Question

An open organ pipe resonated with frequency ` $f_1$' and $2^{nd}$ harmonic. Now one end is closed and the frequency is slowly increased then it resonates with frequency $f_2$ and nth harmonic then

• A. $n=3,f_2=\frac{3}{4}{f}_1$
• B. $n=5,f_2=\frac{3}{4}{f}_1$
• C. $n=3,f_2=\frac{5}{4}{f}_1$
• D. $n=5,f_2=\frac{5}{4}{f}_1$

Answer 1

The correct option is D $n=5,f_2=\frac{5}{4}{f}_1$

Let the length of pipe be "L"

Initial Frequency $f_1=\frac{1}{L}\sqrt{\frac{B}{\rho }}$

Final frequency $f_2=\frac{n}{4L}\sqrt{\frac{B}{\rho }}$

This is just greater than $f_1$ and n must be odd in closed pipes

So $n=5$ and $f_2=\frac{5}{4l}\sqrt{\frac{B}{\rho }}=\frac{5}{4}{f}_1$