A source, approaching with speed 𝑢 towards the open end of a stationary pipe of length 𝐿, is emitting a sound of frequency 𝑓𝑠 . The farther end of the pipe is closed. The speed of sound in air is 𝑣 and 𝑓0 is the fundamental frequency of the pipe. For which of the following combination(s) of 𝑢 and 𝑓𝑠 , will the sound reaching the pipe lead to a resonance? (A) 𝑢 = 0.8𝑣 and 𝑓𝑠 = 𝑓0 (B) 𝑢 = 0.8𝑣 and 𝑓𝑠 = 2𝑓0 (C) 𝑢 = 0.8𝑣 and 𝑓𝑠 = 0.5𝑓0 (D) 𝑢 = 0.5𝑣 and 𝑓𝑠 = 1.5𝑓0

To determine which combination(s) of $u$ and $f_s$ will lead to resonance in the pipe, we need to analyze the situation carefully. Here's a step-by-step breakdown:

1. Understanding the Problem

• Pipe Configuration: The pipe has one open end and one closed end. For such pipes, the fundamental frequency $f_0$ is given by:

  $$f_0=\frac{v}{4L}$$

  where $v$ is the speed of sound in air and $L$ is the length of the pipe.

• Source Movement: The sound source is moving towards the open end of the pipe with speed $u$. This movement affects the frequency observed at the open end due to the Doppler effect.

• Resonance Condition: Resonance occurs when the frequency of the sound wave matches one of the natural frequencies of the pipe. For a pipe with one open and one closed end, the natural frequencies are odd harmonics of the fundamental frequency:

  $$f_n=(2n+1)f_0\text{where }n=0,1,2,\dots$$

2. Doppler Effect Consideration

Since the source is moving towards the stationary pipe, the frequency observed at the open end $f_{\text{observed}}$ is higher than the emitted frequency $f_s$. The Doppler effect formula for this scenario is:

$$f_{\text{observed}}=\frac{v}{v-u}{f}_s$$

where:

• $v$ is the speed of sound,

• $u$ is the speed of the source towards the pipe,

• $f_s$ is the frequency emitted by the source.

3. Resonance Condition

For resonance to occur, the observed frequency $f_{\text{observed}}$ must match one of the natural frequencies of the pipe:

$$f_{\text{observed}}=(2n+1)f_0$$

Substituting the Doppler effect formula:

$$\frac{v}{v-u}{f}_s=(2n+1)f_0$$

Solving for $f_s$:

$$f_s=\frac{(v-u)}{v}(2n+1)f_0$$

4. Analyzing Each Option

Let's analyze each option to see if it satisfies the resonance condition.

Option A: $u=0.8v$ and $f_s=f_0$

Plugging into the equation:

$$f_s=\frac{(v-0.8v)}{v}(2n+1)f_0=0.2(2n+1)f_0$$

Given $f_s=f_0$:

$$f_0=0.2(2n+1)f_0⟹1=0.2(2n+1)⟹2n+1=5⟹n=2$$

Since $n$ must be an integer, this condition is satisfied. Therefore, resonance occurs.

Option B: $u=0.8v$ and $f_s=2f_0$

Plugging into the equation:

$$2f_0=0.2(2n+1)f_0⟹2=0.2(2n+1)⟹2n+1=10⟹n=4.5$$

Since $n$ must be an integer, this condition is not satisfied. Therefore, resonance does not occur.

Option C: $u=0.8v$ and $f_s=0.5f_0$

Plugging into the equation:

$$0.5f_0=0.2(2n+1)f_0⟹0.5=0.2(2n+1)⟹2n+1=2.5⟹n=0.75$$

Since $n$ must be an integer, this condition is not satisfied. Therefore, resonance does not occur.

Option D: $u=0.5v$ and $f_s=1.5f_0$

Plugging into the equation:

$$f_s=\frac{(v-0.5v)}{v}(2n+1)f_0=0.5(2n+1)f_0$$

Given $f_s=1.5f_0$:

$$1.5f_0=0.5(2n+1)f_0⟹1.5=0.5(2n+1)⟹2n+1=3⟹n=1$$

Since $n$ must be an integer, this condition is satisfied. Therefore, resonance occurs.

5. Conclusion

After analyzing each option:

• Option A: Resonance occurs.

• Option B: Resonance does not occur.

• Option C: Resonance does not occur.

• Option D: Resonance occurs.

Final Answer: Options A and D lead to resonance.