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Homework 5 - 4 Solutions

# Homework 5 - 4 Solutions - PHY 317K Homework 6 Solutions 1...

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PHY 317K Homework 6 Solutions 1. D If the restoring force F is proportional to the displacement, but in the opposite direc- tion: F = - kx , where x is displacement (although it does not have to be displacement along the x -axis) and k is a constant. Then, from Newton’s Second Law, it follows that - kx = ma = m d 2 x dt 2 . Letting ω 2 = k/m , this can be rewriten: d 2 x dt 2 + ω 2 x = 0. Any system obeying this differential equation is a simple harmonic oscillator. 2. B For a simple harmonic oscillator, x ( t ) = x m cos( 2 π T t + φ ). If x = 0 at t = 0, φ = π/ 2. Equivalently, we can take x ( t ) = x m sin 2 π T t . The distance from the equilibrium position ( x = 0) is just | x ( t ) | , so this can be evaluated for different values of t . We find: | x (0 . 5 T ) | = | x ( T ) | = | x (1 . 5 T ) | = 0 < | x (1 . 4 T ) | = 0 . 588 x m < | x (0 . 7 T ) | = 0 . 951 x m , so the farthest of these from the equilibrium is t = 0 . 7 T . 3. A If x = x m at t = 0, x ( t ) = x m cos 2 π T t . Then v ( t ) = - 2 π T sin 2 π T t . Then: x (1 . 75 T ) = 0 and v (1 . 75 T ) = 2 πx m T > 0. So, at t = 1 . 75 T , the particle is at x = 0 and is moving in the + x -direction (toward x = + x m ). 4. C If x = x m at t = 0, x ( t ) = x m cos 2 π T t . Then v ( t ) = - 2 π T sin 2 π T t . Then: x (1 . 5 T ) = - x m and v (1 . 5 T ) = 0. So, at t = 1 . 5 T , the particle is at x = - x m and is at rest. 5. E 80 oscillations in 5.0 s means the period T = 5 . 0 s / 80 = 0 . 0625 s, and the frequency is f = 1 /T = 16 Hz. 6. A 50 oscillations in 10.0 s means the period T = 10 . 0 s / 50 = 0 . 200 s, and the angular frequency is f = 2 π/T = 10 π Hz = 31 . 4 rad / s. 7. D The block oscillates between extreme points that are 60 cm apart, so the amplitude is 30 cm. The time to go between extreme points is 0.20 , so the period is 0.40 s, since the period is the time to complete one full cycle. Then the frequency is 1/(0.40 s) = 2.5 Hz. 1

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8. A The frequency is independent of the amplitude. I.e., changing the amplitude has no affect on the period. 9. B x ( t ) = x m cos( ωt + φ ). If x = 0 at t = 0, cos φ = 0; this allows us to solve for φ , but based on this alone there are two solutions: φ = π/ 2 and φ = 3 π/ 2. Luckily we are also told the direction of motion. Since v ( t ) = dx/dt = - ωx m sin( ωt + φ ), at t = 0 we see v = - ωx m sin φ . For the mass to be moving in the negative x -direction, v must be negative, which means φ = π/ 2. That is, if φ = 3 π/ 2, v would be positive. . 10. D When the mass hangs in equilibrium, the upward spring force is equal in magnitude to the downward force of the weight of the object, so Mg = k (4 . 9 × 10 - 2 m). So we can get the natural angular frequency: ω = k M = g 4 . 9 × 10 - 2 m = 9 . 8 m / s 2 4 . 9 × 10 - 2 m = 200 s - 2 = 14 rad / s. 11. A The displacement of the spring is given by: x ( t ) = A cos( ωt + φ ), where ω = k/m . Then v ( t ) = dx/dt = - ωA sin( ωt + φ ). The maximum speed occurs when the | sin( ωt + φ ) | = 1, so it is: = A k/m . 12. E Here the amplitude is 2.0 m and ω = 75 rad / s, so — using the result of the previous problem, the maximum speed is: = 2 . 0 m(75 rad / s) = 150 m / s. 13. C If v m is the maximum speed, we know from two problems ago that v m = A k/m .
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