential energy $$0.4 \mathrm{~J}$$. The amplitude of oscillation is ___... | [] | null | 6 | <p>To solve for the amplitude of oscillation, we start by using the properties of simple harmonic motion (SHM). In SHM, the total energy of the system is conserved and is given by the sum of kinetic energy (KE) and potential energy (PE).</p>
<p>Given:</p>
<ul>
<li>Mass $$m = 0.2 \ \mathrm{kg}$$</li>
<li>Frequency $... | integer | jee-main-2024-online-8th-april-evening-shift | 12,729 |
lv7v4o88 | physics | simple-harmonic-motion | simple-harmonic-motion | <p>A simple pendulum doing small oscillations at a place $$R$$ height above earth surface has time period of $$T_1=4 \mathrm{~s}$$. $$\mathrm{T}_2$$ would be it's time period if it is brought to a point which is at a height $$2 \mathrm{R}$$ from earth surface. Choose the correct relation [$$\mathrm{R}=$$ radius of eart... | [{"identifier": "A", "content": "$$3 \\mathrm{~T}_1=2 \\mathrm{~T}_2$$\n"}, {"identifier": "B", "content": "$$\\mathrm{T}_1=\\mathrm{T}_2$$\n"}, {"identifier": "C", "content": "$$2 \\mathrm{~T}_1=3 \\mathrm{~T}_2$$\n"}, {"identifier": "D", "content": "$$2 \\mathrm{~T}_1=\\mathrm{T}_2$$"}] | ["A"] | null | <p>The time period of a simple pendulum is given by the formula:</p>
<p>$$T = 2\pi \sqrt{\frac{l}{g}}$$</p>
<p>where $$T$$ is the time period, $$l$$ is the length of the pendulum, and $$g$$ is the acceleration due to gravity at the location of the pendulum.</p>
<p>The acceleration due to gravity changes with height ... | mcq | jee-main-2024-online-5th-april-morning-shift | 12,730 |
lvc57pgs | physics | simple-harmonic-motion | simple-harmonic-motion | <p>A particle is doing simple harmonic motion of amplitude $$0.06 \mathrm{~m}$$ and time period $$3.14 \mathrm{~s}$$. The maximum velocity of the particle is _________ $$\mathrm{cm} / \mathrm{s}$$.</p> | [] | null | 12 | <p>For a particle performing simple harmonic motion (SHM), the maximum velocity $$v_{max}$$ can be calculated using the formula:
<p>$v_{max} = A\omega$</p>
<p>where $A$ is the amplitude of the motion and $\omega$ is the angular frequency. The angular frequency $\omega$ is related to the time period $T$ by the formula... | integer | jee-main-2024-online-6th-april-morning-shift | 12,731 |
MFDKspc0MnNCPOLv | physics | simple-harmonic-motion | some-systems-of-executing-shm | A child swinging on a swing in sitting position, stands up, then the time period of the swing will | [{"identifier": "A", "content": "increase "}, {"identifier": "B", "content": "decrease "}, {"identifier": "C", "content": "remains same "}, {"identifier": "D", "content": "increases of the child is long and decreases if the child is short "}] | ["B"] | null | <b>KEY CONCEPT :</b> The time period $$T = 2\pi \sqrt {{\ell \over g}} $$ where
<br><br>$$\ell $$ $$=$$ distance between the point of suspension and the center of mass of the child. This distance decreases when the child stands
<br><br>$$\therefore$$ $$T' < T$$ i.e., the period decreases. | mcq | aieee-2002 | 12,732 |
w8ha9w1H1yXKnYQh | physics | simple-harmonic-motion | some-systems-of-executing-shm | A mass $$M$$ is suspended from a spring of negligible mass. The spring is pulled a little and then released so that the mass executes $$SHM$$ of time period $$T.$$ If the mass is increased by $$m.$$ the time period becomes $${{5T} \over 3}$$. Then the ratio of $${{m} \over M}$$ is | [{"identifier": "A", "content": "$${3 \\over 5}$$ "}, {"identifier": "B", "content": "$${25 \\over 9}$$"}, {"identifier": "C", "content": "$${16 \\over 9}$$"}, {"identifier": "D", "content": "$${5 \\over 3}$$"}] | ["C"] | null | <p>The time period of a simple harmonic motion (SHM) performed by a mass-spring system is given by the formula:</p>
<p>$$T = 2\pi \sqrt{{M \over k}}$$</p>
<p>where:</p>
<ul>
<li>T is the time period,</li>
<li>M is the mass of the object, and</li>
<li>k is the spring constant.</li>
</ul>
<p>We know that if the mass is i... | mcq | aieee-2003 | 12,733 |
WcNVakC4ImT5qVL7 | physics | simple-harmonic-motion | some-systems-of-executing-shm | Two particles $$A$$ and $$B$$ of equal masses are suspended from two massless springs of spring of spring constant $${k_1}$$ and $${k_2}$$, respectively. If the maximum velocities, during oscillation, are equal, the ratio of amplitude of $$A$$ and $$B$$ is | [{"identifier": "A", "content": "$$\\sqrt {{{{k_1}} \\over {{k_2}}}} $$ "}, {"identifier": "B", "content": "$${{{{k_2}} \\over {{k_1}}}}$$ "}, {"identifier": "C", "content": "$$\\sqrt {{{{k_2}} \\over {{k_1}}}} $$ "}, {"identifier": "D", "content": "$${{{{k_1}} \\over {{k_2}}}}$$ "}] | ["C"] | null | Maximum velocity during $$SHM$$ $$ = A\omega = A\sqrt {{k \over m}} $$
<br><br>$$\left[ {\,\,} \right.$$ $$\therefore$$ $$\omega = \sqrt {{k \over m}} $$ $$\left. {\,\,} \right]$$
<br><br>Here the maximum velocity is same and $$m$$ is also same
<br><br>$$\therefore$$ $${A_1}\sqrt {{k_1}} = {A_2}\sqrt {{k_2}} $$
<b... | mcq | aieee-2003 | 12,734 |
sbsnGQLplnIDSD1j | physics | simple-harmonic-motion | some-systems-of-executing-shm | The length of a simple pendulum executing simple harmonic motion is increased by $$21\% $$. The percentage increase in the time period of the pendulum of increased length is | [{"identifier": "A", "content": "$$11\\% $$"}, {"identifier": "B", "content": "$$21\\% $$ "}, {"identifier": "C", "content": "$$42\\% $$"}, {"identifier": "D", "content": "$$10\\% $$"}] | ["D"] | null | <p>The period of a simple pendulum is given by:</p>
<p>$$ T = 2\pi \sqrt{\frac{L}{g}} $$</p>
<p>where:</p>
<ul>
<li>T is the period,</li>
<li>L is the length of the pendulum, and</li>
<li>g is the acceleration due to gravity.</li>
</ul>
<p>Since g is constant, we see that the period T is proportional to the square root... | mcq | aieee-2003 | 12,735 |
65Tec9CL2BozjP1E | physics | simple-harmonic-motion | some-systems-of-executing-shm | The bob of a simple pendulum executes simple harmonic motion in water with a period $$t,$$ while the period of oscillation of the bob is $${t_0}$$ in air. Neglecting frictional force of water and given that the density of the bob is $$\left( {4/3} \right) \times 1000\,\,kg/{m^3}.$$ What relationship between $$t$$ and $... | [{"identifier": "A", "content": "$$t = 2{t_0}$$ "}, {"identifier": "B", "content": "$$t = {t_0}/2$$ "}, {"identifier": "C", "content": "$$t = {t_0}$$ "}, {"identifier": "D", "content": "$$t = 4{t_0}$$ "}] | ["A"] | null | $$t = 2\pi \sqrt {{\ell \over {{g_{eff}}}}} ;\,{t_o}\,\, = 2\pi \sqrt {{\ell \over g}} $$
<br><br><img class="question-image" src="https://imagex.cdn.examgoal.net/9cJCghpFVgj2wXX2Z/rlFjlyIkDfk2ctAXyk2M1v5IhKqTQ/yLr76rdWTkYk7xQv1h3ngS/image.svg" loading="lazy" alt="AIEEE 2004 Physics - Simple Harmonic Motion Question ... | mcq | aieee-2004 | 12,736 |
jbXo9c9iN9BAcsPy | physics | simple-harmonic-motion | some-systems-of-executing-shm | A particle at the end of a spring executes $$S.H.M$$ with a period $${t_1}$$. While the corresponding period for another spring is $${t_2}$$. If the period of oscillation with the two springs in series is $$T$$ then | [{"identifier": "A", "content": "$${T^{ - 1}} = t_1^{ - 1} + t_2^{ - 1}$$ "}, {"identifier": "B", "content": "$${T^2} = t_1^2 + t_2^2$$ "}, {"identifier": "C", "content": "$$T = {t_1} + {t_2}$$ "}, {"identifier": "D", "content": "$${T^{ - 2}} = t_1^{ - 2} + t_2^{ - 2}$$ "}] | ["B"] | null | For first spring, $${t_1} = 2\pi \sqrt {{m \over {{k_1}}}} ,$$
<br><br>For second spring, $${t_2} = 2\pi \sqrt {{m \over {{k_2}}}} $$
<br><br>when springs are in series then, $${k_{eff}} = {{{k_1}{k_2}} \over {{k_1} + {k_2}}}$$
<br><br>$$\therefore$$ $$T = 2\pi \sqrt {{{m\left( {{k_1} + {k_2}} \right)} \over {{k_1}{k_2... | mcq | aieee-2004 | 12,737 |
DwvQomzRTsQjLmej | physics | simple-harmonic-motion | some-systems-of-executing-shm | A particle of mass $$m$$ is attached to a spring (of spring constant $$k$$) and has a natural angular frequency $${\omega _0}.$$ An external force $$F(t)$$ proportional to $$\cos \,\omega t\left( {\omega \ne {\omega _0}} \right)$$ is applied to the oscillator. The time displacement of the oscillator will be proportion... | [{"identifier": "A", "content": "$${1 \\over {m\\left( {\\omega _0^2 + {\\omega ^2}} \\right)}}$$"}, {"identifier": "B", "content": "$${1 \\over {m\\left( {\\omega _0^2 - {\\omega ^2}} \\right)}}$$ "}, {"identifier": "C", "content": "$${m \\over {\\omega _0^2 - {\\omega ^2}}}$$ "}, {"identifier": "D", "content": "$${m ... | ["B"] | null | Given that, initial angular velocity = $${\omega _0}$$
<br><br>and at any instant time t, angular velocity = $$\omega $$
<br><br>So when displacement is x then the resultant acceleration
<br><br>f = $$\left( {\omega _0^2 - {\omega ^2}} \right)x$$
<br><br>So the external force, F = $$m\left( {\omega _0^2 - {\omega ^2}}... | mcq | aieee-2004 | 12,738 |
YeqfjfzgZ2NMPf7Q | physics | simple-harmonic-motion | some-systems-of-executing-shm | The bob of a simple pendulum is a spherical hollow ball filled with water. A plugged hole near the bottom of the oscillating bob gets suddenly unplugged. During observation, till water is coming out, the time period of oscillation would | [{"identifier": "A", "content": "first decrease and then increase to the original value"}, {"identifier": "B", "content": "first increase and then decrease to the original value"}, {"identifier": "C", "content": "increase towards a saturation value"}, {"identifier": "D", "content": "remain unchanged "}] | ["B"] | null | Center of mass of combination of liquid and hollow portion (at position $$\ell $$ ), first goes down (to $$\ell + \Delta \ell $$) and when total water is drained out, center of mass regain its original position (to $$\ell $$),
$$$T = 2\pi \sqrt {{\ell \over g}} $$$
<br><br>$$\therefore$$ $$'T'$$ first increases and ... | mcq | aieee-2005 | 12,739 |
2bzyhJzXtz4R5ytc | physics | simple-harmonic-motion | some-systems-of-executing-shm | An ideal gas enclosed in a vertical cylindrical container supports a freely moving piston of mass $$M.$$ The piston and the cylinder have equal cross sectional area $$A$$. When the piston is in equilibrium, the volume of the gas is $${V_0}$$ and its pressure is $${P_0}.$$ The piston is slightly displaced from the equil... | [{"identifier": "A", "content": "$${1 \\over {2\\pi }}\\,{{A\\gamma {P_0}} \\over {{V_0}M}}$$ "}, {"identifier": "B", "content": "$${1 \\over {2\\pi }}\\,{{{V_0}M{P_0}} \\over {{A^2}\\gamma }}$$ "}, {"identifier": "C", "content": "$${1 \\over {2\\pi }}\\,\\sqrt {{{A\\gamma {P_0}} \\over {{V_0}M}}} $$ "}, {"identifier":... | ["C"] | null | $${{Mg} \over A} = {P_0}$$
<br><br>$$Mg = {P_0}A\,\,\,\,...\left( 1 \right)$$
<br><br>$${P_0}V_0^\gamma = P{V^\gamma }$$
<br><br>$$P = {{{P_0}x_0^\gamma } \over {{{\left( {{x_0} - x} \right)}^y}}}$$
<br><br>Let piston is displaced by distance $$x$$
<br><br>$$Mg - \left( {{{{P_0}x_0^\gamma } \over {{{\left( {{x_0} - x}... | mcq | jee-main-2013-offline | 12,740 |
joTsPVk49bxuHH0a | physics | simple-harmonic-motion | some-systems-of-executing-shm | For a simple pendulum, a graph is plotted between its kinetic energy $$(KE)$$ and potential energy $$(PE)$$ against its displacement $$d.$$ Which one of the following represents these correctly?
<br/>$$(graphs$$ $$are$$ $$schematic$$ $$and$$ $$not$$ $$drawn$$ $$to$$ $$scale)$$ | [{"identifier": "A", "content": "<img src=\"https://app-content.cdn.examgoal.net/fly/@width/image/1l91eit1d/7a6bfae6-634a-4eb5-a695-bd4780392ae7/23d23f10-47d9-11ed-8284-6d7e98c66709/file-1l91eit1e.png?format=png\" data-orsrc=\"https://app-content.cdn.examgoal.net/image/1l91eit1d/7a6bfae6-634a-4eb5-a695-bd4780392ae7/23d... | ["D"] | null | $$K.E = {1 \over 2}k\left( {{A^2} - {d^2}} \right)$$
<br><br>and $$P.E. = {1 \over 2}k{d^2}$$
<br><br>At mean position $$d=0.$$ At extremes positions $$d=A$$ | mcq | jee-main-2015-offline | 12,742 |
0biZwoHVNUIxBTRk | physics | simple-harmonic-motion | some-systems-of-executing-shm | The period of oscillation of a simple pendulum is $$T = 2\pi \sqrt {{L \over g}} $$. Measured value of L is 20.0 cm known to 1 mm accuracy and time for 100 oscillations of the pendulum is found to be 90 s using wrist watch of 1 s resolution. The accuracy in the determination of g is: | [{"identifier": "A", "content": "1 %"}, {"identifier": "B", "content": "5 %"}, {"identifier": "C", "content": "2 %"}, {"identifier": "D", "content": "3 %"}] | ["D"] | null | Given $$T = 2\pi \sqrt {{L \over g}} $$
<br><br>$$ \Rightarrow g = {{4{\pi ^2}L} \over {{T^2}}}$$
<br><br>$$ \Rightarrow g = {{4{\pi ^2}L{n^2}} \over {{t^2}}}$$
<br><br>[ as $$T = {t \over n}$$ ]
<br><br>So, percentage error in $$g$$ =
<br><br>$${{\Delta g} \over g} \times 100 = {{\Delta L} \over L} \times 100 + 2{{\De... | mcq | jee-main-2015-offline | 12,743 |
nac8kY5q9RGKZP0u | physics | simple-harmonic-motion | some-systems-of-executing-shm | A silver atom in a solid oscillates in simple harmonic motion in some direction with a frequency of 10<sup>12</sup>/sec. What is the force constant of the bonds connecting one atom with the other? (Mole wt. of silver = 108 and Avogadro number = 6.02 × 10<sup>23</sup> gm mole<sup>–1</sup>) | [{"identifier": "A", "content": "5.5 N/m"}, {"identifier": "B", "content": "6.4 N/m"}, {"identifier": "C", "content": "7.1 N/m"}, {"identifier": "D", "content": "2.2 N/m"}] | ["C"] | null | 6.02 $$ \times $$ 10<sup>23</sup> atoms of silver = 108 gm
<br><br>1 atoms of silver = $${{108 \times {{10}^{ - 3}}} \over {6.02 \times {{10}^{23}}}}$$ kg
<br><br>For a harmonic oscillator
<br><br>f = $${1 \over {2\pi }}$$ $$\sqrt {{k \over m}} $$
<br><br... | mcq | jee-main-2018-offline | 12,744 |
wIrfCSO0LoC6EEX0BUqZP | physics | simple-harmonic-motion | some-systems-of-executing-shm | An oscillator of mass M is at rest in its equilibrium position in a potential
<br/>V = $${1 \over 2}$$ k(x $$-$$ X)<sup>2</sup>. A particle of mass m comes from right with speed u and collides completely inelastically with M and sticks to it. This process repeats every time the oscillator crosses its equilibrium pos... | [{"identifier": "A", "content": "$${1 \\over {\\sqrt 3 }}$$"}, {"identifier": "B", "content": "$${1 \\over 2}$$"}, {"identifier": "C", "content": "$${2 \\over 3}$$"}, {"identifier": "D", "content": "$${3 \\over {\\sqrt 5 }}$$"}] | ["A"] | null | <p>Potential of the given oscillator is</p>
<p>$$V = {1 \over 2}k{(x - k)^2}$$</p>
<p>Given: M = 10; m = 5, u = 1; k = 1</p>
<p>Initial momentum of the particle of mass m</p>
<p>= mu = m $$\times$$ 5 = 5m</p>
<p>Momentum of (oscillator + particle) after collision = (M + m)</p>
<p>Velocity of oscillator after collision ... | mcq | jee-main-2018-online-16th-april-morning-slot | 12,745 |
hwSXkF7lcurZVhTMigBzk | physics | simple-harmonic-motion | some-systems-of-executing-shm | A cylindrical plastic bottle of negligible mass is filled with 310 ml of water and left floating in a pond with still water. If pressed downward slightly and released, it starts performing simple harmonic motion at angular frequency $$\omega $$. If the radius of the bottle is 2.5 cm then $$\omega $$ is close to – (dens... | [{"identifier": "A", "content": "2.50 rad s<sup>$$-$$1</sup>"}, {"identifier": "B", "content": "3.75 rad s<sup>$$-$$1</sup>"}, {"identifier": "C", "content": "5.00 rad s<sup>$$-$$1</sup>"}, {"identifier": "D", "content": "7.90 rad s<sup>$$-$$1</sup>"}] | ["D"] | null | Restoring force due to pressing the bottle with small
amount x,
<br><br>F = $$ - \left( {\rho Ax} \right)g$$
<br><br>$$ \Rightarrow $$ ma = $$ - \left( {\rho Ax} \right)g$$
<br><br>$$ \Rightarrow $$ a = $$ - \left( {{{\rho Ag} \over m}} \right)x$$
<br><br>$$ \therefore $$ $${{\omega ^2} = {{\rho Ag} \over m}}$$ = $${{{... | mcq | jee-main-2019-online-10th-january-evening-slot | 12,746 |
R70yewfdGLHeUwBVHXqNz | physics | simple-harmonic-motion | some-systems-of-executing-shm | The mass and the diameter of a planet are three times the respective values for the Earth. The period of oscillation of simple pendulum on the Earth is 2 s. The period of oscillation of the same pendulum on the planet would be :
| [{"identifier": "A", "content": "$${{\\sqrt 3 } \\over 2}$$ s"}, {"identifier": "B", "content": "$${3 \\over 2}$$ s"}, {"identifier": "C", "content": "$${2 \\over {\\sqrt 3 }}$$ s"}, {"identifier": "D", "content": "$$2\\sqrt 3 $$ s"}] | ["D"] | null | $$ \because $$ g = $${{GM} \over {{R^2}}}$$
<br><br>$${{{g_p}} \over {{g_e}}}$$ = $${{{M_e}} \over {{M_e}}}{\left( {{{{{\mathop{\rm R}\nolimits} _e}} \over {{R_p}}}} \right)^2}$$ = 3$${\left( {{1 \over 3}} \right)^2}$$ = $${{1 \over 3}}$$
<br><br>Also T $$ \propto $$ $${1 \over {\sqrt g }}$$
<br><br>... | mcq | jee-main-2019-online-11th-january-evening-slot | 12,748 |
CEgJFbJzD7arRv9ZZwXQl | physics | simple-harmonic-motion | some-systems-of-executing-shm | A pendulum is executing simple harmonic motion and its maximum kinetic energy is K<sub>1</sub>. If the length of the pendulum is doubled and it performs simple harmonic motion with the same amplitude as in the first case, its maximum kinetic energy is K<sub>2</sub>. Then : | [{"identifier": "A", "content": "$${K_2}$$ = $${{{K_1}} \\over 2}$$"}, {"identifier": "B", "content": "K<sub>2</sub> = 2K<sub>1</sub>"}, {"identifier": "C", "content": "K<sub>2</sub> = K<sub>1</sub>"}, {"identifier": "D", "content": "K<sub>2</sub> = $${{{K_1}} \\over 4}$$"}] | ["B"] | null | Maximum kinetic energy at lowest point B is given by
<br><br>K = mg$$\ell $$ (1 $$-$$ cos $$\theta $$)
<br><br>where $$\theta $$ = angular amp.
<br><br><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264228/exam_images/hrjmcfhblhhvxx0gmcae.webp" style="max-width: 100%; height: auto;display: block;marg... | mcq | jee-main-2019-online-11th-january-evening-slot | 12,749 |
BYsxQYSbAykOPe28jcSIQ | physics | simple-harmonic-motion | some-systems-of-executing-shm | A rod of mass 'M' and length '2L' is suspended at its middle by a wire. It exhibits torsional oscillations; If two masses each of 'm' are attached at distance 'L/2' from its centre on both sides, it reduces the oscillation frequency by 20%. The value of radio m/M is close to : | [{"identifier": "A", "content": "0.77"}, {"identifier": "B", "content": "0.57"}, {"identifier": "C", "content": "0.37"}, {"identifier": "D", "content": "0.17"}] | ["C"] | null | Initially :
<br><br><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267419/exam_images/l10osugcy8cmollgpksg.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 9th January Evening Slot Physics - Simple Harmonic Motion Question 111 Engli... | mcq | jee-main-2019-online-9th-january-evening-slot | 12,750 |
T22xjk5EZvVaBxf9n77k9k2k5kupnnb | physics | simple-harmonic-motion | some-systems-of-executing-shm | A spring mass system (mass m, spring
constant k and natural length $$l$$) rest in
equilibrium on a horizontal disc. The free end
of the spring is fixed at the centre of the disc.
If the disc together with spring mass system,
rotates about it's axis with an angular velocity
$$\omega $$, (k $$ \gg m{\omega ^2}$$) the rel... | [{"identifier": "A", "content": "$${{m{\\omega ^2}} \\over {3k}}$$"}, {"identifier": "B", "content": "$${{m{\\omega ^2}} \\over k}$$"}, {"identifier": "C", "content": "$${{2m{\\omega ^2}} \\over k}$$"}, {"identifier": "D", "content": "$$\\sqrt {{2 \\over 3}} \\left( {{{m{\\omega ^2}} \\over k}} \\right)$$"}] | ["B"] | null | m$${\omega ^2}$$(l<sub>0</sub> + x) = kx
<br><br>x = $${{m{I_0}{\omega ^2}} \over {k - m{\omega ^2}}}$$
<br><br>For k >> m$${\omega ^2}$$
<br><br>$${x \over {{I_0}}} = {{m{\omega ^2}} \over k}$$ | mcq | jee-main-2020-online-9th-january-evening-slot | 12,751 |
38P6BbDb1DA1trTd3vjgy2xukf3qtlip | physics | simple-harmonic-motion | some-systems-of-executing-shm | A block of mass m attached to a massless spring is performing oscillatory motion of amplitude ‘A’ on
a frictionless horizontal plane. If half of the mass of the block breaks off when it is passing through
its equilibrium point, the amplitude of oscillation for the remaining system become fA. The value of
f is : | [{"identifier": "A", "content": "1"}, {"identifier": "B", "content": "$${1 \\over 2}$$"}, {"identifier": "C", "content": "$$\\sqrt 2 $$"}, {"identifier": "D", "content": "$${1 \\over {\\sqrt 2 }}$$"}] | ["D"] | null | At equilibrium position
<br><br>V<sub>0</sub> = V
<br><br>$${V_0} = {\omega _1}A = \sqrt {{K \over m}} A$$ .....(i)<br><br>$$V = \omega {A^1} = \sqrt {{K \over {{m \over 2}}}} {A^1}$$ .....(ii)<br><br>$$ \therefore $$ $${A^1} = {A \over {\sqrt 2 }}$$ | mcq | jee-main-2020-online-3rd-september-evening-slot | 12,752 |
xEV6aw10Ry8j2kpRUljgy2xukg0bbj8f | physics | simple-harmonic-motion | some-systems-of-executing-shm | When a particle of mass m is attached to a vertical spring of spring constant k and released, its
motion is described by <br/>y(t) = y<sub>0</sub>
sin<sup>2</sup> $$\omega $$t, where 'y' is measured from the lower end of unstretched
spring. Then $$\omega $$ is:
| [{"identifier": "A", "content": "$$\\sqrt {{g \\over {{y_0}}}} $$"}, {"identifier": "B", "content": "$${1 \\over 2}\\sqrt {{g \\over {{y_0}}}} $$"}, {"identifier": "C", "content": "$$\\sqrt {{{2g} \\over {{y_0}}}} $$"}, {"identifier": "D", "content": "$$\\sqrt {{g \\over {2{y_0}}}} $$"}] | ["D"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266008/exam_images/dop75w6vfxnziv56e0zk.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2020 (Online) 6th September Evening Slot Physics - Simple Harmonic Motion Question 95 English Explanation">
<br>... | mcq | jee-main-2020-online-6th-september-evening-slot | 12,753 |
SAxJdNqVo2A6NMP85s1klrgzacs | physics | simple-harmonic-motion | some-systems-of-executing-shm | In the given figure, a mass M is attached to a horizontal spring which is fixed on one side to a rigid support. The spring constant of the spring is k. The mass oscillates on a frictionless surface with time period T and amplitude A. When the mass is in equilibrium position, as shown in the figure, another mass m is ge... | [{"identifier": "A", "content": "$$A\\sqrt {{M \\over {M - m}}} $$"}, {"identifier": "B", "content": "$$A\\sqrt {{{M - m} \\over M}} $$"}, {"identifier": "C", "content": "$$A\\sqrt {{{M + m} \\over M}} $$"}, {"identifier": "D", "content": "$$A\\sqrt {{M \\over {M + m}}} $$"}] | ["D"] | null | Given, initial amplitude = A<br/><br/>Velocity at mean position, v = A$$\omega$$<br/><br/>Applying conservation of momentum at mean position, we get<br/><br/>M<sub>1</sub>v<sub>1</sub> = M<sub>2</sub>v<sub>2</sub><br/><br/>MA$$\omega$$ = (M + m)v'<br/><br/>$$ \Rightarrow v' = {{MA\omega } \over {M + m}} = {{MA\sqrt {{k... | mcq | jee-main-2021-online-24th-february-morning-slot | 12,754 |
7s56qVOX3CoF6bwuSs1klrncgcy | physics | simple-harmonic-motion | some-systems-of-executing-shm | In the given figure, a body of mass M is held between two massless springs, on a smooth inclined plane. The free ends of the springs are attached to firm supports. If each spring has spring constant k, the frequency of oscillation of given body is :<br/><br/>
<img src="data:image/png;base64,UklGRgQPAABXRUJQVlA4IPgOAAAQ... | [{"identifier": "A", "content": "$${1 \\over {2\\pi }}\\sqrt {{{2k} \\over {Mg\\sin \\alpha }}} $$"}, {"identifier": "B", "content": "$${1 \\over {2\\pi }}\\sqrt {{k \\over {Mg\\sin \\alpha }}} $$"}, {"identifier": "C", "content": "$${1 \\over {2\\pi }}\\sqrt {{{2k} \\over M}} $$"}, {"identifier": "D", "content": "$${1... | ["C"] | null | Let T be the time period of oscillation, then<br><br>$$T = 2\pi \sqrt {{M \over {{k_{eq}}}}} $$<br><br>$$\therefore$$ $$T = 2\pi \sqrt {{M \over {2k}}} $$ [$$\because$$ $${k_{eq}} = k + k$$]<br><br><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1kxkk82zr/b235bdc5-c1db-419d-af37-20c5529fafe5/bc2d8370-64... | mcq | jee-main-2021-online-24th-february-evening-slot | 12,755 |
u7GYMMt9l9aVfRHUS61klrnvm02 | physics | simple-harmonic-motion | some-systems-of-executing-shm | The period of oscillation of a simple pendulum is $$T = 2\pi \sqrt {{L \over g}} $$. Measured value of 'L' is 1.0 m from meter scale having a minimum division of 1 mm and time of one complete oscillation is 1.95 s measured from stopwatch of 0.01 s resolution. The percentage error in the determination of 'g' will be : | [{"identifier": "A", "content": "1.30%"}, {"identifier": "B", "content": "1.33%"}, {"identifier": "C", "content": "1.13%"}, {"identifier": "D", "content": "1.03%"}] | ["C"] | null | Given, $$T = 2\pi \sqrt {{L \over g}} $$ .... (i)<br/><br/>where, time period, T = 1.95 s<br/><br/>Length of string, l = 1 m<br/><br/>Acceleration due to gravity = g<br/><br/>Error in time period, $$\Delta$$T = 0.01 s = 10<sup>$$-$$2</sup> s<br/><br/>Error in length, $$\Delta$$L = 1 mm = 1 $$\times$$ 10<sup>$$-$$3</sup... | mcq | jee-main-2021-online-24th-february-evening-slot | 12,756 |
4J2JMncafkZuGsDoO01klt2zt3w | physics | simple-harmonic-motion | some-systems-of-executing-shm | Two identical springs of spring constant '2k' are attached to a block of mass m and to fixed support (see figure). When the mass is displaced from equilibrium position on either side, it executes simple harmonic motion. The time period of oscillations of this system is :<br/><br/><img src="data:image/png;base64,UklGRvQ... | [{"identifier": "A", "content": "$$2\\pi \\sqrt {{m \\over k}} $$"}, {"identifier": "B", "content": "$$\\pi \\sqrt {{m \\over k}} $$"}, {"identifier": "C", "content": "$$2\\pi \\sqrt {{m \\over {2k}}} $$"}, {"identifier": "D", "content": "$$\\pi \\sqrt {{m \\over {2k}}} $$"}] | ["B"] | null | Due to parallel combination K<sub>eff</sub> = 2k + 2k = 4k<br><br>$$\because$$ $$T = 2\pi \sqrt {{m \over {{k_{eff}}}}} $$<br><br>$$ = 2\pi \sqrt {{m \over {4k}}} $$<br><br>$$T = \pi \sqrt {{m \over k}} $$ | mcq | jee-main-2021-online-25th-february-evening-slot | 12,758 |
hD12lNb0C5ziDINBjA1klthpg5z | physics | simple-harmonic-motion | some-systems-of-executing-shm | Assume that a tunnel is dug along a chord of the earth, at a perpendicular distance (R/2) from the earth's centre, where 'R' is the radius of the Earth. The wall of the tunnel is frictionless. If a particle is released in this tunnel, it will execute a simple harmonic motion with a time period : | [{"identifier": "A", "content": "$$2\\pi \\sqrt {{R \\over g}} $$"}, {"identifier": "B", "content": "$${g \\over {2\\pi R}}$$"}, {"identifier": "C", "content": "$${{2\\pi R} \\over g}$$"}, {"identifier": "D", "content": "$${1 \\over {2\\pi }}\\sqrt {{g \\over R}} $$"}] | ["A"] | null | <picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266658/exam_images/zun8khvyufh7t9b4xo32.webp"><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264515/exam_images/jhpcteln5ehnlo5lrvo7.webp"><source media="(max-wid... | mcq | jee-main-2021-online-26th-february-morning-slot | 12,759 |
fRNuoiDVSnesqdH7At1klunj0s9 | physics | simple-harmonic-motion | some-systems-of-executing-shm | Given below are two statements :<br/><br/>Statement I : A second's pendulum has a time period of 1 second.<br/><br/>Statement II : It takes precisely one second to move between the two extreme positions.<br/><br/>In the light of the above statements, choose the correct answer from the options given below : | [{"identifier": "A", "content": "Both Statement I and Statement II are false"}, {"identifier": "B", "content": "Statement I is false but Statement II is true"}, {"identifier": "C", "content": "Both Statement I and Statement II are true"}, {"identifier": "D", "content": "Statement I is true but Statement II is false"}] | ["B"] | null | As we know time period of second’s penduklum is 2 sec, so statement (1) is incorrect.
<br><br>Time taken by particle performing SHM between two
extreme position is half of the time period.
<br><br>Here, T = 2 sec.
<br><br>So, time = 2/2 = 1 sec | mcq | jee-main-2021-online-26th-february-evening-slot | 12,761 |
rijiXi6kO44NZgT6mc1klunnhpm | physics | simple-harmonic-motion | some-systems-of-executing-shm | Time period of a simple pendulum is T. The time taken to complete $${5 \over 8}$$ oscillations starting from mean position is $${\alpha \over \beta }T$$. The value of $$\alpha$$ is _________. | [] | null | 7 | <picture><source media="(max-width: 1319px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263899/exam_images/n06belmbt6uhyxiksqjr.webp"><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267397/exam_images/jiy9gt4bip3dapgpzorj.webp"><source media="(max-wi... | integer | jee-main-2021-online-26th-february-evening-slot | 12,762 |
rCX9fIpzCJjNkEfGtf1kmhntg12 | physics | simple-harmonic-motion | some-systems-of-executing-shm | Time period of a simple pendulum is T inside a lift when the lift is stationary. If the lift moves upwards with an acceleration g/2, the time period of pendulum will be : | [{"identifier": "A", "content": "$$\\sqrt 3 T$$"}, {"identifier": "B", "content": "$$\\sqrt {{2 \\over 3}} T$$"}, {"identifier": "C", "content": "$${T \\over {\\sqrt 3 }}$$"}, {"identifier": "D", "content": "$$\\sqrt {{3 \\over 2}} T$$"}] | ["B"] | null | When lift is stationary<br><br>$$T = 2\pi \sqrt {{L \over g}} $$<br><br>A pseudo force will act downwards when lift is moving upwards.<br><br>$$ \therefore $$ $${g_{eff}} = g + {g \over 2} = {{3g} \over 2}$$<br><br>$$ \therefore $$ New time period<br><br>$$T' = 2\pi \sqrt {{L \over {{g_{eff}}}}} $$<br><br>$$T' = 2\pi \... | mcq | jee-main-2021-online-16th-march-morning-shift | 12,763 |
tVpFKsoxk44JyaLDHZ1kmj46n3a | physics | simple-harmonic-motion | some-systems-of-executing-shm | Consider two identical springs each of spring constant k and negligible mass compared to the mass M as shown. Fig. 1 shows one of them and Fig. 2 shows their series combination. The ratios of time period of oscillation of the two SHM is T<sub>b</sub>/T<sub>a</sub> = $$\sqrt x $$, where value of x is ___________. (Round... | [] | null | 2 | $${T_a} = 2\pi \sqrt {{M \over k}} $$
<br><br>$${K_{e{q_{series}}}} = {{{k_1}{k_2}} \over {{k_1} + {k_2}}} = {k \over 2}$$<br><br>$${T_b} = 2\pi \sqrt {{M \over {k/2}}} = 2\pi \sqrt {{{2M} \over k}} $$<br><br>$${T_b} = \sqrt 2 {T_a}$$<br><br>$${{{T_b}} \over {{T_a}}} = \sqrt 2 $$<br><br>$$ \therefore $$ x = 2 | integer | jee-main-2021-online-17th-march-morning-shift | 12,764 |
mp5rBNVEt1QwcsVSFr1kmkapnhb | physics | simple-harmonic-motion | some-systems-of-executing-shm | Two particles A and B of equal masses are suspended from two massless springs of spring constants K<sub>1</sub> and K<sub>2</sub> respectively. If the maximum velocities during oscillations are equal, the ratio of the amplitude of A and B is | [{"identifier": "A", "content": "$${{{K_1}} \\over {{K_2}}}$$"}, {"identifier": "B", "content": "$$\\sqrt {{{{K_1}} \\over {{K_2}}}} $$"}, {"identifier": "C", "content": "$${{{K_2}} \\over {{K_1}}}$$"}, {"identifier": "D", "content": "$$\\sqrt {{{{K_2}} \\over {{K_1}}}} $$"}] | ["D"] | null | $$ \because $$ $${V_{\max }} = A\omega $$<br><br>Given, $${\omega _1}{A_1} = {\omega _2}{A_2}$$<br><br>We know that $$\omega = \sqrt {{K \over m}} $$<br><br>$$ \therefore $$ $$\sqrt {{{{k_1}} \over m}} {A_1} = \sqrt {{{{k_2}} \over m}} {A_2}$$<br><br>$$ \Rightarrow $$ $${{{A_1}} \over {{A_2}}} = \sqrt {{{{k_2}} \over ... | mcq | jee-main-2021-online-17th-march-evening-shift | 12,765 |
1krsvn5pi | physics | simple-harmonic-motion | some-systems-of-executing-shm | T<sub>0</sub> is the time period of a simple pendulum at a place. if the length of the pendulum is reduced to $${1 \over {16}}$$ times of its initial value, the modified time period is : | [{"identifier": "A", "content": "4 T<sub>0</sub>"}, {"identifier": "B", "content": "$${1 \\over {4}}$$ T<sub>0</sub>"}, {"identifier": "C", "content": "T<sub>0</sub>"}, {"identifier": "D", "content": "8$$\\pi$$ T<sub>0</sub>"}] | ["B"] | null | $$T = 2\pi \sqrt {{l \over g}} $$<br><br>$$T' = {{{T_0}} \over 4}$$ | mcq | jee-main-2021-online-22th-july-evening-shift | 12,766 |
1krun9poi | physics | simple-harmonic-motion | some-systems-of-executing-shm | In the reported figure, two bodies A and B of masses 200 g and 800 g are attached with the system of springs. Springs are kept in a stretched position with some extension when the system is released. The horizontal surface is assumed to be frictionless. The angular frequency will be ____________ rad/s when k = 20 N/m.<... | [] | null | 10 | $$\omega = \sqrt {{{{k_{eq}}} \over \mu }} $$<br><br>$$\mu$$ = reduced mass<br><br>springs are in series connection<br><br>$${k_{eq}} = {{{k_1}{k_2}} \over {{k_1} + {k_2}}}$$<br><br>$${k_{eq}} = {{k \times 4k} \over {5k}} = {{4k} \over 5}$$<br><br>$${k_{eq}} = {{4 \times 20} \over 5}$$ N/m = 16 N/m<br><br>$$\mu = {{... | integer | jee-main-2021-online-25th-july-morning-shift | 12,767 |
1kth5pmux | physics | simple-harmonic-motion | some-systems-of-executing-shm | A particle of mass 1 kg is hanging from a spring of force constant 100 Nm<sup>$$-$$1</sup>. The mass is pulled slightly downward and released so that it executes free simple harmonic motion with time period T. The time when the kinetic energy and potential energy of the system will become equal, is $${T \over x}$$. The... | [] | null | 8 | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264953/exam_images/eegpdnjnkggnfqmki1wd.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 31st August Morning Shift Physics - Simple Harmonic Motion Question 60 English Explanation 1"><br>... | integer | jee-main-2021-online-31st-august-morning-shift | 12,768 |
1ktjobzfj | physics | simple-harmonic-motion | some-systems-of-executing-shm | A bob of mass 'm' suspended by a thread of length l undergoes simple harmonic oscillations with time period T. If the bob is immersed in a liquid that has density $${1 \over 4}$$ times that of the bob and the length of the thread is increased by 1/3<sup>rd</sup> of the original length, then the time period of the simpl... | [{"identifier": "A", "content": "T"}, {"identifier": "B", "content": "$${3 \\over 2}$$T"}, {"identifier": "C", "content": "$${3 \\over 4}$$T"}, {"identifier": "D", "content": "$${4 \\over 3}$$T"}] | ["D"] | null | $$T = 2\pi \sqrt {l/g} $$<br><br>When bob is immersed in liquid<br><br>mg<sub>eff</sub> = mg $$-$$ Buoyant force<br><br>mg<sub>eff</sub> = mg $$-$$ v$$\sigma$$g ($$\sigma$$ = density of liquid)<br><br>$$ = mg - v{\rho \over 4}g$$<br><br>$$ = mg - {{mg} \over 4} = {{3mg} \over 4}$$<br><br>$$\therefore$$ $${g_{eff}} = {... | mcq | jee-main-2021-online-31st-august-evening-shift | 12,769 |
1ktmp4ops | physics | simple-harmonic-motion | some-systems-of-executing-shm | A mass of 5 kg is connected to a spring. The potential energy curve of the simple harmonic motion executed by the system is shown in the figure. A simple pendulum of length 4 m has the same period of oscillation as the spring system. What is the value of acceleration due to gravity on the planet where these experiments... | [{"identifier": "A", "content": "10 m/s<sup>2</sup>"}, {"identifier": "B", "content": "5 m/s<sup>2</sup>"}, {"identifier": "C", "content": "4 m/s<sup>2</sup>"}, {"identifier": "D", "content": "9.8 m/s<sup>2</sup>"}] | ["C"] | null | From the potential energy curve,<br/><br/>$${U_{\max }} = {1 \over 2}k{A^2}$$<br/><br/>$$10 = {1 \over 2}k{(2)^2}$$<br/><br/>$$\Rightarrow$$ k = 5 N/m<br/><br/>The length of the simple pendulum, L = 4 m<br/><br/>Time period of spring,<br/><br/>$$T = 2\pi \sqrt {{k \over m}} $$<br/><br/>Time period of simple pendulum,<b... | mcq | jee-main-2021-online-1st-september-evening-shift | 12,770 |
1l54813lr | physics | simple-harmonic-motion | some-systems-of-executing-shm | <p>A body is performing simple harmonic with an amplitude of 10 cm. The velocity of the body was tripled by air jet when it is at 5 cm from its mean position. The new amplitude of vibration is $$\sqrt{x}$$ cm. The value of x is _____________.</p> | [] | null | 700 | <p>$$v = \omega \sqrt {{A^2} - {y^2}} $$</p>
<p>$$ \Rightarrow 3\omega \sqrt {{{10}^2} - {5^2}} = \omega \sqrt {{{(A')}^2} - {5^2}} $$</p>
<p>$$ \Rightarrow 9 \times 75 = {(A')^2} - 25$$</p>
<p>$$ \Rightarrow A' = \sqrt {28 \times 25} $$ cm</p>
<p>$$ \Rightarrow x = 700$$</p> | integer | jee-main-2022-online-29th-june-morning-shift | 12,771 |
1l57q9zaw | physics | simple-harmonic-motion | some-systems-of-executing-shm | <p>The displacement of simple harmonic oscillator after 3 seconds starting from its mean position is equal to half of its amplitude. The time period of harmonic motion is :</p> | [{"identifier": "A", "content": "6 s"}, {"identifier": "B", "content": "8 s"}, {"identifier": "C", "content": "12 s"}, {"identifier": "D", "content": "36 s"}] | ["D"] | null | <p>Time taken by the harmonic oscillator to move from mean position to half of amplitude is $${T \over {12}}$$</p>
<p>So, $${T \over {12}}$$ = 3</p>
<p>T = 36 sec.</p> | mcq | jee-main-2022-online-27th-june-morning-shift | 12,772 |
1l58bke0g | physics | simple-harmonic-motion | some-systems-of-executing-shm | <p>Time period of a simple pendulum in a stationary lift is 'T'. If the lift accelerates with $${g \over 6}$$ vertically upwards then the time period will be :</p>
<p>(Where g = acceleration due to gravity)</p> | [{"identifier": "A", "content": "$$\\sqrt {{6 \\over 5}} T$$"}, {"identifier": "B", "content": "$$\\sqrt {{5 \\over 6}} T$$"}, {"identifier": "C", "content": "$$\\sqrt {{6 \\over 7}} T$$"}, {"identifier": "D", "content": "$$\\sqrt {{7 \\over 6}} T$$"}] | ["C"] | null | $T^{\prime}=2 \pi \sqrt{\frac{I}{g_{\text {eff }}}}$
<br/><br/>$T^{\prime}=2 \pi \sqrt{\frac{I}{g+\frac{g}{6}}}=2 \pi \sqrt{\frac{6 l}{7 g}}$
<br/><br/>$\Rightarrow T^{\prime}=\sqrt{\frac{6}{7}} T$ | mcq | jee-main-2022-online-26th-june-morning-shift | 12,773 |
1l5bbmix6 | physics | simple-harmonic-motion | some-systems-of-executing-shm | <p>Two massless springs with spring constants 2 k and 9 k, carry 50 g and 100 g masses at their free ends. These two masses oscillate vertically such that their maximum velocities are equal. Then, the ratio of their respective amplitudes will be :</p> | [{"identifier": "A", "content": "1 : 2"}, {"identifier": "B", "content": "3 : 2"}, {"identifier": "C", "content": "3 : 1"}, {"identifier": "D", "content": "2 : 3"}] | ["B"] | null | <p>$${\omega _1}{A_1} = {\omega _2}{A_2}$$</p>
<p>$$ \Rightarrow {{{A_1}} \over {{A_2}}} = {{{\omega _2}} \over {{\omega _1}}}$$</p>
<p>$$ = \sqrt {{{{k_2}} \over {{m_2}}}} \times \sqrt {{{{m_1}} \over {{k_1}}}} = \sqrt {{{9k} \over {100}} \times {{50} \over {2k}}} = {3 \over 2}$$</p> | mcq | jee-main-2022-online-24th-june-evening-shift | 12,774 |
1l6dydu1z | physics | simple-harmonic-motion | some-systems-of-executing-shm | <p><img src="data:image/png;base64,UklGRsYLAABXRUJQVlA4ILoLAABwgwCdASoAAw4BP4G61WW2LiunIdRKEsAwCWlu/D65FnYnZ18fsX/g/B7h02f39d4G4jJ+v4Xu7f47/67/QRPVTKaxAyE/HZyEBJXARlN0cP/bGK2FPhPx2chASWIGO2jNo4erUw7OQgJLEDIQmPoNYQhAyE/HZyEBJXJWRuHZyEBJYgZCfZ7ePxpV0RAyE/HZyEBJYarzNLZyEBJYgZCfjO7MM7sw7OQgJLEDIT7jauz2CEDIT8dnIP/zKLmd2YdnI... | [{"identifier": "A", "content": "$$\n\\frac{\\mathrm{T}_{1}}{\\mathrm{~T}_{2}}=\\frac{3}{\\sqrt{2}}\n$$"}, {"identifier": "B", "content": "$$\n\\frac{\\mathrm{T}_{1}}{\\mathrm{~T}_{2}}=\\sqrt{\\frac{3}{2}}\n$$"}, {"identifier": "C", "content": "$$\n\\frac{\\mathrm{T}_{1}}{\\mathrm{~T}_{2}}=\\sqrt{\\frac{2}{3}}\n$$"}, {... | ["A"] | null | <p>Both the springs are in parallel combination in both the diagrams so</p>
<p>$${T_1} = 2\pi \sqrt {{{3m} \over {2k}}} $$</p>
<p>and $${T_2} = 2\pi \sqrt {{m \over {3k}}} $$</p>
<p>So, $${{{T_1}} \over {{T_2}}} = {3 \over {\sqrt 2 }}$$</p> | mcq | jee-main-2022-online-25th-july-morning-shift | 12,775 |
1l6i44yn2 | physics | simple-harmonic-motion | some-systems-of-executing-shm | <p>As per given figures, two springs of spring constants $$k$$ and $$2 k$$ are connected to mass $$m$$. If the period of oscillation in figure (a) is $$3 \mathrm{s}$$, then the period of oscillation in figure (b) will be $$\sqrt{x}~ s$$. The value of $$x$$ is ___________.</p>
<p><img src="data:image/png;base64,UklGRkIU... | [] | null | 2 | <p>For case (a),</p>
<p>$${K_{eq}} = {{2K} \over 3}$$</p>
<p>For case (b),</p>
<p>$${K_{eq}} = 3K$$</p>
<p>$$\because$$ $$T = 2\pi \sqrt {{m \over K}} $$</p>
<p>$$\therefore$$ $${{{T_a}} \over {{T_b}}} = \sqrt {{{{K_b}} \over {{K_a}}}} $$</p>
<p>$${3 \over {{T_b}}} = \sqrt {{{3K \times 3} \over {2K}}} = {3 \over {\sqr... | integer | jee-main-2022-online-26th-july-evening-shift | 12,776 |
1l6jjeoyt | physics | simple-harmonic-motion | some-systems-of-executing-shm | <p>A mass $$0.9 \mathrm{~kg}$$, attached to a horizontal spring, executes SHM with an amplitude $$\mathrm{A}_{1}$$. When this mass passes through its mean position, then a smaller mass of $$124 \mathrm{~g}$$ is placed over it and both masses move together with amplitude $$A_{2}$$. If the ratio $$\frac{A_{1}}{A_{2}}$$ i... | [] | null | 16 | <p>$$(0.9){A_1}\sqrt {{K \over {0.9}}} = (0.9 + 0.124){A_2}\sqrt {{K \over {0.9 + 0.124}}} $$</p>
<p>$${{{A_1}} \over {{A_2}}} = \sqrt {{{0.9 + 0.124} \over {0.9}}} $$</p>
<p>$$ = \sqrt {{{1.024} \over {0.9}}} $$</p>
<p>$$ = {\alpha \over {\alpha - 1}}$$</p>
<p>$$\alpha = 16$$</p> | integer | jee-main-2022-online-27th-july-morning-shift | 12,777 |
1l6nrcg6d | physics | simple-harmonic-motion | some-systems-of-executing-shm | <p>Assume there are two identical simple pendulum clocks. Clock - 1 is placed on the earth and Clock - 2 is placed on a space station located at a height h above the earth surface. Clock - 1 and Clock - 2 operate at time periods 4 s and 6 s respectively. Then the value of h is -</p>
<p>(consider radius of earth $$R_{E}... | [{"identifier": "A", "content": "1200 km"}, {"identifier": "B", "content": "1600 km"}, {"identifier": "C", "content": "3200 km"}, {"identifier": "D", "content": "4800 km"}] | ["C"] | null | <p>$$T \propto \sqrt {1/g} $$</p>
<p>$$ \Rightarrow {{{T_1}} \over {{T_2}}} = \sqrt {{{{g_2}} \over {{g_1}}}} = {R \over {R + h}}$$</p>
<p>$${4 \over 6} = {R \over {R + h}}$$</p>
<p>$$ \Rightarrow h = R/2$$</p>
<p>$$ = 3200$$ km</p> | mcq | jee-main-2022-online-28th-july-evening-shift | 12,778 |
1l6ntvsk9 | physics | simple-harmonic-motion | some-systems-of-executing-shm | <p>The potential energy of a particle of mass $$4 \mathrm{~kg}$$ in motion along the x-axis is given by $$\mathrm{U}=4(1-\cos 4 x)$$ J. The time period of the particle for small oscillation $$(\sin \theta \simeq \theta)$$ is $$\left(\frac{\pi}{K}\right) s$$. The value of $$\mathrm{K}$$ is _________.</p> | [] | null | 2 | <p>$$U = 4(1 - \cos 4x)$$</p>
<p>$$ \Rightarrow F = - {{dU} \over {dx}} = - (4)(4\sin 4x)$$</p>
<p>$$ = - 16\sin 4x$$</p>
<p>as small x</p>
<p>$$F = - 16(4x) = - 64x \equiv - kx$$</p>
<p>$$T = 2\pi \sqrt {{m \over k}} = 2\pi \sqrt {{4 \over {64}}} = {\pi \over 2}$$</p>
<p>$$ \Rightarrow K = 2$$</p> | integer | jee-main-2022-online-28th-july-evening-shift | 12,779 |
1l6p4qe4d | physics | simple-harmonic-motion | some-systems-of-executing-shm | <p>The time period of oscillation of a simple pendulum of length L suspended from the roof of a vehicle, which moves without friction down an inclined plane of inclination $$\alpha$$, is given by :</p> | [{"identifier": "A", "content": "$$2 \\pi \\sqrt{\\mathrm{L} /(\\mathrm{g} \\cos \\alpha)}$$"}, {"identifier": "B", "content": "$$2 \\pi \\sqrt{\\mathrm{L} /(\\mathrm{g} \\sin \\alpha)}$$"}, {"identifier": "C", "content": "$$2 \\pi \\sqrt{\\mathrm{L} / \\mathrm{g}}$$"}, {"identifier": "D", "content": "$$2 \\pi \\sqrt{\... | ["A"] | null | <p>$$\left| {{g_{eff}}} \right| = \left| {\overline g - \overline a } \right|$$</p>
<p>$$ \Rightarrow {g_{eff}} = g\cos \theta $$</p>
<p>$$ \Rightarrow T = 2\pi \sqrt {{l \over {{g_{eff}}}}} $$</p>
<p>$$ = 2\pi = \sqrt {{L \over {g\cos \theta }}} $$</p> | mcq | jee-main-2022-online-29th-july-morning-shift | 12,780 |
1l6rivslm | physics | simple-harmonic-motion | some-systems-of-executing-shm | <p>The metallic bob of simple pendulum has the relative density 5. The time period of this pendulum is $$10 \mathrm{~s}$$. If the metallic bob is immersed in water, then the new time period becomes $$5 \sqrt{x}$$ s. The value of $$x$$ will be ________.</p> | [] | null | 5 | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l7mjet29/4f2391a8-16f2-4745-a196-f0a748033206/49212110-2be0-11ed-9545-a32c92221434/file-1l7mjet2a.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l7mjet29/4f2391a8-16f2-4745-a196-f0a748033206/49212110-2be0-11ed-9545-a32c92221434/fi... | integer | jee-main-2022-online-29th-july-evening-shift | 12,781 |
1ldnwmm5h | physics | simple-harmonic-motion | some-systems-of-executing-shm | <p>Choose the correct length (L) versus square of the time period ($$\mathrm{T}^{2}$$) graph for a simple pendulum executing simple harmonic motion.</p> | [{"identifier": "A", "content": "<img src=\"https://app-content.cdn.examgoal.net/fly/@width/image/1ldod7o1a/8ee3a526-5f84-422a-b7fa-fea9da4c323a/e96128e0-a3aa-11ed-bed3-417993225459/file-1ldod7o1b.png?format=png\" data-orsrc=\"https://app-content.cdn.examgoal.net/image/1ldod7o1a/8ee3a526-5f84-422a-b7fa-fea9da4c323a/e96... | ["D"] | null | The relationship between the length of a pendulum and its period is given by the equation:
<br/><br/>$$T = 2\pi \sqrt {{L \over g}} $$
<br/><br/>$$ \Rightarrow $$ $$\mathrm{T}^{2} = \frac{4\pi^2}{g} \cdot \mathrm{L}$$
<br/><br/>where T is the period of the pendulum, g is the acceleration due to gravity, and L is the l... | mcq | jee-main-2023-online-1st-february-evening-shift | 12,782 |
1ldpmsxyi | physics | simple-harmonic-motion | some-systems-of-executing-shm | <p>In the figure given below, a block of mass $$M=490 \mathrm{~g}$$ placed on a frictionless table is connected with two springs having same spring constant $$\left(\mathrm{K}=2 \mathrm{~N} \mathrm{~m}^{-1}\right)$$. If the block is horizontally displaced through '$$\mathrm{X}$$' $$\mathrm{m}$$ then the number of compl... | [] | null | 20 | $\mathrm{K_eff}=\mathrm{K}+\mathrm{K}$ $=2 \mathrm{k}$ (As both springs are in use in parallel)
<br/><br/>$=2 \times 2=4 \mathrm{~N} / \mathrm{m}$
<br/><br/>$$
\mathrm{m}=490 \mathrm{gm}
$$
<br/><br/>$$
\begin{aligned}
& =0.49 \mathrm{~kg}
\end{aligned}
$$
<br/><br/>$\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{m}}{\mathrm... | integer | jee-main-2023-online-31st-january-morning-shift | 12,783 |
ldqvx3oy | physics | simple-harmonic-motion | some-systems-of-executing-shm | <p>For a simple harmonic motion in a mass spring system shown, the surface is frictionless. When the mass of the block is $1 \mathrm{~kg}$, the angular frequency is $\omega_{1}$. When the mass block is $2 \mathrm{~kg}$ the angular frequency is $\omega_{2}$. The ratio $\omega_{2} / \omega_{1}$ is</p>
<p><img src="data:i... | [{"identifier": "A", "content": " $1 / \\sqrt{2}$"}, {"identifier": "B", "content": "$1 / 2$"}, {"identifier": "C", "content": "2"}, {"identifier": "D", "content": "$\\sqrt{2}$"}] | ["A"] | null | <p>$$\omega = \sqrt {{K \over m}} \Rightarrow \omega \propto {1 \over {\sqrt m }}$$</p>
<p>$${{{\omega _2}} \over {{\omega _1}}} = \sqrt {{{{m_1}} \over {{m_2}}}} = \sqrt {{1 \over 2}} $$</p> | mcq | jee-main-2023-online-30th-january-evening-shift | 12,784 |
1lduh9rsm | physics | simple-harmonic-motion | some-systems-of-executing-shm | <p>T is the time period of simple pendulum on the earth's surface. Its time period becomes $$x$$ T when taken to a height R (equal to earth's radius) above the earth's surface. Then, the value of $$x$$ will be :</p> | [{"identifier": "A", "content": "4"}, {"identifier": "B", "content": "$$\\frac{1}{2}$$"}, {"identifier": "C", "content": "2"}, {"identifier": "D", "content": "$$\\frac{1}{4}$$"}] | ["C"] | null | At surface of earth time period<br/><br/>
$$
\mathrm{T}=2 \pi \sqrt{\frac{\ell}{\mathrm{g}}}
$$<br/><br/>
At height $\mathrm{h}=\mathrm{R}$<br/><br/>
$$
\begin{aligned}
& \mathrm{g}^{\prime}=\frac{\mathrm{g}}{\left(1+\frac{\mathrm{h}}{\mathrm{R}}\right)^2}=\frac{\mathrm{g}}{4} \\\\
& \therefore \,\mathrm{xT}=2 \pi \sqr... | mcq | jee-main-2023-online-25th-january-morning-shift | 12,785 |
1ldyetbqi | physics | simple-harmonic-motion | some-systems-of-executing-shm | <p><img src="data:image/png;base64,UklGRkgcAABXRUJQVlA4IDwcAACQjQCdASoAA6kAPm0wlUgkIqIhItBsMIANiWlu6B/I++LZbLeZf8V2q/479c/Mn8a+cfxf9v/wv/H5LvXX/G9DP5D9xf3/9q9uH8r3y/MfUC/If6f/v/zF4GKz//R9QLvV/yf8D6q/zPmP9mP+z7gH9C/r3/c9d/954V33f/cfsF8AX9C/vf7Ce8B/Z/t/6Nvqf2GP2M9Nj2Jfux/////8VJhT525fIiuerPkN2kCJNHEALcFzlf5auggMTAAWk4OXy... | [] | null | 5 | Both the springs are in parallel so net spring constant is $K_{\text {net }}=K_{1}+K_{2}=40 \mathrm{~N} / \mathrm{m}$
<br/><br/>
So $T=2 \pi \sqrt{\frac{m}{K_{n e t}}}$
<br/><br/>
$$
\begin{aligned}
& =2 \pi \sqrt{\frac{2}{40}} \\\\
& =\frac{\pi}{\sqrt{5}}
\end{aligned}
$$
<br/><br/>
$\therefore x=5$ | integer | jee-main-2023-online-24th-january-morning-shift | 12,786 |
1lh25bgwb | physics | simple-harmonic-motion | some-systems-of-executing-shm | <p>A mass $$m$$ is attached to two strings as shown in figure. The spring constants of two springs are $$\mathrm{K}_{1}$$ and $$\mathrm{K}_{2}$$. For the frictionless surface, the time period of oscillation of mass $$m$$ is :</p>
<p><img src="data:image/png;base64,UklGRhocAABXRUJQVlA4IA4cAAAwkwCdASoAA7sAPm0ylkgkIqIhJPD... | [{"identifier": "A", "content": "$$2\\pi \\sqrt {{m \\over {{K_1} + {K_2}}}} $$"}, {"identifier": "B", "content": "$$2\\pi \\sqrt {{m \\over {{K_1} - {K_2}}}} $$"}, {"identifier": "C", "content": "$${1 \\over {2\\pi }}\\sqrt {{{{K_1} + {K_2}} \\over m}} $$"}, {"identifier": "D", "content": "$${1 \\over {2\\pi }}\\sqrt ... | ["A"] | null | Since, both spring are in parallel
<br/><br/>Combination, therefore, $K_{\text {eq }}=K_1+K_2$
<br/><br/>Time period of oscillation, $T=2 \pi \sqrt{\frac{m}{K_{\text {eq }}}} $
<br/><br/>$\Rightarrow T=2 \pi \sqrt{\frac{m}{K_1+K_2}}$
<br/><br/>Option (a) is correct. | mcq | jee-main-2023-online-6th-april-morning-shift | 12,788 |
jaoe38c1lsd8owqi | physics | simple-harmonic-motion | some-systems-of-executing-shm | <p>The time period of simple harmonic motion of mass $$M$$ in the given figure is $$\pi \sqrt{\frac{\alpha M}{5 k}}$$, where the value of $$\alpha$$ is _________.</p>
<p><img src="data:image/png;base64,UklGRvQWAABXRUJQVlA4IOgWAACQNQGdASonAgADP4HA2GQ2MSymorLp2sAwCWlu/B24omHnZ17/tD/tvVTdtsNsO/2/ZdTqdsPzKv/+Htv36B3nn00///... | [] | null | 12 | <p>$$\mathrm{k}_{\mathrm{eq}}=\frac{2 \mathrm{k} \cdot \mathrm{k}}{3 \mathrm{k}}+\mathrm{k}=\frac{5 \mathrm{k}}{3}$$</p>
<p>Angular frequency of oscillation $$(\omega)=\sqrt{\frac{\mathrm{k}_{\mathrm{eq}}}{\mathrm{m}}}$$</p>
<p>$$(\omega)=\sqrt{\frac{5 \mathrm{k}}{3 \mathrm{~m}}}$$</p>
<p>Period of oscillation $$(\tau)... | integer | jee-main-2024-online-31st-january-evening-shift | 12,789 |
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