This is the most general way we can start the motion. (We cannot specify the acceleration with which it started, true, because that is determined by the spring, once we specify xo )Now let us calculate A and B. We start with the equation Since we shall later need the velocity also, we differentiate x and obtain U=-woA sin wof WoB cos wot. These expressions are valid for all t, but we have special knowledge about x and v at 1=0. So if we put t=0 into these equations, on the left we get xo and vo because that is what x and v are at t= 0: also we know that the cosine of zero is unity, and the sine of zero is zero. Therefore we get A·1+B.0=A U=-ω0A·0+coB·l=coB So for this particular case we find that A=xo From these values of A and B, we can get a and a if we wish. That is the end of our solution, but there is one physically interesting thing to check, and that is the conservation of energy. Since there are no frictional losses, energy ought to be conserved. Let us use the formula x=acos(aot+△) the Now let us find out what the kinetic energy T is, and what the potential energy U is. The potential energy at any moment is tkx, where x is the displacement and k is the constant of the spring. If we substitute for x, using our expression above, we get U= dkx= fka-cos(wot A) Of course the potential energy is not constant; the potential never becomes negative naturally--there is always some energy in the spring, but the amount of energy fluctuates with x. The kinetic energy, on the other hand, is imu, and by sub stituting for v we get T=imu= gmaoa sin(wot +4). Now the kinetic energy is zero when x is at the maximum, because then there is no velocity; on the other hand, it is maximal when x is passing through zero, because then it is moving fastest. This variation of the kinetic energy is just the opposite of that of the potential energy. But the total energy ought to be a constant. If we note that k T+U cos2(aot+△)+sin2(ot+△)=mo The energy is dependent on the square of the amplitude; if we have twice the amplitude, we get an oscillation which has four times the energy. The average potential energy is half the maximum and, therefore, half the total, and the average kinetic energy is likewise half the total energy 21-5 Forced oscillations Next we shall discuss the forced harmonic oscillator, i. e, one in which there is an external driving force acting. The equation then is the following md x/dt=-kx+F(