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accident in the units, so that we happen to have &/mn = 1. Thus we sha]l try to |
solve the equation |
đuy (dt = —ø. (9.12) |
To proceed, we must know what „ is, but oŸ course we know that the velocity is |
the rate of change of the position. |
--- Trang 184 --- |
9-5 Meaning of the dynamical equations |
Now let us try to analyze Just what Eq. (9.12) means. Suppose that at a given |
time ý the object has a certain velocity „; and position z. What is the velocity |
and what is the position at a slightly later time ý + c? If we can answer this |
question our problem is solved, for then we can start with the given condition and |
compute how 1% changes for the first instant, the next instant, the next instant, |
and so on, and in this way we gradually evolve the motion. To be speciffic, let us |
suppose that at the time ý = Ö we are given that z = 1 and ø„ =0. Why does |
the object move at all? Because there is a ƒforce on it when it is at any position |
except z = 0. lÝz >0, that force is upward. Therefore the velocity which 1s |
zero starts to change, because of the law of motion. Ônce it starts to build up |
some velocity the object starts to move up, and so on. Now at any time #, IÍ € is |
very small, we may express the position at time £ + e in terms of the position at |
time ý and the velocity at time £ to a very good approximation as |
z(t + €) = z(t) + cuz(Ð). (9.13) |
The smaller the c, the more accurate this expression is, but it is still usefully |
accurate even 1Ý e is not vanishingly smaill. Now what about the velocity? In |
order to get the velocity later, the velocity at the time # + c, we need to know |
how the velocity changes, the øccelerai#ion. And how are we going to find the |
acceleration? That is where the law of dynamics comes in. The law of dynamics |
tells us what the acceleration is. It says the acceleration is —z. |
0z(t + €) = 0x(É) + eax(£) (9.14) |
= 0„z(f) — ez(f). (9.15) |
Equation (9.14) is merely kinematics; it says that a velocity changes because of |
the presence of acceleration. But Eq. (9.15) is đựụnøœmics, because it relates the |
acceleration to the force; it says that at this particular time for this particular |
problem, you can replace the acceleration by —z(#). Therefore, if we know both |
the z and 0 at a given time, we know the acceleration, which tells us the new |
velocity, and we know the new position——this is how the machinery works. The |
velocity changes a little bit because of the force, and the position changes a little |
bit because of the velocity. |
--- Trang 185 --- |
9-6 Numerical solution of the equations |
Now let us really solve the problem. Suppose that we take e = 0.100 sec. After |
we do all the work iŸ we fnd that this is not small enough we may have to go back |
and do it again with e = 0.010 sec. Starting with our initial value z(0) = 1.00, |
what is (0.1)? It is the old position #(0) plus the velocity (which is zero) tỉmes |
0.10 sec. Thus z(0.1) is still 1.00 because it has not yet started to move. But |
the new velocity at 0.10 sec will be the old velocity ø(0) = 0 plus e times the |
acceleration. The acceleration is —#(0) = —1.00. Thus |
(0.1) = 0.00 — 0.10 x 1.00 = —0.10. |
Now at 0.20 sec |
+(0.2) = z(0.1) + eo(0.1) |
= 1.00 — 0.10 x 0.10 = 0.99 |
0(0.2) = 0(0.1) + ea(0.1) |
= —0.10 — 0.10 x 1.00 = —0.20. |
And so, on and on and on, we can calculate the rest of the motion, and that is just |
what we shall do. However, for practical purposes there are some little trieks by |
which we can increase the accuracy. IÝ we continued this calculation as we have |
started it, we would fnd the motion only rather crudely because e —= 0.100 sec |
is rather crude, and we would have to go to a very small interval, say e = 0.01. |
Then to go through a reasonable total time interval would take a lot of cycles |
of computation. So we shall organize the work in a way that will increase the |
precision of our calculations, using the same coarse interval e = 0.10 sec. 'This |
can be done iŸ we make a subtle improvement in the technique of the analysis. |
Notice that the new position is the old position plus the time interval e times |
the velocity. But the velocity œhenŸ The velocity at the beginning of the time |
interval is one velocity and the velocity at the end of the time interval is another |
velocity. Our improvement is to use the velocity halftuau betueen. ]Ý we know |
the speed now, but the speed is changing, then we are not goỉng to get the right |
answer by going at the same speed as now. We should use some speed between |
the “now” speed and the “then” speed at the end of the interval. "The same |
considerations also apply to the velocity: to compute the velocity changes, we |
should use the acceleration midway between the two times at which the velocity |
1s to be found. Thus the equations that we shall actually use will be something |
--- Trang 186 --- |
Table 9-1 |
Solution of du„/dt = —z |
Interval: e = 0.10 sec |
£ % U„ đ„ |
0.0 1.000 0.000 | —1.000 |
—0.050 |
0.1 0.995 —0.995 |
—0.150 |
0.2 0.980 —0.980 |
—0.248 |
0.3 0.955 —0.955 |
—0.343 |
0.4 0.921 —0.921 |
—0.435 |
0.5 0.877 —0.877 |
—0.523 |
0.6 0.825 —0.825 |
—0.605 |
0.7 0.764 —0.764 |
—0.682 |
0.8 0.696 —0.696 |
—0.751 |
0.9 0.621 —0.621 |
—0.814 |
1.0 0.540 —0.540 |
—0.868 |
1.1 0.453 —0.453 |
—0.913 |
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