Doppler Effect and Resonance in Closed Pipe with Moving Source

To determine which combinations of the source speed $u$ and emitted frequency $f_s$ will produce resonance in the pipe, we analyze the Doppler effect and the natural frequencies of a closed pipe.

1. Resonant Frequencies of a Closed Pipe

For a pipe of length $L$ open at one end and closed at the other, the natural frequencies (resonant frequencies) are given by odd multiples of the fundamental frequency $f_0$: $f_0 = \frac{v}{4L}$ $f_{\text{res}} = (2n + 1) f_0 \quad \text{for } n = 0, 1, 2, 3, \dots$ Thus, the possible frequencies that can cause resonance are $f_0, 3f_0, 5f_0, 7f_0, \dots$

2. Apparent Frequency due to the Doppler Effect

The source is approaching the stationary pipe with a speed $u$. The apparent frequency $f'$ of the sound wave entering the open end of the pipe is given by: $f' = f_s \left( \frac{v}{v - u} \right)$

For resonance to occur, the apparent frequency $f'$ must match one of the resonant frequencies of the pipe: $f' = (2n + 1) f_0 \quad \text{where } n \in \{0, 1, 2, \dots\}$

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3. Evaluation of Options

• Option (A): $u = 0.8v$ and $f_s = f_0$ $f' = f_0 \left( \frac{v}{v - 0.8v} \right) = f_0 \left( \frac{v}{0.2v} \right) = 5f_0$ Since $5f_0 = (2 \cdot 2 + 1)f_0$, $f'$ is an odd multiple of $f_0$. Resonance occurs.

• Option (B): $u = 0.8v$ and $f_s = 2f_0$ $f' = 2f_0 \left( \frac{v}{v - 0.8v} \right) = 2f_0 \left( 5 \right) = 10f_0$ Since $10f_0$ is an even multiple of $f_0$, resonance does not occur.

• Option (C): $u = 0.8v$ and $f_s = 0.5f_0$ $f' = 0.5f_0 \left( \frac{v}{v - 0.8v} \right) = 0.5f_0 \left( 5 \right) = 2.5f_0$ Since $2.5f_0$ is not an odd integer multiple of $f_0$, resonance does not occur.

• Option (D): $u = 0.5v$ and $f_s = 1.5f_0$ $f' = 1.5f_0 \left( \frac{v}{v - 0.5v} \right) = 1.5f_0 \left( \frac{v}{0.5v} \right) = 1.5f_0 (2) = 3f_0$ Since $3f_0 = (2 \cdot 1 + 1)f_0$, $f'$ is an odd multiple of $f_0$. Resonance occurs.

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Conclusion

The correct options are (A) and (D).