A tuning fork of frequency ' $n$ ' is held near the open end of tube which is closed at the other end and…

A tuning fork of frequency ' $n$ ' is held near the open end of tube which is closed at the other end and the length are adjusted until resonance occurs. The first resonance occurs at length $\mathrm{L}_1$ and immediate next resonance occurs at length $\mathrm{L}_2$. The speed of sound in air is
  1. $\mathrm{n}\left(\mathrm{L}_2-\mathrm{L}_1\right)$
  2. $\frac{\mathrm{n}\left(\mathrm{L}_2-\mathrm{L}_1\right)}{2}$
  3. $2 \mathrm{n}\left(\mathrm{L}_2-\mathrm{L}_1\right)$
  4. $\frac{\mathrm{n}\left(\mathrm{L}_2+\mathrm{L}_1\right)}{2}$

Solution

For first resonance $\mathrm{L}_1=\frac{\lambda}{4}$ For second resonance $\mathrm{L}_2=\frac{3 \lambda}{4}$ $\begin{aligned} & \therefore \mathrm{L}_2-\mathrm{L}_1=\frac{\lambda}{2} \text { or } \lambda=2\left(\mathrm{~L}_2-\mathrm{L}_1\right) \\ & \mathrm{V}=\mathrm{n} \lambda=2 \mathrm{n}\left(\mathrm{L}_2-\mathrm{L}_1\right) \end{aligned}$

Asked in: MHT CET 2021 (22 Sep Shift 2)

Practice more Waves and Sound questions on Aicharya