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. $n =3 ; f _{2}=\frac{3 f _{1}}{4}$
• B. $n =3 ; f _{2}=\frac{5 f _{1}}{4}$
• C. $n =5 ; f _{2}=\frac{3 f _{1}}{4}$
• D. $n =5 ; f _{2}=\frac{5 f _{1}}{4}$

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, $f_{1}=\frac{2 v }{2 l }=\frac{ v }{ l }$
For closed pipe, frequency of $n ^{\text {th }}$ harmonic, $f _{2}=\frac{(2 n -1) v }{41}$
$\Longrightarrow f _{2}=\frac{(2 n -1) f _{1}}{4}$
Put $n =3$ in equation (2) we get $f _{2}=\frac{(2 \times 3-1) f _{1}}{4}=\frac{5 f _{1}}{4}$
Put $n =5$ in equation (2) we get $f _{2}=\frac{(2 \times 5-1) f _{1}}{4}=\frac{9 f _{1}}{4}$