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 ..... $m/s$
Medium
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(b) Distance between two consecutive nodes
$ = \frac{\lambda }{2} = 46 - 16 = 30$
$ \Rightarrow \lambda = 60\,cm = 0.6m$
$\therefore $ $v = n\lambda = 500 \times 0.6 = 300$$m/s$.
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