wave_mechanics critical_thinking Question 34
Question: An open pipe is in resonance in its 2nd harmonic with tuning fork of frequency$ f _{1} $ . Now it is closed at one end. If the frequency of the tuning fork is increased slowly from $ f _{1} $ then again a resonance is obtained with a frequency$ f _{2} $ . If in this case the pipe vibrates $ n^{th} $ harmonics then
[IIT-JEE (Screening) 2005]
Options:
A) $ n=3, $
$ f _{2}=\frac{3}{4}f _{1} $
B) $ n=3, $
$ f _{2}=\frac{5}{4}f _{1} $
C) $ n=5, $
$ f _{2}=\frac{5}{4}f _{1} $
D) $ n=5, $
$ f _{2}=\frac{3}{4}f _{1} $
Show Answer
Answer:
Correct Answer: C
Solution:
Open pipe resonance frequency $ f _{1}=\frac{2v}{2L} $ Closed pipe resonance frequency $ f _{2}=\frac{nv}{4L} $
$ f _{2}=\frac{n}{4}f _{1} $ (where n is odd and $ f _{2}>f _{1} $ ) \ n = 5
BETA
