## College Physics (4th Edition)

Published by McGraw-Hill Education

# Chapter 11 - Problems - Page 429: 30

#### Answer

We can rank the waves in order of maximum transverse speed, from largest to smallest: $e \gt a \gt b = d \gt c$

#### Work Step by Step

The maximum transverse speed $v_m = A~\omega = A~(2\pi~f)$. The frequency is the number of cycles per second. Let $t$ be the total time shown on each graph. We can count the number of cycles in each graph to find an expression for the frequency. The amplitude is the maximum distance each point moves away from the center during each cycle. We can count the units from the center of the wave to find an expression for the amplitude for each graph. We can find an expression for the maximum transverse speed for each graph. (a) $\omega =2\pi f = \frac{(2\pi)(5)}{t} = \frac{10\pi}{t}$ $A = 3~units$ $v_m = A~\omega = (3)(\frac{10\pi}{t}) = \frac{30\pi}{t}$ (b) $\omega =2\pi f = \frac{(2\pi)(2)}{t} = \frac{4\pi}{t}$ $A = 4~units$ $v_m = A~\omega = (4)(\frac{4\pi}{t}) = \frac{16\pi}{t}$ (c) $\omega =2\pi f = \frac{(2\pi)(2)}{t} = \frac{4\pi}{t}$ $A = 2~units$ $v_m = A~\omega = (2)(\frac{4\pi}{t}) = \frac{8\pi}{t}$ (d) $\omega =2\pi f = \frac{(2\pi)(4)}{t} = \frac{8\pi}{t}$ $A = 2~units$ $v_m = A~\omega = (2)(\frac{8\pi}{t}) = \frac{16\pi}{t}$ (e) $\omega =2\pi f = \frac{(2\pi)(4)}{t} = \frac{8\pi}{t}$ $A = 4~units$ $v_m = A~\omega = (4)(\frac{8\pi}{t}) = \frac{32\pi}{t}$ We can rank the waves in order of maximum transverse speed, from largest to smallest: $e \gt a \gt b = d \gt c$

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