A closed pipe and an open pipe have their first overtones identical in frequency. Their lengths are in the…

A closed pipe and an open pipe have their first overtones identical in frequency. Their lengths are in the ratio
  1. $1:2$
  2. $2:3$
  3. $3:4$
  4. $4:5$

Solution

If it is given that First overtone of closed pipe = First overtone of open pipe $\Rightarrow 3 \left( \frac{\nu}{4 l_{1}} \right) = 2 \left( \frac{\nu}{2 l_{2}} \right)$; where $l_{1}$ and $l_{2}$ are the lengths of closed and open organ pipes hence $\frac{l_{1}}{l_{2}} = \frac{3}{4}$

Asked in: MHT CET Full Test 2

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