Math 302, Week 8: Oscillations T y eq Y y = y eq + Y mg Figure : Simple harmonic motion. At equilibrium the string is of total length y eq. During the motion we let Y be the extension beyond equilibrium, so that the total length is y = y eq + Y. Recall vertical oscillations of a ss on a light elastic string (see figure. Let y be the downward displacement of the ss m from the point O where the string is fixed. Then we have mÿ = mg T where T is the tension in the string, given by T = k(y l, where l is the natural length of the string (i.e. its unstretched length. Thus mÿ = mg k(y l. Now if the string is resting in equilibrium, the weight equals the tension, so if y eq is the equilibrium value of y, T eq = k(y eq l = mg, which on rearrangement gives y eq = mg + l. Now suppose that the k ss is pulled down and released. We will assume that the string reins taut for all t (which is so if it is not stretched further than y eq l beyond equilibrium. Let Y be the extension of the string beyond equilibrium: y = y eq + Y. Then the equation of motion becomes Since y eq is constant, m d2 dt 2 (y eq + Y = mg k(y eq + Y l, mÿ = mg k(y eq l ky = ky, since the equilibrium y eq satisfies k(y eq l = mg. The general solution of Ÿ = k m Y is Y (t = A cos(ωt + δ, where A, δ are constants determined by the initial conditions and ω 2 = k/m. Here A is the amplitude of the oscillation, ω is its angular frequency and we have that its period T = 2π/ω. Another way of approaching this problem is to note that the tension T = T eq + ky, since the stretch of the string Y beyond the equilibrium tension T eq provides an additional force of ky. Newton s 2nd law then gives since at equilibrium T eq = mg. mÿ = mg T = mg T eq ky = (mg T eq ky = ky,
2 5 0 5 20 25 ω = µ / - 20 0 5 0 5 20 25 ω = µ -0-20 Forced oscillations Figure 2: Forced simple harmonic oscillator. When the forcing frequency µ is distinct from the oscillator frequency ω the system reins bounded. When µ = ω we obtain resonance in which the amplitude of the oscillation increases with t. Suppose that we add a periodic component of forcing to the ss, say f(t = f 0 cos(µt. Then the equations of motion (for when the string reins taut and there are no collisions between the ss and the other objects become Ÿ + k m Y = f(t m = f 0 cos(µt. ( m Recall that to solve such linear second order differential equations, we first solve the homogeneous equation Ÿ + k Y = 0 for the complementary functions and then add on the particular integral. m Thus with ω 2 = k/m we get Y (t = A cos(ωt + δ + Y P I. Now if µ ω we y try a PI of the form Y P I = C cos(µt, which gives so that C = f 0 m(ω 2 µ 2 f 0 m cos(µt = µ2 C cos(µt + ω 2 C cos(µt, and general solution is Y (t = A cos(ωt + δ + f 0 cos(µt. (2 m(ω 2 µ 2 On the other hand, when µ = ω the particular integral becomes Y P I = B t sin(ωt. Substituting into the differential equation ( we obtain f 0 m cos(ωt = 2ωB cos(ωt B ω 2 t sin(ωt + B ω 2 t sin(ωt = 2ωB cos(ωt For example, if the ss carries a charge, we could impose a periodic electric field. 2
so that B = f 0. Thus in this case we have 2mω Y (t = A cos(ωt + δ + f 0 2mω t sin(ωt. Note that after a certain time the oscillations have grown so large that the string does not rein taut during the motion and the solution no longer applies. Compound oscillators Now consider a system of two sses M, m on two elastic strings as follows. Mass M is suspended from O on a string natural length l and with constant k. The ss M is attached to the ss m by a string of natural length l 2 and constant k 2. We let Y be the distance of m below its equilibrium, and the distance of M below its equilibrium (see figure 3. As usual we will Figure 3: Compound pendulum of two ss with two elastic strings. For i =, 2, Y i is the extension of string i beyond equilibrium. assume that the motion is such that the strings rein taut for all time. motion are, using Newton s second law, The equations of mÿ = T 2 + mg T MŸ2 = Mg T 2 where In equilibrium T = T,eq + k Y, T 2 = T 2,eq + k 2 ( Y. T,eq = T 2,eq + mg, T 2,eq = Mg. 3
Thus we have since T 2,eq + mg T,eq = 0. Similarly, mÿ = T 2 + mg T = T 2,eq + k 2 ( Y + mg T,eq k Y = T 2,eq + mg T,eq + k 2 ( Y k Y = k 2 ( Y k Y, since Mg T 2,eq = 0. To sumrize: We seek solutions of this system of the form ( Y Y = = MŸ2 = Mg T 2 = Mg T 2,eq k 2 ( Y = k 2 ( Y, mÿ = (k + k 2 Y + k 2 MŸ2 = k 2 Y k 2. ( C C 2 (3 cos(ωt + δ. (4 For such a solution we have Ÿ = ω 2 C cos(ωt + δ and Ÿ2 = ω 2 C 2 cos(ωt + δ. Thus if (4 is a solution of (3 ω 2 mc cos(ωt + δ = (k + k 2 C cos(ωt + δ + k 2 C 2 cos(ωt + δ ω 2 MC 2 cos(ωt + δ = k 2 C cos(ωt + δ k 2 C 2 cos(ωt + δ. Since this is true for all t, we have, rearranging (mω 2 k k 2 C + k 2 C 2 = 0 k 2 C + (Mω 2 k 2 C 2 = 0. If this is to provide a non-trivial solution Y 0, i.e. (C, C 2 T 0 we must have (mω 2 k k 2 (Mω 2 k 2 k 2 2 = 0. Set q = ω 2. Then we obtain the following quadratic for the possible values of q: This has two roots; are they real? They are if Mmq 2 (k 2 m + M(k + k 2 q + k k 2 = 0. (k 2 m + (k + k 2 M 2 4Mmk k 2. 4
But k 2 m + (k + k 2 M k 2 m + k M so that (k 2 m + (k + k 2 M 2 4Mmk k 2 (k 2 m + k M 2 4Mmk k 2 = (k 2 m k M 2 0, so the roots are real. Lets call them q, q +. Since k k 2 > 0, we see that both real roots are positive, by noting that q satisfies q = k k 2 + Mmq 2 k 2 m + M(k + k 2. Since we have show that q is real, the right-hand side is positive, and hence both q, q + > 0. Thus the four possible values of ω, namely ± q +, ± q are all real. We only need only concern ourselves with ω 2 +, ω 2. Now we y find the possible C, C 2 for each value of q = ω 2. When ω 2 = q we have C k 2 = C 2 (k 2 Mω 2 = C 2 (k 2 Mq. Hence one possible solution is, for any real B, ( k2 Mq Y = B k 2 cos( q t + δ. When ω 2 = q + we have C k 2 = C 2 (k 2 Mω 2 = C 2 (k 2 Mq +, giving another possible solution, for any real B +, ( k2 Mq + Y + = B k + 2 cos( q + t + δ +. The general solution is any linear combination of these two solutions: for any real α, β ( k2 Mq Y (t = α k 2 cos( ( k2 Mq + q t + δ + β k 2 cos( q + t + δ + The frequencies ω = q, ω + = q + are called the eigenfrequencies of the system, and the Y, Y + are the respective norl modes associated with these frequencies. Illustrative example Suppose we have M = 2, m = 3 and k = k 2 =. Then we have 3Ÿ = 2Y + 2Ÿ2 = Y 5
Trying a solution we obtain Y = ( Y = ( A B cos(ωt + δ 3ω 2 A = 2A + B, A + (2ω 2 B = 0, Setting q = ω 2 we obtain 6q 2 7q + = 0, so q =, /6. ω = / 6, ω + =. This gives the eigenfrequencies For ω = we obtain A = B and so the norl modes for w + are any multiple of ( Y + = cos(t + δ +. For ω = / 6 we find that 3A = 2B and hence the norl modes for w are any multiple of ( 2 Y = cos( t + δ 3. 6 The general solution is thus, for any real α, β, ( Y (t = α cos(t + δ + + β ( 2 3 cos( t 6 + δ. For the norl mode Y (t = Y (t we see that Y (t = (t for all t. This means that in this mode the particles oscillate in antiphase (figure 4. For q = q = / 6, the particles oscillate Y = 0 = 0 Figure 4: Norl mode q = q + = for the compound pendulum of two ss with two elastic strings where particle oscillate in antiphase: Y = in phase, but the amplitude of is one and a half times that of Y (figure 5. In figure 6 we show typical plots of these two modes and the general solution. 6
Y = 0 = 0 Figure 5: Norl mode q = q = / 6 for the compound pendulum of two ss with two elastic strings where particle oscillate in phase, but with different amplitudes Y = 2Y 3 2 Example: Two sses joined by three springs Consider the system of two identical sses of ss m linked by three identical springs of stiffness k (see figure 7. The equations of motion are mẍ = T 2 T (5 mẍ 2 = T 3 T 2, (6 where T = T eq + kx, T 2 = T eq + k(x 2 x and T 3 = T eq kx 2. Thus we obtain This y be rewritten as where x = (x, x 2 T and mẍ = 2kx + kx 2 (7 mẍ 2 = kx 2kx 2, (8 ẍ = Ax, (9 ( 2σ σ A = σ 2σ Now the eigenvalues of A are σ and 3σ: They are the roots of ( 2σ λ σ det = λ 2 4σλ + 3σ 2 = 0. σ 2σ λ For λ = 3σ we y take an eigenvector v = (, T and for λ = σ we y take v 2 = (, T. Now note that if ( P =, then P = ( /2 /2 /2 /2 7.,
Y 0.5-5 -0-5 5 0 5 ω = -0.5 - = -Y 3 = 3Y /2 2 Y ω = (/6-5 -0-5 5 0 5 - -2 /2-3 4 Y 2-5 -0-5 5 0 5-2 general solution -4 Figure 6: First plot: Norl mode q = q + =, antiphase; Second plot: Norl mode q = q = / 6, in phase. Third plot, a typical solution constructed from combinations of the two modes. and P AP = diag(3σ, σ = D. Thus from (9 we have, and so Now define Y = (Y, T = P x, to obtain which expands to Thus ẍ = P DP x, P ẍ = DP x. Ÿ = DY, Ÿ = 3σY Ÿ 2 = σ. Y = α cos(ω t + δ, = α 2 cos(ω 2 t + δ 2, where α, α 2, δ, δ 2 are constants and ω = 3σ, ω 2 = σ. Now we use x = P Y to obtain x = Y + = α cos(ω t + δ + α 2 cos(ω 2 t + δ 2 x 2 = Y + = α cos(ω t + δ + α 2 cos(ω 2 t + δ 2. 8
Figure 7: Two identical sses between 3 identical springs Notice that we y also write this as x = v α cos(ω t + δ + v 2 α 2 cos(ω 2 t + δ 2. Such a decomposition works in general: Suppose an oscillator system has x = (x,..., x n T and satisfies ẍ = Ax, (0 where A is a real n n trix with distinct eigenvalues then the general solution is x(t = n α k v k cos( λ k t + δ k, k= where v k is an eigenvector associated with the eigenvalue λ k of A and the α k, δ k are constants to be found from the initial conditions. To prove this, we diagonalise A as above: P AP = D where D = diag(λ,..., λ n and P is de up from eigenvectors as columns. From (0 we obtain ẍ = P DP x, so that as above we set Y = P x to obtain P ẍ = P P DP x = DP x so that Ÿ = DY. This gives Ÿk = λ k Y k for k =,..., n which has general solution Y k (t = α k cos( λ k t + δ k for k =,..., n. Then we finally have x = P Y = (v v 2,, v n }{{} trix of column vectors v k α cos( λ t + δ α 2 cos( λ 2 t + δ 2. α n cos( λ n t + δ n = n α k v k cos( λ k t + δ k. k= 9
Example: Compound pendulum Consider the compound pendulum shown in figure 8. Two sses, ss m, M, are supported by two light inextensible strings both of length a. We will assume that the strings rein taut for all time. From Newton s 2nd law for ss m: Figure 8: Compound pendulum: two sses on light inextensible strings length a swinging in the vertical plane. Similarly, for the ss M, Geometrically, mẍ = T 2 sin φ T sin θ mÿ = mg + T 2 cos φ T cos θ. Mẍ 2 = T 2 sin φ Mÿ 2 = Mg T 2 cos φ. x = a sin θ, y = a cos θ, x 2 = x + a sin φ, y 2 = y + a cos φ. We assume that oscillations are sll: θ, φ rein sll and sin θ θ, cos θ and similarly for the angle φ. This gives x aθ, y a, x 2 a(θ + φ, y 2 2a. 0
Putting these approxitions into the above equations of motion yields, the approxition for sll angles: From (2 and (4 we obtain (for this approxition Hence from ( and (3 we have We y rewrite this as ( 0 Ma Ma θ = T 2 φ T θ ( 0 = mg + T 2 T (2 Ma( θ + φ = T 2 φ (3 0 = Mg T 2. (4 T 2 = Mg, T = Mg + mg. θ = Mgφ (M + mgθ (5 Ma( θ + φ = Mgφ. (6 d 2 dt 2 ( θ φ = ( (M + mg Mg 0 Mg ( θ φ Hence ( d 2 θ dt 2 φ ( 0 = Ma Ma ( (M + mg Mg 0 Mg ( θ φ ( θ = A φ where g (M + m Mg g g (M + m A = (M + m The characteristic equation for the eigenvalues λ of A reads that is: λ 2 2g ( g 2 (M + mλ + (M + m Mg λ 2 2g (M + mλ + g2 This gives two real and positive solutions λ ± = and so the eigenfrequencies are ω ± = g (M + m = 0. 2 ( (m + M ± M(m + M ( g (M + m = 0,, g ( (m + M ± M(m + M.
When m = M we obtain ω ± = 2g ± 2g. a For ω = ω + we have an eigenvector (, / 2 T and when ω = ω we have the eigenvector (, / 2 T. Hence the two norl modes are ( ( ( θ = φ cos 2g ± 2 2g t + δ ±. 2 a ± ± Notice that the angular velocities of these norl modes are ( ( θ 2g ± 2 ( ( 2g = φ a ± sin 2g ± 2 2g t + δ ±, 2 a from which we see that for ω + either both θ, φ are increasing, or they are both decreasing, but for ω if one angle is increasing the other is decreasing. 2