minus sign to remind us that it pulls back). Thus the mass times the acceleration For simplicity, suppose it happens(or we change our unit of time measurement that the ratio k /m=1. We shall first study the equation Later we shall come back to Eq(21.2)with the k and m explicitly present. We have already analyzed Eq.(. 3)in detail numerically; when we first introduced the subject of mechanics we solved this equation(see Eq 9.12)to find the motion. By numerical integration we found a curve(fig. 9-4)which showed that if m was initially displaced, but at rest, it would come down and go through zero; we did not then follow it any farther, but of course we know that it just keeps going up and down-it oscillates. When we calculated the motion numer- ically, we found that it went through the equilibrium point at t=1.570.The length of the whole cycle is four times this long, or to= 6.28"sec. "This was found numerically, before we knew much calculus. We assume that in the meantime the Mathematics Department has brought forth a function which, when differ entiated twice, is equal to itself with a minus sign. (There are, of course, ways of getting at this function in a direct fashion, but they are more complicated than thready knowing what the answer is. )The function is x cos t. If we differentiate sin t and d2x/dr2 t=-x. The function x= cos t starts, at t=0, with x= 1, and no initial velocity; that was the situation with which we started when we did our numerical work. Now that we know that x= cos t, we can calculate a precise value for the time at which it should pass x=0. The answer is t= T/2, or 1.57108. We were wrong in the last figure because of the errors of numerical analysis, but it was very close Now to go further with the original problem, we restore the time units to real seconds. What is the solution then? First of all, we might think that we can get the constants k and m in by multiplying cos t by something. So let us try the equation x= A cos t; then we find dx/dt =-A sin t, and dx/dr2 cos t=-x. Thus we discover to our horror that we did not succeed in solving Eq.(21.2), but we got Eq. (21. 3)again! That fact illustrates one of the most important properties of linear differential equations: if we multiply a solution of the equation by any constant, it is again a solution. The mathematical reason for this is clear. If x is a solution, and we multiply both sides of the equation, say by 4 we see that all derivatives are also multiplied by A, and therefore Ax is just as good a solution of the original equation as x was. The physics of it is the following If we have a weight on a spring, and pull it down twice as far, the force is twice as much, the resulting acceleration is twice as great, the velocity it acquires in a ne is twice as great, the distance covered in a given time is twice as great but it has to cover twice as great a distance in order to get back to the origin because it is pulled down twice as far. So it takes the same time to get back to the origin, irrespective of the initial displacement. In other words, with a linear equa tion, the motion has the same time pattern, no matter how"strong"it is Son That was the wrong thing to do-it only taught us that we can multiply the tion by anything, and it satisfies the same equation, but not a different equation After a little cut and try to get to an equation with a different constant multiplying x,we find that we must alter the scale of time. In other words, Eq.(21. 2)has a solution of the form x= cos wot (t is important to realize that in the present case, wo is not an angular velocity of a spinning body, but we run out of letters if we are not allowed to use the same letter for more than one thing. The reason we put a subscript"0"on w is that we are omegas before long; let us remember tha efers to the atural motion of this oscillator. Now