A train moves towards a stationary observer with speed 34 m/s. The train sounds a whistle and its fr

Q: Why is there a minus sign in the denominator when the source moves towards the observer?

When the source moves towards you, it 'chases' its own sound waves, compressing the wavelength. A smaller wavelength means a higher frequency. Mathematically, a minus sign in the denominator makes the denominator smaller, which increases the overall value of the fraction, representing this higher frequency!

Q: What if the train was moving away from the observer?

If the train moves away, it stretches the sound waves behind it, increasing the wavelength and lowering the frequency. In the formula, the minus sign in the denominator would change to a plus sign: $f' = f \left(\frac{v}{v + v_s}\right)$.

Q: Does the actual frequency of the whistle change?

No! The actual frequency $f$ emitted by the whistle remains absolutely constant. The change in frequency is purely 'apparent'—it is only registered differently by the observer due to the relative motion between the source and the observer.

Q: Why did the speed of sound not change when the train slowed down?

The speed of sound depends solely on the properties of the medium (like temperature, pressure, and composition of air), not on the speed of the source emitting the sound.

Q: How did you know 323 and 306 are divisible by 17?

A great trick is to look at their difference! $323 - 306 = 17$. If two numbers are divisible by a common factor, their difference must also be divisible by that factor. Since their difference is exactly 17, we immediately test if 17 divides both, which it beautifully does!

Question

A train moves towards a stationary observer with speed $34	ext{ m/s}$. The train sounds a whistle and its frequency registered by the observer is $f_1$. If the train's speed is reduced to $17	ext{ m/s}$, the frequency registered is $f_2$. If the speed of sound is $340	ext{ m/s}$, then the ratio $\frac{f_1}{f_2}$ is

Options:

(A) $\frac{18}{19}$

(B) $\frac{1}{2}$

(C) $2$

(D) $\frac{19}{18}$

Sigma Percentile · Accepted Answer

### Introduction to the Doppler Effect

Have you ever stood by the side of a road as an ambulance sped past with its siren blaring?

You probably noticed a distinct shift in the pitch of the siren—high and piercing as it rushed toward you, then suddenly dropping to a lower, flatter tone the moment it passed.

This fascinating phenomenon is known as the **Doppler Effect**, and it is one of the most elegant demonstrations of wave mechanics in everyday life.

In this problem, we explore a classic scenario: a train blowing its whistle as it speeds toward a stationary observer.

By analyzing how the registered frequency changes when the train slows down, we will unlock the mathematical beauty behind this shift in pitch.

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### The Physics of Compressed Waves

To understand why the frequency shifts, let us visualize the sound waves propagating through the air.

Imagine the train is stationary and blowing its whistle.

It emits spherical wavefronts that spread out evenly in all directions, like ripples on a pond.

An observer standing nearby receives these wavefronts at regular intervals, registering the actual frequency of the whistle, $f$.

Now, let the train start moving toward the observer.

As the train emits a wavefront, it immediately begins chasing after it.

By the time it emits the next wavefront, it has moved closer to the first one.

This causes the wavefronts in front of the train to bunch up, or compress.

Because the distance between successive wavefronts (the wavelength) is shortened, the observer receives them at a faster rate.

This results in a higher registered frequency, $f'$.

Mathematically, this relationship is captured by the master formula:

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

Here, $v$ is the speed of sound in air, and $v_s$ is the speed of the source (the train) moving toward the observer.

The minus sign in the denominator is crucial—it makes the denominator smaller, which in turn makes the overall fraction larger, representing the increase in frequency.

---

### Analyzing Case 1: The High-Speed Approach

Let us apply our master formula to the first scenario described in the problem.

The train is moving toward the stationary observer at a speed of $v_{s1} = 34	ext{ m/s}$.

The speed of sound in air is given as $v = 340	ext{ m/s}$.

Substituting these values into our Doppler formula, we get the first registered frequency, $f_1$:

$$f_1 = f \left(\frac{340}{340 - 34}ight)$$

Subtracting $34$ from $340$ in the denominator yields $306$:

$$f_1 = f \left(\frac{340}{306}ight)$$

This is our first key equation.

Notice that we do not need to know the actual frequency $f$ of the whistle; it will elegantly cancel out later!

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### Analyzing Case 2: The Decelerated Approach

Now, the train slows down.

Its speed is reduced to $v_{s2} = 17	ext{ m/s}$, which is exactly half of its original speed.

The speed of sound in air remains unchanged at $v = 340	ext{ m/s}$.

Let us set up the equation for the new registered frequency, $f_2$:

$$f_2 = f \left(\frac{340}{340 - 17}ight)$$

Subtracting $17$ from $340$ in the denominator gives us $323$:

$$f_2 = f \left(\frac{340}{323}ight)$$

This is our second key equation.

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### The Elegant Cancellation and Ratio Calculation

We are asked to find the ratio of the first registered frequency to the second registered frequency, $\frac{f_1}{f_2}$.

Let us divide our expression for $f_1$ by our expression for $f_2$:

$$\frac{f_1}{f_2} = \frac{f \left(\frac{340}{306}ight)}{f \left(\frac{340}{323}ight)}$$

Look at how beautifully the terms cancel out!

The unknown actual frequency $f$ disappears from both the numerator and denominator.

Similarly, the speed of sound term, $340$, cancels out as well.

This leaves us with a simple reciprocal fraction:

$$\frac{f_1}{f_2} = \frac{323}{306}$$

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### Simplifying the Fraction

At first glance, the fraction $\frac{323}{306}$ might look intimidating to simplify.

But here is a clever mathematical trick: let us look at the difference between the numerator and the denominator:

$$323 - 306 = 17$$

If two numbers share a common factor, their difference must also be divisible by that factor.

Since the difference is exactly $17$, let us test if $17$ divides both numbers!

Dividing the numerator by $17$:

$$\frac{323}{17} = 19$$

Dividing the denominator by $17$:

$$\frac{306}{17} = 18$$

It works perfectly!

Thus, the simplified ratio is:

$$\frac{f_1}{f_2} = \frac{19}{18}$$

This matches **Option (d)** perfectly.

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### Conclusion & Key Takeaways

This problem beautifully illustrates how a complex physical phenomenon like the Doppler Effect can be distilled into a simple, elegant ratio of integers.

By understanding the physical mechanism of wave compression and applying the correct mathematical framework, we solved the problem with ease.

Always remember to check the relative direction of motion to ensure you use the correct sign in the Doppler formula, and keep an eye out for clever arithmetic shortcuts to save precious time during exams!

Select Answer:

Visualized Solution

Visualizing the Setup

The Doppler Effect Formula

Setting up Case 1

Calculating $f_{1}$

Setting up Case 2

Calculating $f_{2}$

Forming the Ratio $\frac{f _{1}}{f _{2}}$

Simplifying the Ratio

Reducing the Fraction

Matching the Option

The Way Forward

The Sigma Insight: Doppler Effect

Introduction to the Doppler Effect

The Physics of Compressed Waves

Analyzing Case 1

Analyzing Case 2

The Elegant Cancellation and Ratio Calculation

Simplifying the Fraction

Conclusion & Key Takeaways

Similar Questions

A train is moving on a straight track with speed $20 ms^{- 1}$ . It is blowing its whistle at the frequency of $1000 Hz$ . 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 = $320 ms^{- 1}$ )

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 $300 m/s$ .)

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 $= 300 m/s$ )

A whistle giving out $450 Hz$ approaches a stationary observer at a speed of $33 m/s$ . The frequency heard by the observer (in $Hz$ ) is (Speed of sound $= 330 m/s$ )