Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 11 - Oscillations and Waves - General Problems - Page 325: 64

Answer

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Work Step by Step

a. Find the frequency form the length and g. $$f=\frac{1}{2 \pi}\sqrt{g/L}=\frac{1}{2 \pi}\sqrt{(9.80m/s^2)/0.72m}=0.59Hz$$ b. See Figure 11-12. When a pendulum of length L is released from rest, at an angle of $\theta_0$ away from the vertical, the mass m is at a height of $L(1-cos \theta_0)$ above the lowest point. Use energy conservation to relate the potential energy at the starting, maximum height of the pendulum, and the kinetic energy at the lowest point. Let the lowest point represent zero gravitational potential energy. $$PE_i=KE_f$$ $$mg L(1-cos \theta_0)=\frac{1}{2}mv_{max}^2$$ $$v_{max}=\sqrt{2gL(1-cos \theta_0)}$$ $$v_{max}=\sqrt{2(9.80m/s^2)(0.72m)(1-cos 12^{\circ})}=0.56m/s$$ c. The total energy is the PE at the start, or the KE at the lowest point. $$E=\frac{1}{2}mv_{max}^2=\frac{1}{2}(0.295kg)(0.5553m/s)^2=0.045J$$
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