Questions
An open pipe is in resonance in its 2nd harmonic with tuning fork of frequency f1. Now it is closed at one end. If the frequency of the tuning fork is increased slowly from
f1 then again a resonance is obtained with a frequency f2 . If in this case the pipe vibrates nth harmonics then
detailed solution
Correct option is C
Open pipe resonance frequency f1=2v2LClosed pipe resonance frequency f2=nv4Lf2=n4f1 (where n is odd and f2>f1 ) ∴ n=5Talk 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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