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Question

The air column in an open organ pipe is made to vibrate in its first overtone by a tuning fork of frequency 680 Hz. Speed of sound in air is 340 m/s and 105 N/m2 is the mean pressure at any point in the pipe. If 10 N/m2 is the maximum amplitude of pressure variation, then

A
Length of organ pipe is 1 m.
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B
Amplitude of pressure variation at x=13 is 8.6 N/m2.
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C
Pressure at the end of the pipe is 105 N/m2.
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D
Maximum pressure in the pipe is (105+10) N/m2.
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Solution

The correct option is D Maximum pressure in the pipe is (105+10) N/m2.
Given that,
Frequency of tuning fork (f)=680 Hz
Speed of the sound (V)=340 m/s
Mean Pressure (P0)=105 N/m2
Max. amplitude of pressure (Po)=10 N/m2

Modes of vibration of air column in an open organ pipe are given by
fn=nv2L
For first overtone,
f2=vL
λ=L

From the data given in the question,
f=f2
680=340L
L=340680=0.5 m

Stationary wave can be represented as
Δp=Δposin(kx)cos(ωt)
Δp=10sin(4πx)cos(ωt) [k=2πλ]

For amplitude of pressure variation at x=13,

Δp(x)=|10sin(4πx)|
Δp(x)=10sin(4π3)
Δp(x)=53=8.6 N/m2

As pressure node is formed at the end of the pipe, the pressure at the end of pipe is equal to the mean pressure.
pend=105 N/m2

Maximum pressure in the pipe is
pmax=po+Δpo=(105+10) N/m2
Options B, C, D are correct.

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