Percentage Change in Pipe Length for Resonance with Moving Tuning Fork

To find the required percentage change in the length of the pipe for resonance to occur with the moving tuning fork, we analyze the system using the Doppler effect and the condition for standing waves in organ pipes.

Step 1: Initial Resonance Condition

Let the original frequency of the stationary tuning fork be $f_0$ and the initial length of the pipe be $L_0$.

For an organ pipe (whether open at both ends or closed at one end), the resonant frequency $f$ of any harmonic is inversely proportional to its length $L$: $f = \frac{n v}{4L} \quad \text{or} \quad f = \frac{n v}{2L}$

where $v = 320\text{ ms}^{-1}$ is the speed of sound in air and $n$ is an integer representing the harmonic.

Thus, for a given mode of resonance: $f \cdot L = \text{constant}$

Initially, when the fork is stationary: $f_0 \cdot L_0 = C \implies L_0 = \frac{C}{f_0}$

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Step 2: Frequency of Sound Received from the Moving Tuning Fork

When the tuning fork moves with a speed $u = 2\text{ ms}^{-1}$ parallel to the open end of the pipe, the frequency of sound entering the pipe is modified due to the Doppler effect.

The maximum shift in frequency occurs when the component of velocity along the line joining the source to the pipe opening is maximum (i.e., when the fork is approaching the line of sight, $\theta \to 0^\circ$): $f' = f_0 \left( \frac{v}{v - u} \right)$

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Step 3: New Pipe Length for Resonance

For the air column inside the pipe to resonate with the modified frequency $f'$, the length of the pipe must be changed to a new length $L'$ such that: $f' \cdot L' = C$

Substituting $C = f_0 L_0$: $f' \cdot L' = f_0 \cdot L_0$

$L' = L_0 \left( \frac{f_0}{f'} \right)$

Substituting the Doppler-shifted frequency $f'$ into the expression: $L' = L_0 \left( \frac{v - u}{v} \right) = L_0 \left( 1 - \frac{u}{v} \right)$

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Step 4: Percentage Change in Length

The magnitude of the change in length of the pipe is: $\Delta L = |L_0 - L'| = L_0 \left( \frac{u}{v} \right)$

The percentage change required in the length of the pipe is: $\text{Percentage Change} = \left( \frac{\Delta L}{L_0} \right) \times 100\% = \left( \frac{u}{v} \right) \times 100\%$

Substituting the given values $u = 2\text{ ms}^{-1}$ and $v = 320\text{ ms}^{-1}$: $\text{Percentage Change} = \left( \frac{2}{320} \right) \times 100\% = \frac{1}{160} \times 100\% = \frac{5}{8}\% = 0.625\%$

Thus, the required minimum percentage change in the length of the pipe is $0.625\%$ (or $0.63\%$).