An observer standing on a railway crossing receives frequency of 2.2 kHz and 1.8 kHz when the train

Q: Why does the frequency increase when the train is approaching? Doesn't the speed of sound stay constant?

Yes, the speed of sound relative to the air remains completely constant. However, because the train is moving forward while emitting sound, each successive wave crest is emitted from a position closer to the observer than the previous one. This bunches the wavefronts closer together in space, reducing the wavelength and increasing the frequency heard by the observer.

Q: Why did we use a minus sign in the denominator for the approaching case?

In the Doppler formula, when the source moves toward a stationary observer, the apparent frequency must increase ($f' > f$). To make the fraction $\frac{v}{v \pm v_T}$ greater than $1$, we must make the denominator smaller than the numerator. Therefore, we subtract $v_T$ from $v$ in the denominator.

Q: Can we solve this problem if the speed of sound in air is not given?

No. If you look at the simplified ratio $\frac{f'}{f''} = \frac{v + v_T}{v - v_T}$, the value of $v_T$ directly depends on the speed of sound $v$. Without knowing $v$, we would have one equation with two unknowns ($v$ and $v_T$), making it impossible to find a unique numerical value for $v_T$.

Q: What is Componendo and Dividendo, and when can I use this shortcut?

Componendo and Dividendo is an algebraic rule stating that if $\frac{a}{b} = \frac{c}{d}$, then $\frac{a-b}{a+b} = \frac{c-d}{c+d}$. You can use this shortcut in any Doppler problem where a source passes a stationary observer symmetrically, as it directly relates the ratio of the observed frequencies to the ratio of the velocities: $\frac{f' - f''}{f' + f''} = \frac{v_T}{v}$.

Q: Does the wind speed affect the apparent frequencies in this problem?

Yes, if there is wind blowing, it changes the effective speed of sound relative to the ground. If wind blows with velocity $w$ in the direction of the sound propagation, the speed of sound $v$ is replaced by $(v + w)$ in the formulas. Since no wind is mentioned here, we assume the air is still.

Question

An observer standing on a railway crossing receives frequency of $2.2	ext{ kHz}$ and $1.8	ext{ kHz}$ when the train approaches and recedes from the observer. Find the velocity of the train. (The speed of the sound in air is $300	ext{ m/s}$.)

Sigma Percentile · Accepted Answer

### The Magic of the Doppler Effect

Have you ever stood by a highway or a railway track and noticed how the pitch of a passing vehicle's horn suddenly drops as it zooms past you?

That thrilling shift in pitch is not an illusion—it is a fundamental physical phenomenon known as the **Doppler Effect**.

When a source of sound moves relative to a medium, the wavefronts in front of it are bunched together, while those behind it are stretched out.

To a stationary observer, this compression of wavefronts translates to a higher frequency (higher pitch) as the source approaches, and a lower frequency (lower pitch) as it recedes.

In this problem, we are standing right at a railway crossing, acting as the stationary observer.

We hear the train's whistle at a frequency of $2.2	ext{ kHz}$ as it approaches us, and then at $1.8	ext{ kHz}$ as it moves away.

Using the speed of sound in air, which is given as $300	ext{ m/s}$, our mission is to determine the exact velocity of this train.

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### Setting Up the Mathematical Model

To translate this physical story into mathematics, we use the general Doppler Effect formula:

$$f_{	ext{app}} = f \left( \frac{v \pm v_O}{v \mp v_S} ight)$$

where:
- $f_{	ext{app}}$ is the apparent frequency heard by the observer.
- $f$ is the actual frequency of the whistle emitted by the train.
- $v$ is the speed of sound in air ($300	ext{ m/s}$).
- $v_O$ is the velocity of the observer.
- $v_S$ is the velocity of the source (the train, $v_T$).

Since we are standing still on the crossing, our velocity $v_O = 0$. This simplifies our general formula significantly!

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### Case I: The Approach

As the train rushes toward us, the sound waves are compressed.

This means the apparent frequency $f'$ must be higher than the actual frequency $f$.

To make the fraction larger, we must decrease the denominator. Thus, we use the minus sign in the denominator:

$$f' = f \left( \frac{v}{v - v_T} ight)$$

Substituting our known values ($f' = 2.2	ext{ kHz}$ and $v = 300	ext{ m/s}$):

$$2.2 = f \left( \frac{300}{300 - v_T} ight) \quad 	ext{--- (Equation 1)}$$

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### Case II: The Recession

Once the train crosses us and begins to recede, the sound waves are stretched out behind it.

Consequently, the apparent frequency $f''$ drops below the actual frequency $f$.

To make the fraction smaller, we must increase the denominator. Thus, we use the plus sign in the denominator:

$$f'' = f \left( \frac{v}{v + v_T} ight)$$

Substituting our known values ($f'' = 1.8	ext{ kHz}$ and $v = 300	ext{ m/s}$):

$$1.8 = f \left( \frac{300}{300 + v_T} ight) \quad 	ext{--- (Equation 2)}$$

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### The Elegant Elimination

We now have a system of two equations with two variables: the actual frequency $f$ and the train's velocity $v_T$.

Since the problem does not ask for the actual frequency $f$, we can eliminate it completely by dividing Equation 1 by Equation 2:

$$\frac{2.2}{1.8} = \frac{f \left( \frac{300}{300 - v_T} ight)}{f \left( \frac{300}{300 + v_T} ight)}$$

Notice how beautifully the unknown frequency $f$ and the speed of sound factor of $300$ in the numerator cancel out!

This leaves us with a clean, simple algebraic ratio:

$$\frac{22}{18} = \frac{300 + v_T}{300 - v_T}$$

Reducing the fraction on the left-hand side by dividing both the numerator and denominator by $2$:

$$\frac{11}{9} = \frac{300 + v_T}{300 - v_T}$$

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### Solving the Algebra

Now, let's cross-multiply to solve for $v_T$:

$$11(300 - v_T) = 9(300 + v_T)$$

Expanding both sides:

$$3300 - 11v_T = 2700 + 9v_T$$

Let's group the constant terms on the left and the $v_T$ terms on the right:

$$3300 - 2700 = 9v_T + 11v_T$$

$$600 = 20v_T$$

Dividing both sides by $20$ gives us our final result:

$$v_T = \frac{600}{20} = 30	ext{ m/s}$$

To put this into perspective, $30	ext{ m/s}$ is equivalent to $108	ext{ km/h}$—a perfectly realistic speed for an express train!

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### The JEE Shortcut: Componendo and Dividendo

In highly competitive exams like JEE, speed is just as important as accuracy.

Whenever you have a symmetric Doppler scenario where a source passes a stationary observer, you can use a brilliant algebraic shortcut.

Let's write the ratio of the two apparent frequencies:

$$\frac{f'}{f''} = \frac{v + v_T}{v - v_T}$$

Applying the rule of **Componendo and Dividendo** (i.e., if $\frac{a}{b} = \frac{c}{d}$, then $\frac{a - b}{a + b} = \frac{c - d}{c + d}$):

$$\frac{f' - f''}{f' + f''} = \frac{(v + v_T) - (v - v_T)}{(v + v_T) + (v - v_T)} = \frac{2v_T}{2v} = \frac{v_T}{v}$$

This gives us a direct, elegant formula for the source velocity:

$$v_T = v \left( \frac{f' - f''}{f' + f''} ight)$$

Let's verify our answer using this shortcut:

$$v_T = 300 \left( \frac{2.2 - 1.8}{2.2 + 1.8} ight) = 300 \left( \frac{0.4}{4.0} ight) = 300 	imes 0.1 = 30	ext{ m/s}$$

Both methods yield the exact same result! This shortcut is a powerful tool to keep in your exam toolkit.

Animated Solution for Physics - Waves: An observer standing on a railway crossing receives frequency of $2.2 kHz$ and $1.8 kHz$ when the train approaches and recedes from the observer. Find the velocity of the train. (The speed of the sound in air is $300 m/s$ .)

Enter Numerical Value:

Visualized Solution

Visualizing the Doppler Effect

The Doppler Effect Formula

Apparent Frequency during Approach

Substituting Values for Case I

Apparent Frequency during Recession

Substituting Values for Case II

Eliminating the Unknown Frequency $f$

Simplifying the Ratio

Cross-Multiplying to Solve for $v_{T}$

Grouping Like Terms

Calculating the Velocity of the Train

Shortcut Formula for Symmetric Doppler

Verifying with the Shortcut Formula

The Sigma Insight: Doppler Effect

The Magic of the Doppler Effect

Setting Up the Mathematical Model

Case I

Case II

The Elegant Elimination

Solving the Algebra

The JEE Shortcut

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