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 resonance again occurs in n${}^{th}$ 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

The correct option is B $n=3$; $f_2=\frac{5f_1}{4}$

Let the length of the pipe be $l$ and speed of sound in air be $v$.

For open pipe, frequency of second harmonic, $f_1=\frac{2v}{2l}=\frac{v}{l}$ ......(1)

For closed pipe, frequency of $n^{th}$ harmonic, $f_2=\frac{(2n-1)v}{4l}$

$⟹$ $f_2=\frac{(2n-1)f_1}{4}$ ......(2)

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 $f_2=\frac{(2\times 5-1)f_1}{4}=\frac{9f_1}{4}$

Hence option B is correct.