Animated Solution for Physics - Waves: A source, approaching with speed $u$ towards the open end of a stationary pipe of length $L$, is emitting a sound of frequency $f_s$. The farther end of the pipe is closed. The speed of sound in air is $v$ and $f_0$ is the fundamental frequency of the pipe. For which of the following combination(s) of $u$ and $f_s$, will the sound reaching the pipe lead to a resonance ?

Imagine you are standing near a railway track, and a train approaches you while blowing its horn. You've probably noticed that the pitch of the horn sounds higher as the train comes towards you and drops suddenly as it passes by. This phenomenon is known as the Doppler Effect.

In this problem, we have a similar setup, but instead of a human ear, the observer is the open end of a stationary pipe. A sound source is moving towards this pipe with a velocity $u$, emitting a frequency $f_s$. Because the source is chasing its own sound waves, the wavefronts get compressed in the forward direction. This compression leads to a shorter effective wavelength and, consequently, a higher frequency reaching the pipe.

We can quantify this apparent frequency $f$ using the standard Doppler formula for a moving source and a stationary observer:

Here, $v$ is the speed of sound in air. The minus sign in the denominator is crucial—it mathematically represents the source moving towards the observer, which decreases the denominator and increases the overall frequency $f$.

Now, let's shift our focus to the pipe itself. The problem states that the farther end of the pipe is closed. This boundary condition is extremely important. At the closed end, the air molecules cannot move, creating a displacement node. At the open end, the air molecules are free to oscillate maximally, creating a displacement antinode.

Because of these strict boundary conditions, a closed pipe cannot resonate at just any frequency. It is very picky! It will only resonate if the incoming sound wave can form a perfect standing wave inside it. Mathematically, this happens only when the length of the pipe $L$ is an odd multiple of a quarter wavelength ($\lambda /4$).

In terms of frequency, this means a closed pipe only supports odd harmonics of its fundamental frequency $f_0$. Therefore, for resonance to occur, the apparent frequency $f$ must satisfy:

If the incoming frequency is an even multiple (like $2f_0$ or $4f_0$) or a fractional multiple (like $2.5f_0$), the waves will destructively interfere, and no resonance will build up.

Armed with our Doppler formula and our resonance condition, we are now ready to interrogate each option to see which ones pass the test.

Evaluating Option (A): We are given $u=0.8v$ and $f_s=f_0$. Let's plug these into our Doppler equation:

Since $5$ is an odd integer, this frequency perfectly matches the 5th harmonic of the pipe. Resonance will occur.

Evaluating Option (B): Here, $u=0.8v$ and $f_s=2f_0$. Substituting these values:

The number $10$ is an even integer. As we established, a closed pipe strictly forbids even harmonics. No resonance.

Evaluating Option (C): For this option, $u=0.8v$ and $f_s=0.5f_0$. Let's calculate:

The multiplier $2.5$ is a fraction. It does not correspond to any harmonic mode of the pipe. No resonance.

Evaluating Option (D): Finally, we have $u=0.5v$ and $f_s=1.5f_0$. Let's see what happens:

The number $3$ is an odd integer, corresponding to the 3rd harmonic (or the first overtone). Resonance will occur.

This problem is a beautiful illustration of how JEE Advanced weaves multiple concepts together. You cannot solve it just by knowing the Doppler effect, nor can you solve it just by knowing the physics of organ pipes. You must synthesize the two: the moving source dictates the input frequency, and the geometry of the pipe dictates the acceptable frequencies.

By carefully applying both principles, we confidently conclude that only the combinations in (A) and (D) will lead to a resonant standing wave.