Find the frequency of minimum distance between compression & rarefaction of a wire. If the length of the wire is $1m$ & velocity of sound in air is $360\, m/s$ .... $sec^{-1}$
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(a) The minimum distance between compression and rarefaction of the wire
$l = \frac{\lambda }{4}$
$\therefore $ Wave length $\lambda = 4l$
Now by $v = n\lambda \Rightarrow n = \frac{{360}}{{4 \times 1}} = 90\,{\sec ^{ - 1}}.$
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