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Chapter 9: Problem 165

Velocity of sound in air is \(320 \mathrm{~m} / \mathrm{s}\). A pipe closed at one end has a length of \(1 \mathrm{~m}\). Neglecting end corrections, the air column in the pipe can resonate for sound at frequency (A) \(80 \mathrm{~Hz}\) (B) \(240 \mathrm{~Hz}\) (C) \(320 \mathrm{~Hz}\) (D) \(400 \mathrm{~Hz}\)

Short Answer

Expert verified
The resonant frequency of the air column in the pipe is calculated as \( 160 Hz \) which is not part of the given options. Thus, there seems to be an issue with the given exercise.

Step by step solution

01

Identifying the Given Information

In this exercise, the velocity of sound in air, \( v \), is given as \( 320 m/s \). The length of the pipe, \( L \), is given as \( 1 m \). So, these are the values that we need to use in the formula for calculating the resonant frequency.
02

Applying the Formula

Now, applying all the values to the formula \( f = v/(2L) \), we replace \( v \) with \( 320 m/s \) and \( L \) with \( 1 m \). So the formula becomes \( f = 320/(2*1) \).
03

Calculating the Resonant Frequency

Calculating the above expression, we have \( f = 320/2 \) which gives us \( f = 160 Hz \). So the resonant frequency of the air column in the pipe is \( 160 Hz \). However, none of the given options match with this value. Hence there may be a mistake in the given exercise.

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