Wave Mechanics Question 1106
Question: The length of two open organ pipes are l and $ (l+\Delta l) $ respectively. Neglecting end correction, the frequency of beats between them will be approximately
[MP PET 1994; BHU 1995]
Options:
A) $ \frac{v}{2l} $
B) $ \frac{v}{4l} $
C) $ \frac{v\Delta l}{2l^{2}} $
D) $ \frac{v\Delta l}{l} $ (Here v is the speed of sound)
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Answer:
Correct Answer: C
Solution:
$ {\lambda _{1}}=2l,{\lambda _{2}}=2l+2\Delta l $
Therefore $ n _{1}=\frac{v}{2l} $ and $ n _{2}=\frac{v}{2l+2\Delta l} $
Therefore No. of beats $ =n _{1}-n _{2}=\frac{v}{2}( \frac{1}{l}-\frac{1}{l+\Delta l} )=\frac{v\Delta l}{2l^{2}} $
BETA
