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Questions  

The stationary wave y=2a sin kx cos ωt in a closed organ pipe is the result of the superposition of y=a sin(ωtkx) and 

a
y=−acos(ω t+kx)
b
y=−asin(ω t+kx)
c
y=asin(ω t+kx)
d
y=acos(ω t+kx)

detailed solution

Correct option is B

In closed organ pipe. If  yincident=asin(ωt−kx)then  yreflected=asin(ωt+kx+π)=−asin(ωt+kx)Superimposition of these two waves give the required stationary wave.

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A tube of certain diameter and of length 48 cm is open at both ends. Its fundamental frequency of resonance is found to be 320 Hz. The velocity of sound in air is 320 m/s. One end of the tube is now closed, considering the effect of end correction, calculate the lowest frequency of resonance for the tube (in Hz). 


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