Forced oscillations 61 SOLUTION. In this case the extension of the spring is (if = cos The kinetic energy is T = 1mr2o2, and the frequency is therefore w = V[F(++1)/mrl]. PROBLEM 5. Find the frequency of oscillations of the pendulum shown in Fig. 2 (§5), whose point of support carries a mass M1 and is free to move horizontally. SOLUTION. For < 1 the formula derived in $14, Problem 3 gives T Hence PROBLEM 6. Determine the form of a curve such that the frequency of oscillations of a particle on it under the force of gravity is independent of the amplitude. SOLUTION. The curve satisfying the given condition is one for which the potential energy of a particle moving on it is U = 1ks2, where s is the length of the arc from the position of equilibrium. The kinetic energy T = 1ms2, where m is the mass of the particle, and the fre- quency is then w = (k/m) whatever the initial value of S. In a gravitational field U = mgy, where y is the vertical co-ordinate. Hence we have 1ks2 = mgy or y = w2s2/2g. But ds2 = dx2+dy², whence dy = SV[(g/2w2y)-1] dy. The integration is conveniently effected by means of the substitution y = g(1-cos E)/4ww, which yields x = g(s+sin 5)/4w2. These two equations give, in parametric form, the equation of the required curve, which is a cycloid. $22. Forced oscillations Let us now consider oscillations of a system on which a variable external force acts. These are called forced oscillations, whereas those discussed in §21 are free oscillations. Since the oscillations are again supposed small, it is implied that the external field is weak, because otherwise it could cause the displacement x to take too large values. The system now has, besides the potential energy 1kx2, the additional potential energy Ue(x, t) resulting from the external field. Expanding this additional term as a series of powers of the small quantity x, we have Ue(x, t) 12 x[dUe/dx]x_0. The first term is a function of time only, and may therefore be omitted from the Lagrangian, as being the total time derivative of another function of time. In the second term - [dUe/dx]x_0 is the external “force” acting on the system in the equilibrium position, and is a given function of time, which we denote by F(t). Thus the potential energy involves a further term -xF(t), and the Lagrangian of the system is L = (22.1) The corresponding equation of motion is m+kx = F(t) or (22.2) where we have again introduced the frequency w of the free oscillations. The general solution of this inhomogeneous linear differential equation with constant coefficients is x = xo+x1, where xo is the general solution of 62 Small Oscillations §22 the corresponding homogeneous equation and X1 is a particular integral of the inhomogeneous equation. In the present case xo represents the free oscillations discussed in $21. Let us consider a case of especial interest, where the external force is itself a simple periodic function of time, of some frequency y: F(t) = f cos(yt+)). (22.3) We seek a particular integral of equation (22.2) in the form X1 = b cos(yt+B), with the same periodic factor. Substitution in that equation gives b = f/m(w2-r2); adding the solution of the homogeneous equation, we obtain the general integral in the form (22.4) The arbitrary constants a and a are found from the initial conditions. Thus a system under the action of a periodic force executes a motion which is a combination of two oscillations, one with the intrinsic frequency w of the system and one with the frequency y of the force. The solution (22.4) is not valid when resonance occurs, i.e. when the fre- quency y of the external force is equal to the intrinsic frequency w of the system. To find the general solution of the equation of motion in this case, we rewrite (22.4) as x = a where a now has a different value. Asy->w, the second term is indetermin- ate, of the form 0/0. Resolving the indeterminacy by L’Hospital’s rule, we have x = acos(wt+a)+(f/2mw)tsin(wt+B). = (22.5) Thus the amplitude of oscillations in resonance increases linearly with the time (until the oscillations are no longer small and the whole theory given above becomes invalid). Let us also ascertain the nature of small oscillations near resonance, when y w+E with E a small quantity. We put the general solution in the com- plex form = A exp(iwt) exp[i(w+t)) = [A+B exp(iet)]exp(ist) (22.6) Since the quantity A+B exp(iet) varies only slightly over the period 2n/w of the factor exp(iwt), the motion near resonance may be regarded as small oscillations of variable amplitude.t Denoting this amplitude by C, we have = A B exp(iet)|. Writing A and B in the form a exp(ix) and b exp(iB) respectively, we obtain (22.7) t The “constant” term in the phase of the oscillation also varies. §22 Forced oscillations 63 Thus the amplitude varies periodically with frequency E between the limits |a-b a+b. This phenomenon is called beats. The equation of motion (22.2) can be integrated in a general form for an arbitrary external force F(t). This is easily done by rewriting the equation as or = (22.8) where s=xtiwx (22.9) is a complex quantity. Equation (22.8) is of the first order. Its solution when the right-hand side is replaced by zero is $ = A exp(iwt) with constant A. As before, we seek a solution of the inhomogeneous equation in the form $ = A(t) exp(iwt), obtaining for the function A(t) the equation À(t) = F(t) exp(-iwt)/m. Integration gives the solution of (22.9): & = - (22.10) where the constant of integration so is the value of $ at the instant t = 0. This is the required general solution; the function x(t) is given by the imagin- ary part of (22.10), divided by w.t The energy of a system executing forced oscillations is naturally not con- served, since the system gains energy from the source of the external field. Let us determine the total energy transmitted to the system during all time, assuming its initial energy to be zero. According to formula (22.10), with the lower limit of integration - 00 instead of zero and with ( - 00) = 0, we have for t 00 exp(-iwt)dt| The energy of the system is E = 1m(x2+w2x2)= = 1ME2. (22.11) Substituting we obtain the energy transferred (22.12) t The force F(t) must, of course, be written in real form. 64 Small Oscillations §22 it is determined by the squared modulus of the Fourier component of the force F(t) whose frequency is the intrinsic frequency of the system. In particular, if the external force acts only during a time short in com- parison with 1/w, we can put exp(-iwt) Ill 1. Then This result is obvious: it expresses the fact that a force of short duration gives the system a momentum I F dt without bringing about a perceptible displacement. PROBLEMS PROBLEM 1. Determine the forced oscillations of a system under a force F(t) of the follow- ing forms, if at time t = 0 the system is at rest in equilibrium (x = x = 0): (a) F = Fo, a constant, (b) F = at, (c) F = Fo exp(-at), (d) F = Fo exp(-at) cos Bt. SOLUTION. (a) x = (Fo/mw2)(1-cos wt). The action of the constant force results in a dis- placement of the position of equilibrium about which the oscillations take place. (b) x = (a/mw3)(wt-sin wt). (c) x = - cos wt +(a/w) sin wt]. (d) x = wt + sin wt + +exp(-at)[(wpta2-B2) cos Bt-2aB sin This last case is conveniently treated by writing the force in the complex form F=Foexp(-ati)t]. PROBLEM 2. Determine the final amplitude for the oscillations of a system under a force which is zero for t<0, Fot/T for 0 <t<<, and Fo for t > T (Fig. 24), if up to time t = 0 the system is at rest in equilibrium. F Fo , T FIG. 24 SOLUTION. During the interval 0<+<T the oscillations are determined by the initial condition as x = (Fo/mTw3)(wt-sin wt). For t > T we seek a solution in the form =c1w(t-T)+c2 sin w(t - T)+Fo/mw2 The continuity of x and x at t = T gives C1 = -(Fo/mTw3) sin wT, C2 = (Fo/mTw3 X X (1 - cos wT). The amplitude is a = = (2Fo/mTw3) sin twT. This is the smaller, the more slowly the force Fo is applied (i.e. the greater T). PROBLEM 3. The same as Problem 2, but for a constant force Fo which acts for a finite time T (Fig. 25).