Question

The velocity of sound in air is $332m{s}^{-1}$. The length of a closed pipe whose frequency of second overtone is $332Hz,$ will be :

• A. 0.51 m
• B. 0.75 m
• C. 1.25 m
• D. 1.75 m

Answer 1

The correct option is C 1.25 m

For a closed pipe the frequency is given by

$\nu =(2n+1){\nu }_0=(2n+1)\frac{v}{4l}$

where, ${\nu }_0$ is fundamental frequency, v is velocity of sound in air and n is an integer such as n = 0,1,2,3.........

The second overtone frequency is

$\nu =\left[2\right(2)+1]\frac{v}{4l}$

$\nu =\frac{5v}{4l}$

$l=\frac{5v}{4\nu }$

$l=\frac{5\left(332\right)}{4\left(332\right)}$

$l=\frac{5}{4}=1.25m$