Question

A stationary tuning fork is in resonance with an air column in a pipe. If the tuning fork is moved with a speed of $2m{s}^{-1}$ in front of the open end of the pipe and parallel to it, the length of the pipe should be changed for the resonance to occur with the moving tuning fork. If the speed of sound in air is $320m{s}^{-1}$, the smallest value of the percentage change required in the length of the pipe is ____________.

Answer 1

Step 1. given data

Speed of tuning fork $v=2m{s}^{-1}$

The speed of sound in air is $v_s=320m{s}^{-1}$

Step 2. Finding the percentage change required in the length

For an open pipe, length $L_1$ frequency is given by, $f=\frac{n{v}_s}{4L_1}$

For fundamental frequency $n=1$, we have

$f=\frac{v_s}{4L_1}$

When the tuning fork is moved with velocity $v$, Due to the Doppler effect the frequency becomes,

$$\begin{gathered} f_d=\frac{v_s}{v_s-v}f \\ =\frac{320}{320-2}\frac{320}{4L_1} \end{gathered}$$

For resonance to occur the length of the pipe is changed to$L_2$ such that,

$$\begin{gathered} f_d=\frac{320}{320-2}\frac{320}{4L_1}=\frac{320}{4L_2} \\ \Rightarrow L_2=\frac{318}{320}{L}_1 \end{gathered}$$

the percentage change in length is,

$$\begin{gathered} \frac{L_1-L_2}{L_1}\times 100=\frac{L_1-{\displaystyle \frac{318}{320}}{L}_1}{L_1}\times 100 \\ =\frac{2}{320}\times 100=0.625\% \end{gathered}$$

Hence, the answer is $0.625$