## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson

# Chapter 15 - Oscillations - Exercises and Problems - Page 418: 56

#### Answer

$v = 0.65~m/s$

#### Work Step by Step

We can find the period for a full swing as: $T = 2\pi~\sqrt{\frac{L}{g}}$ $T = 2\pi~\sqrt{\frac{0.90~m}{9.80~m/s^2}}$ $T = 1.9~s$ Since it is only half a swing from one handhold to the next, the time from one handhold to the next is $\frac{T}{2}$ which is $0.95~s$. We can find the amplitude of the motion as: $\frac{A}{L} = sin(\theta)$ $A = L~sin(\theta)$ $A = (0.90~m)~sin(20^{\circ})$ $A = 0.31~m$ Since the distance from one handhold to the next is $2A$, the distance from one handhold to the next is 0.62 meters. We can find the speed of forward motion as: $v = \frac{distance}{time}$ $v = \frac{0.62~m}{0.95~s}$ $v = 0.65~m/s$

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