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Question

An air column vibrates in a pipe closed at one end in its third overtone due to a tuning fork of frequency 595 Hz. Po is the mean pressure at any point in the pipe and ΔPo is the maximum amplitude of pressure variation. Speed of sound in air is 340 m/s and end correction has a negligible effect. If the amplitude of pressure variation at the middle of the pipe is ΔPon, then n is:
[Given cos(595π340)=12 ]

A
2
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B
1
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C
3
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D
4
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Solution

The correct option is A 2
Given, speed of sound (v)=340 m/s
Frequency, f=595 Hz
So, wavelength (λ)=vf=340595 m

Modes of vibration of an air column in a closed organ pipe are given by

fn=(2n1)v4L

Now, for third overtone,
f4=7v4L

From the data given in the question,

L=7v4f4=7×3404×595=1 m

The general equation of pressure variation for a stationary sound wave in a closed organ pipe is given by
Δp=Δpocos(kx)sin(ωt)

So, pressure amplitude =Δpocos(2πλx)

Pressure amplitude at the middle of the pipe

=∣ ∣Δpocos(2π(340/595)×12)∣ ∣

=Δpocos(595π340)=Δpo2

Comparing this with Δp0n,
we get, n=2.

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