Ocio

Problem 21.1 In Active Example 21.1, suppose that the pulley has radius R = 100 mm and its moment of inertia is I = 0.005 kg-m2 . The mass m = 2 kg, and the spring constant is k = 200 N/m. If the mass is displaced downward from its equilibrium position and released, what are the period and frequency of the resulting vibration?
k

R

Solution: From Active Example 21.1 we have
ω= k I m+ 2 R =(200 N/m) (2 kg) + (0.005 kg-m2 ) (0.1 m)2 = 8.94 rad/s

m

Thus τ= 2π 2π = = 0.702 s, ω 8.94 rad/s f = 1.42 Hz. f = 1 = 1.42 Hz. τ

τ = 0.702 s,

Problem 21.2 In Active Example 21.1, suppose that the pulley has radius R = 4 cm and its moment of inertia is I = 0.06 kg-m 2. The suspended object weighs 5 N, and the spring constant is k = 10 N/m. The system is initially at rest in itsequilibrium position. At t = 0, the suspended object is given a downward velocity of 1 m/s. Determine the position of the suspended object relative to its equilibrium position as a function of time.
k

R

m

Solution: From Active Example 21.1 we have
ω= k m+ I R2 = (10 N/m) 5N 9.81 m/s2 + (0.06 kg-m 2 ) (0.04 m) 2 = 0.51 rad/s

The general solution is x = A sin ωt + B cos ωt, v = Aω cos ωt −Bω sin ωt.

Putting in the initial conditions, we have x(t = 0) = B = 0 ⇒ B = 0, v(t = 0) = Aω = (1 ft/s) ⇒ A = Thus the equation is x = 1.96 sin 0.51t (m ). 1 m/s = 1.96 m . 0.51 rad/s

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Problem 21.3 The mass m = 4 kg. The spring is unstretched when x = 0. The period of vibration of the mass is measured and determined to be 0.5 s. The mass is displaced to the position x = 0.1 m and released from rest at t = 0. Determine its position at t = 0.4 s.x k m

Solution: Knowing the period, we can find the natural frequency
ω= The general solution is x = A sin ωt + B cos ωt, v = Aω cos ωt − Bω sin ωt. 2π 2π = = 12.6 rad/s. τ 0.5 s

Putting in the initial conditions, we have x(t = 0) = B = 0.1 m ⇒ B = 0.1 m, v(t = 0) = Aω = 0 ⇒ A = 0. Thus the equation is x = (0.1 m) cos(12.6 rad/s t) At the time t = 0.4 s, we find x = 0.0309 m.

Problem21.4 The mass m = 4 kg. The spring is unstretched when x = 0. The frequency of vibration of the mass is measured and determined to be 6 Hz. The mass is displaced to the position x = 0.1 m and given a velocity dx/dt = 5 m/s at t = 0. Determine the amplitude of the resulting vibration.

x k m

Solution: Knowing the frequency, we can find the natural frequency ω = 2πf = 2π(6 Hz) = 37.7 rad/s. Thegeneral solution is x = A sin ωt + B cos ωt, v = Aω cos ωt − Bω sin ωt.

Putting in the initial conditions, we have x(t = 0) = B = 0.1 m ⇒ B = 0.1 m, v(t = 0) = Aω = 5 m/s ⇒ A = 5 m/s = 0.133 m. 37.7 rad/s

The amplitude of the motion is given by A2 + B 2 = (0.1 m)2 + (0.133 m)2 = 0.166 m.

Amplitude = 0.166 m.

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c 2008 Pearson Education South Asia Pte Ltd. All rights reserved. Thispublication is protected by Copyright and permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording or likewise.

Problem 21.5 The mass m = 4 kg and the spring constant is k = 64 N/m. For vibration of the spring-mass oscillator relative to itsequilibrium position, determine (a) the frequency in Hz and (b) the period.
k x

Solution: Since the vibration is around the equilibrium position,
we have k = m 1 cycle 64 N/m = 4 rad/s 4 kg 2π rad

m
20

(a)

ω=

= 0.637 Hz

(b)

τ =

2π 2π = = 1.57 s ω 4 rad/s

Problem 21.6 The mass m = 4 kg and the spring constant is k = 64 N/m. The spring is unstretched when x = 0. At t...