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.