## College Physics (4th Edition)

Published by McGraw-Hill Education

# Chapter 12 - Problems - Page 464: 39

#### Answer

The beat frequency is $2.3~Hz$

#### Work Step by Step

At a temperature of $20.0^{\circ}C$, the speed of sound in the air is $343~m/s$. We can find the speed of sound in the air at a temperature of $25.0^{\circ}C$: $v = 331+0.6~T$ $v = 331+0.6~(25.0^{\circ}C)$ $v = 346~m/s$ We can write an expression for the frequency of the pipe at the front: $f_f = \frac{343~m/s}{\lambda}$ We can write an expression for the frequency of the pipe at the back: $f_b = \frac{346~m/s}{\lambda}$ We can divide the two expressions to find $f_b$: $\frac{f_b}{f_f} = \frac{\frac{346~m/s}{\lambda}}{\frac{343~m/s}{\lambda}}$ $f_b = \frac{346~m/s}{343~m/s}~f_f$ $f_b = \frac{346~m/s}{343~m/s}~(264.0~Hz)$ $f_b = 266.3~Hz$ Since the beat frequency is the difference between the two frequencies, the beat frequency is $2.3~Hz$

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