Question

An open pipe resonates with a tuning fork of frequency $ 500 \,Hz $ . It is observed that two successive nodes are formed at distances $ 16 $ and $ 46 \,cm $ from the open end. The speed of sound in air in the pipe is

• A. 230 m/s
• B. 300 m/s
• C. 320 m/s
• D. 360 m/s

Answer 1

Answer: (B)

Position of first node $=16\,cm$
$\frac{\lambda}{2}+e=16\,cm$ $\dots (i)$
Where, $e$= end correction
Position of second node $=46\,cm$
$\frac{\lambda}{2}+\frac{\lambda}{2}+e=46\,cm$ $\dots (ii)$
From Eqs. $(i)$ and $(ii)$, $\frac{\lambda}{2}=30\,cm$
$\lambda=60\,cm$
$=\frac{60}{100}m$
$\therefore $ Speed of sound, $V=n \lambda$
$=500\times \frac{60}{100}$ $(\because$ $n=500\,Hz$, Given)
$=300\, m/s$