A siren placed at a railway platform is emitting sound of frequency 5 kHz. A passenger sitting in a

Q: Why did we use a plus sign in the numerator of the Doppler formula?

We use a plus sign because the observer (the train) is moving towards the stationary source. As the observer approaches the source, they encounter wave crests more frequently, which increases the observed frequency. Mathematically, a plus sign in the numerator increases the value of $f'$.

Q: Do we need to know the actual speed of sound in air to solve this problem?

No! That is the beauty of this problem. By expressing the speed of each train as a fraction of the speed of sound ($v_A/v$ and $v_B/v$), the speed of sound $v$ completely cancels out when we take the ratio $v_B/v_A$.

Q: What if the passenger was sitting in a train that was moving away from the platform?

If the train were moving away, the passenger would encounter wave crests less frequently, resulting in a lower observed frequency. The formula would then use a minus sign: $f' = f \left( \frac{v - v_o}{v} \right)$.

Q: Why does the frequency of the siren remain constant at 5 kHz?

The source frequency is a fundamental characteristic of the source itself (the siren's physical vibration mechanism) and does not change regardless of how the observer moves. Only the *apparent* frequency heard by the observer changes due to relative motion.

Q: Can we solve this using the wavelength of sound instead of frequency?

Yes, but it is much more tedious. Since the source is stationary, the wavelength of sound in the air remains unchanged at $\lambda = v/f$. The moving observer perceives a different relative speed of sound $v' = v + v_o$, leading to an apparent frequency $f' = v'/\lambda = (v + v_o)/(v/f) = f(v + v_o)/v$, which brings us back to the same formula.

Question

A siren placed at a railway platform is emitting sound of frequency $5	ext{ kHz}$. A passenger sitting in a moving train $A$ records a frequency of $5.5	ext{ kHz}$, while the train approaches the siren. During his return journey in a different train $B$ he records a frequency of $6.0	ext{ kHz}$ while approaching the same siren. The ratio of the velocity of train $B$ to that of train $A$ is

Options:

(A) $\frac{242}{252}$

(B) 2

(C) $\frac{5}{6}$

(D) $\frac{11}{6}$

Sigma Percentile · Accepted Answer

### The Magic of Relative Motion

Imagine standing on a railway platform. A train approaches, blowing its horn. As it rushes past you, the pitch of the horn suddenly drops. This everyday phenomenon is the **Doppler Effect**, one of the most beautiful and intuitive concepts in wave mechanics.

In this problem, we are exploring a variation of this classic setup. Instead of a moving source and a stationary observer, we have a **stationary source** (the siren on the platform) and a **moving observer** (the passenger in the train).

Let's dive deep into the physics of why this happens and how we can use simple mathematics to unlock the ratio of the speeds of two different trains.

### The Physics of a Moving Observer

Why does the frequency change when you move towards a sound source?

When a source is stationary, it emits sound waves that propagate outward in concentric spheres. The distance between successive wave crests—the wavelength $\lambda$—is constant in all directions and is given by:

$$\lambda = \frac{v}{f}$$

where $v$ is the speed of sound in air, and $f$ is the natural frequency of the source.

If you are also stationary, these wave crests pass by you at a rate of $f$ crests per second. But what happens if you start running towards the source with a speed $v_o$?

Because you are moving towards the incoming waves, the relative speed of the waves with respect to you increases to:

$$v_{	ext{relative}} = v + v_o$$

Since the wavelength $\lambda$ in the air remains unchanged, the rate at which you encounter these wave crests—which is the apparent frequency $f'$—increases:

$$f' = \frac{v_{	ext{relative}}}{\lambda} = \frac{v + v_o}{v/f} = f \left( \frac{v + v_o}{v} ight)$$

This is the master formula that governs our entire problem!

### Analyzing Train A

Let's apply our master formula to the first part of the passenger's journey on Train $A$.

The siren emits a frequency of $f = 5	ext{ kHz}$. The passenger in Train $A$, which is approaching the platform at speed $v_A$, hears a frequency of $f'_A = 5.5	ext{ kHz}$.

Substituting these values into our Doppler formula:

$$5.5 = 5 \left( \frac{v + v_A}{v} ight)$$

To solve for the ratio of the train's speed to the speed of sound, we divide both sides by $5$:

$$\frac{5.5}{5} = 1 + \frac{v_A}{v}$$

$$1.1 = 1 + \frac{v_A}{v}$$

Subtracting $1$ from both sides gives us a very clean result:

$$\frac{v_A}{v} = 0.1$$

This tells us that Train $A$ is moving at exactly $10\%$ of the speed of sound!

### Analyzing Train B

Now, let's look at the return journey in Train $B$.

The source frequency is still $f = 5	ext{ kHz}$, but Train $B$ is moving at a different speed $v_B$. The passenger now hears an even higher frequency of $f'_B = 6.0	ext{ kHz}$.

Using the same Doppler formula for Train $B$:

$$6.0 = 5 \left( \frac{v + v_B}{v} ight)$$

Dividing both sides by $5$:

$$\frac{6.0}{5} = 1 + \frac{v_B}{v}$$

$$1.2 = 1 + \frac{v_B}{v}$$

Subtracting $1$ from both sides:

$$\frac{v_B}{v} = 0.2$$

This means Train $B$ is moving at $20\%$ of the speed of sound!

### The Elegant Finale

We are asked to find the ratio of the velocity of Train $B$ to that of Train $A$:

$$	ext{Ratio} = \frac{v_B}{v_A}$$

We don't need to know the actual speed of sound $v$ because we can simply divide our two results:

$$\frac{v_B/v}{v_A/v} = \frac{0.2}{0.1}$$

$$\frac{v_B}{v_A} = 2$$

The speed of sound cancels out perfectly, leaving us with a simple, elegant ratio of **2**.

This means Train $B$ is traveling exactly twice as fast as Train $A$! This corresponds to option **(b)**.

Select Answer:

Visualized Solution

Understanding the Physical Setup

The Doppler Effect Formula

Setting up the Equation for Train $A$

Simplifying the Equation for Train $A$

Setting up the Equation for Train $B$

Simplifying the Equation for Train $B$

Calculating the Ratio of Velocities

Exploring Further Variations

The Sigma Insight: Doppler Effect

The Magic of Relative Motion

The Physics of a Moving Observer

Analyzing Train A

Analyzing Train B

The Elegant Finale

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