Question

An open pipe is in resonance in its $2^{\text {nd }}$ 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 $n^{\text {th }}$ harmonics then

• 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. $n=5, f_{2}=\frac{3}{4} f_{1}$

Answer 1

Answer: (C)

Open pipe resonance frequency $f_{1}=\frac{2 v}{2 L}$
Closed pipe resonance frequency $f_{2}=\frac{n v}{4 L}$
$f_{2}=\frac{n}{4} f_{1} $ (where $ n $ is odd and $ f_{2}>f_{1}) $
$\therefore n=5$