## College Physics (4th Edition)

Published by McGraw-Hill Education

# Chapter 12 - Problems - Page 464: 40

#### Answer

(a) The tension is $85.6~N$ (b) $v = 432.3~m/s$ (c) The frequency of the slide whistle is $335.0~Hz$. (d) The length of the tube is $0.256~m$

#### Work Step by Step

(a) We can find the tension $F$: $\sqrt{\frac{F}{\mu}} = v$ $\sqrt{\frac{F}{\mu}} = \lambda~f$ $\sqrt{\frac{F}{m/L}} = 2L~f$ $\frac{F}{m/L} = (2L~f)^2$ $\frac{F~L}{m} = 4L^2~f^2$ $F = 4L~f^2~m$ $F = (4)(0.655~m)(330.0~Hz)^2~(0.300\times 10^{-3}~kg)$ $F = 85.6~N$ The tension is $85.6~N$ (b) We can find the wave speed on the string: $v = \lambda~f$ $v = 2L~f$ $v = (2)(0.655~m)(330.0~Hz)$ $v = 432.3~m/s$ (c) The beat frequency of $5~Hz$ is the difference between the frequency of the string and the frequency of the slide whistle. Since the frequency of the slide whistle is higher, the frequency of the slide whistle is $335.0~Hz$. (d) We can find the length of the tube: $\lambda~f = v$ $4L~f = v$ $L = \frac{v}{4f}$ $L = \frac{343~m/s}{(4)(335.0~Hz)}$ $L = 0.256~m$ The length of the tube is $0.256~m$

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