Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 14 - Oscillations - Exercises and Problems - Page 403: 38

Answer

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Work Step by Step

We know that the average speed is given by $$v_{\rm avg}=\dfrac{\Delta x}{\Delta t}$$ And for a particle that undergoes a simple harmonic motion, during a one-half period $T/2$, it moves from $+A\rightarrow -A$. This means that $\Delta x=-A-(+A)=-2A$ and we can neglect the negative sign here. Thus, $$v_{\rm avg}=\dfrac{2A}{\frac{1}{2}T}=\dfrac{4A}{T}$$ We know that the period $T$ is given by $\omega=2\pi f=2\pi/T$ Hence, $T=2\pi/\omega$ $$v_{\rm avg} =\dfrac{4A}{\frac{2\pi}{\omega}}=\dfrac{2 A\omega}{\pi }$$ Recalling that $v_{\rm max}=A\omega$, $$v_{\rm avg} =\dfrac{2 v_{\rm max}}{\pi }$$ Therefore, $$\boxed{v_{\rm max}=\dfrac{\pi v_{\rm avg}}{2}}$$
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