A train approaching a hill at a speed of 40 km/h sounds a whistle of frequency 580 Hz when it is at

Q: Why does the wind speed add to the speed of sound in the forward direction but subtract in the return direction?

Sound waves travel through the medium (air). If the air itself is moving (wind), the velocity of the sound relative to the ground is the vector sum of the velocity of sound in still air and the velocity of the wind. Thus, when traveling with the wind, the speeds add; when traveling against the wind, they subtract.

Q: Why is the hill considered a stationary observer and then a stationary source?

The hill does not move relative to the ground. When the sound first reaches the hill, the hill acts as an observer receiving the frequency $f'$. Since the hill simply reflects the sound waves without absorbing or modifying their frequency, it effectively acts as a stationary source emitting that same frequency $f'$ back towards the train.

Q: How do we choose the signs in the Doppler formula when wind is present?

First, calculate the effective speed of sound $v_{\text{eff}} = v \pm v_w$ along the direction of propagation. Once $v_{\text{eff}}$ is determined, use the standard Doppler sign conventions: subtract the source speed in the denominator if the source moves towards the observer, and add the observer speed in the numerator if the observer moves towards the source.

Q: Why is there a slight difference between $620\text{ Hz}$ and $621.43\text{ Hz}$ in the final answer?

The $620\text{ Hz}$ answer is obtained by strictly applying the Doppler effect step-by-step for each leg of the journey. The $621.43\text{ Hz}$ answer comes from an approximate combined formula used in some textbooks that simplifies the wind term in the denominator. Both are generally accepted, but the step-by-step method is physically more accurate.

Q: Can we solve this problem using relative velocity directly?

Yes, but you must be extremely careful because the wind velocity is relative to the ground, and the train's velocity is also relative to the ground. It is always safest to define a ground-based coordinate system and find the effective speed of sound relative to the ground first.

Question

A train approaching a hill at a speed of $40	ext{ km/h}$ sounds a whistle of frequency $580	ext{ Hz}$ when it is at a distance of $1	ext{ km}$ from a hill. A wind with a speed of $40	ext{ km/h}$ is blowing in the direction of motion of the train. Find
(a) the frequency of the whistle as heard by an observer on the hill,
(b) the distance from the hill at which the echo from the hill is heard by the driver and its frequency.
(Velocity of sound in air = $1200	ext{ km/h}$)

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine standing on a massive, immovable hill.

In the distance, a train is rushing towards you at a speed of $v_s = 40	ext{ km/h}$.

At the same time, a powerful wind is blowing in the exact same direction at $v_w = 40	ext{ km/h}$.

When the train is exactly $1	ext{ km}$ away, it sounds a whistle of frequency $f = 580	ext{ Hz}$.

Our goal is to understand how the wind and the motion of the train combine to alter the frequency of the sound waves as they travel to the hill, and then as they reflect back to the moving train.

Let's break this down into two distinct phases: the forward journey of the sound waves, and the return journey (the echo).

### Phase 1: The Forward Journey (Part a)

When the sound waves travel from the train to the hill, they are moving in the same direction as the wind.

Therefore, the wind assists the sound waves, and the **effective speed of sound** in this direction is:

$$v_{	ext{eff, forward}} = v + v_w = 1200 + 40 = 1240	ext{ km/h}$$

Now, let's apply the **Doppler Effect** formula.

The observer on the hill is stationary ($v_o = 0$), while the source (the train) is moving towards the observer with speed $v_s = 40	ext{ km/h}$.

The apparent frequency $f'$ heard by the observer on the hill is given by:

$$f' = f \left(\frac{v_{	ext{eff, forward}}}{v_{	ext{eff, forward}} - v_s}ight)$$

Substituting the values:

$$f' = 580 \left(\frac{1240}{1240 - 40}ight) = 580 \left(\frac{1240}{1200}ight) \approx 599.33	ext{ Hz}$$

This is the frequency of the sound waves that reach the hill.

Since the hill reflects the sound waves without changing their frequency, the hill now acts as a stationary source emitting sound of frequency $f' = 599.33	ext{ Hz}$.

### Phase 2: Finding the Distance of the Echo (Part b)

Let $x$ be the distance of the train from the hill when the driver hears the echo.

During the time $t$ it takes for the sound to travel to the hill and reflect back to the train, the train has traveled a distance of $1 - x$ at a speed of $v_s = 40	ext{ km/h}$.

Thus, the time taken by the train is:

$$t = \frac{1 - x}{40}$$

During this same time interval, the sound wave travels $1	ext{ km}$ to the hill with an effective speed of $1240	ext{ km/h}$, and then travels $x	ext{ km}$ back to the train.

On the return journey, the sound is traveling against the wind, so its effective speed is:

$$v_{	ext{eff, return}} = v - v_w = 1200 - 40 = 1160	ext{ km/h}$$

Therefore, the total time taken by the sound is:

$$t = \frac{1}{1240} + \frac{x}{1160}$$

Equating the two expressions for time:

$$\frac{1 - x}{40} = \frac{1}{1240} + \frac{x}{1160}$$

Multiplying the entire equation by $40$ to simplify:

$$1 - x = \frac{40}{1240} + \frac{40x}{1160} = \frac{1}{31} + \frac{x}{29}$$

Rearranging the terms to solve for $x$:

$$1 - \frac{1}{31} = x \left(1 + \frac{1}{29}ight)$$

$$\frac{30}{31} = x \left(\frac{30}{29}ight)$$

Canceling $30$ from both sides yields:

$$x = \frac{29}{31} \approx 0.935	ext{ km}$$

So, the driver hears the echo when the train is approximately **$0.935	ext{ km}$** away from the hill.

### Phase 3: Frequency of the Echo Heard by the Driver

Now, let's find the frequency of the echo heard by the driver.

The hill acts as a stationary source emitting $f' = 599.33	ext{ Hz}$.

The driver (observer) is moving towards the hill with speed $v_o = 40	ext{ km/h}$.

Since the sound waves are traveling from the hill to the train, they are moving against the wind, so the effective speed of sound is $v_{	ext{eff, return}} = 1160	ext{ km/h}$.

Using the Doppler formula for a moving observer and stationary source:

$$f'' = f' \left(\frac{v_{	ext{eff, return}} + v_o}{v_{	ext{eff, return}}}ight)$$

$$f'' = 599.33 \left(\frac{1160 + 40}{1160}ight) = 599.33 \left(\frac{1200}{1160}ight) = 620	ext{ Hz}$$

*Note on Textbook Variations:* Some textbooks use a combined approximation formula:

$$f'' = f \left[\frac{v - v_w + v_o}{v - v_w - v_s}ight]$$

Substituting the values into this formula gives:

$$f'' = 580 \left[\frac{1200 - 40 + 40}{1200 - 40 - 40}ight] = 580 \left[\frac{1200}{1120}ight] \approx 621.43	ext{ Hz}$$

Both methods are conceptually rich, but the step-by-step physical derivation yielding **$620	ext{ Hz}$** is mathematically precise and physically rigorous.

Visualized Solution

Visualizing the Setup

Effective Speed of Sound (Forward Journey)

Frequency Heard on the Hill (Part a)

Calculating $f^{'}$

Setting up the Echo Time Equation (Part b)

Solving for Distance $x$

Frequency of the Echo Heard by the Driver

Summary and Key Takeaways

The Sigma Insight: Doppler Effect

Analyzing the Setup

Phase 1

Phase 2

Phase 3

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