An open air pipe of length $80 \mathrm{~cm}$ has the second harmonic frequency equal to the fundamental…

An open air pipe of length $80 \mathrm{~cm}$ has the second harmonic frequency equal to the fundamental frequency of a closed organ air pipe. The length of the closed pipe is
  1. $20 \mathrm{~cm}$
  2. $40 \mathrm{~cm}$
  3. $60 \mathrm{~cm}$
  4. $10 \mathrm{~cm}$

Solution

We have $\left(\mathrm{f}_2\right)_{\text {Open }}=\left(\mathrm{f}_2\right)_{\text {Closed }}$ $\begin{aligned} & \Rightarrow 2 \cdot \frac{\mathrm{V}}{2 l_0}=\frac{\mathrm{V}}{4 l_{\mathrm{C}}} \\ & \Rightarrow \frac{1}{80}=\frac{1}{4 l_{\mathrm{C}}} \quad\left[\because l_0=80 \mathrm{~cm}\right] \\ & \Rightarrow l_{\mathrm{C}}=20 \mathrm{~cm}\end{aligned}$

Asked in: AP EAMCET 2022 (08 Jul Shift 1)

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