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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

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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)=5√3=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.

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)=5√3=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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