A sound source S is moving along a straight track with speed v and is emitting sound of frequency

Q: Why doesn't the frequency stay constant while the source is approaching?

Because the observer is at a finite distance from the track, the angle between the track and the line of sight continuously changes. This means the component of velocity towards the observer is not constant, causing the frequency to continuously drop.

Q: What if the observer was standing exactly on the track?

If the observer were on the track, the angle would be zero until the source hits them. The frequency would be a constant high value, and then instantly drop to a constant low value after crossing. This would look like a step function (Graph a).

Q: Why is the frequency exactly $\nu_o$ at $t_0$?

At $t_0$, the source is at the closest point, meaning its velocity is perpendicular to the line of sight. The component of velocity towards the observer is zero, so there is no Doppler shift at that exact instant.

Q: Why does the curve flatten out at the ends?

When the source is very far away (at $t \to -\infty$ or $t \to \infty$), the line of sight is almost parallel to the track. The velocity component becomes almost equal to the full speed $v$, so the frequency approaches a constant maximum or minimum asymptote.

Q: Could it be Graph (c) or (d)?

Graphs (c) and (d) show the frequency returning to $\nu_o$ after a long time. But we know that as the source moves far away, it is receding at speed $v$, so the frequency must stay lower than $\nu_o$. Thus, it cannot return to $\nu_o$.

Question

A sound source S is moving along a straight track with speed $v$ and is emitting sound of frequency $
u_o$ (see figure). An observer is standing at a finite distance, at the point O, from the track. The time variation of frequency heard by the observer is best represented by (Here, $t_0$ represents the instant when the distance between the source and observer is minimum.)

Options:

(A) Graph (a)

(B) Graph (b)

(C) Graph (c)

(D) Graph (d)

Sigma Percentile · Accepted Answer

### Setting the Scene: The Train and the Observer

Imagine you are standing a little distance away from a straight railway track. A train is coming towards you, blowing its whistle. Let's draw this setup to see exactly how the sound reaches you.

We place the observer at a perpendicular distance $D$ from the track. The source $S$ moves with a constant velocity $v$ along the track. At a specific instant, $t = t_0$, the source is exactly at the closest point to the observer.

### The Doppler Effect: It's All About the Line of Sight

Now, according to the Doppler effect, the frequency you hear depends on the velocity of the source along the **line of sight**. It's not just the speed of the train, but how fast it's coming directly towards you.

The general formula for observed frequency is:

$$
u = 
u_o \left( \frac{v_{	ext{sound}}}{v_{	ext{sound}} - v_{	ext{source, line of sight}}} ight)$$

We need to find the component of the source's velocity along the line connecting the source to the observer.

### The Approach: High but Dropping

When the train is far away and approaching ($t < t_0$), the angle $	heta$ between the track and the line of sight is small. The component of velocity towards you is $v \cos	heta$. Because it's moving towards you, the frequency you hear is higher than the original frequency.

$$
u_{	ext{obs}} = 
u_o \left( \frac{v_{	ext{sound}}}{v_{	ext{sound}} - v \cos	heta} ight)$$

But here is the catch. As the train gets closer, the angle $	heta$ keeps increasing towards $90^\circ$. This means $\cos	heta$ decreases, and the velocity component towards you drops. Consequently, the denominator $(v_{	ext{sound}} - v \cos	heta)$ increases, and the high frequency you hear starts to continuously decrease.

### The Crossing: A Moment of Truth

Exactly at time $t_0$, the train is right in front of you. The angle is exactly $90^\circ$, making $\cos 90^\circ = 0$.

At this very instant, the train is neither coming towards you nor moving away. The velocity component towards the observer is zero. So, you hear the exact original frequency:

$$
u_{	ext{obs}} = 
u_o \left( \frac{v_{	ext{sound}}}{v_{	ext{sound}} - 0} ight) = 
u_o$$

### The Departure: Fading Away

After crossing you ($t > t_0$), the train starts moving away. Now, the velocity component is directed away from you, given by $v \cos	heta'$.

$$
u_{	ext{obs}} = 
u_o \left( \frac{v_{	ext{sound}}}{v_{	ext{sound}} + v \cos	heta'} ight)$$

The frequency drops below the original value. As it goes further, the angle $	heta'$ decreases, meaning $\cos	heta'$ increases. This causes the denominator to grow larger, making the frequency drop even more, approaching a constant lower value.

### The Grand Finale: Plotting the Curve

So, if we plot this, the frequency starts high, smoothly and continuously decreases, crosses the original frequency exactly at $t_0$, and then flattens out at a lower value.

This represents a smooth, monotonically decreasing curve. Looking at our options, Graph (b) perfectly matches this physical behavior.

Select Answer:

Visualized Solution

Visualizing the Setup

The Doppler Effect Formula

Approaching Source ( $t < t_{0}$ )

Variation During Approach

At the Closest Point ( $t = t_{0}$ )

Receding Source ( $t > t_{0}$ )

The Final Graph

The Sigma Insight: Doppler Effect

Setting the Scene

The Doppler Effect

The Approach

The Crossing

The Departure

The Grand Finale

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