A train is moving on a straight track with speed 20 ms^-1. It is blowing its whistle at the frequenc

Q: Why does the frequency change when the train passes?

As the train moves towards you, it "chases" its own sound waves, compressing them and increasing the frequency (pitch). When it moves away, the waves are stretched out, lowering the frequency.

Q: Why do we use $c - v_s$ for approaching and $c + v_s$ for receding?

The denominator represents the effective speed of the wave relative to the source. When approaching, the source reduces the distance between wave crests, effectively subtracting its speed from the wave speed ($c - v_s$). When receding, it adds to the distance ($c + v_s$).

Q: Can we just find the average of the two frequencies?

No! The Doppler effect is not a linear relationship. The increase in frequency when approaching is mathematically greater than the decrease in frequency when receding for the same speed.

Q: Why is the percentage change calculated relative to $f_0$ and not $f_1$?

In standard physics problems, "percentage change in frequency" due to a passing object is normalized against the true, unshifted frequency of the source ($f_0$) to provide a consistent baseline, unless specified otherwise.

Q: What if the observer was also moving?

You would use the general Doppler formula: $f' = f_0 \left( \frac{c \pm v_o}{c \mp v_s} \right)$. The top signs apply when the observer and source are moving towards each other, and the bottom signs when moving apart.

Question

A train is moving on a straight track with speed $20	ext{ ms}^{-1}$. It is blowing its whistle at the frequency of $1000	ext{ Hz}$. The percentage change in the frequency heard by a person standing near the track as the train passes him is close to (speed of sound = $320	ext{ ms}^{-1}$)

Options:

(A) 6\%

(B) 12\%

(C) 18\%

(D) 24\%

Sigma Percentile · Accepted Answer

### The Symphony of the Passing Train

Imagine you are standing right next to a railway track. A train is approaching you at a speed of $20 	ext{ m/s}$, blowing its whistle at a frequency of $1000 	ext{ Hz}$.

This is a classic real-world manifestation of the **Doppler Effect**. As the train moves, it alters the way sound waves reach your ears.

### The Approach: Compressed Waves

When the source of the sound is moving towards you, the sound waves get compressed. This means the crests of the waves are closer together, leading to a higher pitch.

According to the Doppler effect, the apparent frequency $f_1$ you hear is given by the formula:

$$f_1 = f_0 \left( \frac{c}{c - v_s} ight)$$

Here, $c$ is the speed of sound ($320 	ext{ m/s}$), $v_s$ is the speed of the source ($20 	ext{ m/s}$), and $f_0$ is the original frequency ($1000 	ext{ Hz}$).

Substituting our values, we get:

$$f_1 = 1000 \left( \frac{320}{320 - 20} ight)$$

We will leave this in its raw form for now to make our final calculation more elegant.

### The Departure: Stretched Waves

Now, the train zooms past you and starts moving away. Notice how the sound suddenly drops in pitch?

That is because the sound waves are now stretching out behind the train. For a receding source, the denominator in our Doppler formula changes.

$$f_2 = f_0 \left( \frac{c}{c + v_s} ight)$$

This positive sign in the denominator ensures the apparent frequency $f_2$ is lower than the original frequency.

Plugging in our values again:

$$f_2 = 1000 \left( \frac{320}{320 + 20} ight)$$

### The Grand Difference

The question asks for the percentage change in the frequency heard as the train passes. This requires us to find the difference between $f_1$ and $f_2$, and then calculate its percentage relative to the original frequency $f_0$.

Let's calculate the difference, $\Delta f$.

$$\Delta f = f_1 - f_2 = f_0 c \left( \frac{1}{c - v_s} - \frac{1}{c + v_s} ight)$$

Taking the common denominator, we get a neat algebraic expression:

$$\Delta f = f_0 \left( \frac{2 c v_s}{c^2 - v_s^2} ight)$$

To find the percentage change, we divide this difference by $f_0$ and multiply by $100$.

$$	ext{Percentage Change} = \frac{\Delta f}{f_0} 	imes 100 = \frac{2 c v_s}{c^2 - v_s^2} 	imes 100$$

Notice how beautifully $f_0$ cancels out! **The percentage change is completely independent of the original frequency of the whistle.**

### Final Calculation

Now for the final calculation phase. Let's substitute our numerical values into the simplified expression.

$$	ext{Percentage Change} = \frac{2 	imes 320 	imes 20}{320^2 - 20^2} 	imes 100$$

In the denominator, instead of squaring large numbers, we can use the difference of squares identity, $(a-b)(a+b)$, to make our lives easier.

$$= \frac{12800}{(320 - 20)(320 + 20)} 	imes 100$$

$$= \frac{12800}{300 	imes 340} 	imes 100$$

Cancel the zeros carefully. We are left with:

$$= \frac{12800}{102000} 	imes 100 = \frac{1280}{102} \approx 12.54\%$$

Looking at our options, $12.54\%$ is closest to **$12\%$**. Therefore, option (b) is our correct answer.

This is a beautiful and highly scoring application of the Doppler effect!

Select Answer:

Visualized Solution

Visualizing the Setup

Doppler Effect: Approaching Source

Substituting Values for $f_{1}$

The Train Passes By

Doppler Effect: Receding Source

Substituting Values for $f_{2}$

Formula for Percentage Change

Calculating the Difference $Δ f$

Setting up the Percentage Equation

Simplifying the Expression

Final Arithmetic

Conclusion

The Way Forward

The Sigma Insight: Doppler Effect

The Symphony of the Passing Train

The Approach

The Departure

The Grand Difference

Final Calculation

Similar Questions

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 $300 m/s$ .)

An observer moves towards a stationary source of sound, with a velocity one-fifth of the velocity of sound. What is the percentage increase in the apparent frequency?

A stationary source emits sound waves of frequency $500 Hz$ . Two observers moving along a line passing through the source detect sound to be of frequencies $480 Hz$ and $530 Hz$ . Their respective speeds are in $ms^{- 1}$ , (Take, speed of sound $= 300 m/s$ )