Questions
The length of two open organ pipes are l and respectively. Neglecting end correction, the frequency of beats between them will be approximately
(Here v is the speed of sound)
detailed solution
Correct option is C
λ1=2l, λ2=2l+2Δl ⇒n1=v2l and n2=v2l+2Δl⇒ No. of beats =n1−n2=v21l−1l+Δl=vΔl2l2Talk to our academic expert!
Similar Questions
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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