Q. An open pipe resonates with a tuning fork of frequency 500 Hz. It is observed that two successive nodes are formed at distance 16 and 46 cm from the open end. The speed of sound in air in the pipe is
A) $260\ ms^{- 1}$
B) $300\ ms^{- 1}$
C) $320\ ms^{- 1}$
D) $360\ ms^{- 1}$
Explanation
(b) Position of first node=16cm
$\frac{\lambda}{2} + e = 16cm$
Where e =end correction
Position of second node=46cm
$\frac{\lambda}{2} + \frac{\lambda}{2} + e = 46cm$
Dividing Eq. (ii) by Eq.(i)
$\frac{\lambda}{2} = 30\ cm$
$\lambda = 60\ cm = \frac{60}{100}m$
∴ speed of sound v=v$\lambda$
$= 500 \times \frac{60}{100} = 300ms^{- 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.