Question

The speed of sound in air is $320m{s}^{-1}$. The length of a closed pipe is $1m$. Neglecting the end correction, the resonant frequency for the pipe will be

• A. $80Hz$
• B. $240Hz$
• C. $320Hz$
• D. $400Hz$

Answer 1

Answer: (A)

$80Hz$

The resonant frequency for the closed pipe is given by
$\nu =(2n+1)\frac{v}{4l}$

For n=0, the fundamental resonant frequency(minimum allowed frequency) ${\nu }_0$ is ${\nu }_0=\frac{v}{4l}$

where, v is the speed of sound in air and l is the length of closed pipe.
Therefore,

$n{u}_0=\frac{320}{4\left(1\right)}$

$n{u}_0=80Hz$