Question

An open pipe is in resonance in 2nd harmonic with frequency $f_1.$ Now one end of the tube is closed and frequency is increased to $f_2$ such that the resonance again occurs in nth harmonic. Choose the correct option.

• A. $n=3;f_2=\frac{3f_1}{4}$
• B. $n=3;f_2=\frac{5f_1}{4}$
• C. $n=5;f_2=\frac{3f_1}{4}$
• D. $n=5;f_2=\frac{5f_1}{4}$

Answer 1

Answer: (B)

Let the length of the pipe be 1 and speed of sound in air be v.
For open pipe, frequency of second harmonic, $f_1=\frac{2v}{2l}=\frac{v}{l}$
For closed pipe, frequency of $n^{\text{th }}$ harmonic, $f_2=\frac{(2n-1)v}{41}$
$⟹f_2=\frac{(2n-1)f_1}{4}$
Put $n=3$ in equation (2) we get $f_2=\frac{(2\times 3-1)f_1}{4}=\frac{5f_1}{4}$
Put $n=5$ in equation (2) we get $f_2=\frac{(2\times 5-1)f_1}{4}=\frac{9f_1}{4}$