A train is moving at a constant speed of v = 90 km/h on a straight level track. From a railway stati

Q: Why isn't the heard duration just the actual duration $\tau$?

Because the source is moving! The sound emitted at the end of the honk has a shorter distance to travel than the sound emitted at the beginning, so it arrives 'sooner' relative to the first sound wave, compressing the duration.

Q: Is this the same as the Doppler effect for frequency?

Yes, it's fundamentally the same phenomenon. Just as the waves are compressed in space (increasing frequency), the time interval between the first and last wave is also compressed (decreasing duration).

Q: Why do we use $x - v\tau$ for the second distance?

The train starts honking at distance $x$. The honk lasts for time $\tau$, during which the train travels a distance $v\tau$ towards the station. So, its new distance is $x - v\tau$.

Q: Does the speed of sound $c$ change because the train is moving?

No, the speed of sound in a medium depends only on the properties of the medium (like temperature and wind). The velocity of the source does not affect the speed of the sound waves once they are emitted.

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

If the train were moving away, the second distance would be $x + v\tau$. The sound from the end of the honk would have to travel further, arriving later, and the heard duration would be longer than $\tau$.

Question

A train is moving at a constant speed of $v = 90 	ext{ km/h}$ on a straight level track. From a railway station P on the track, there is a village Q at a distance $y = 1.2 	ext{ km}$ in a direction perpendicular to the track. When the engine E is $x = 1.6 	ext{ km}$ away from the station, the driver honks a beep of horn of duration $	au = 44 	ext{ s}$. Calculate durations of the honking $	au_P$ and $	au_Q$ heard at the station and in the village. Speed of sound in still air is $c = 350 	ext{ m/s}$ and there is no wind.

Sigma Percentile · Accepted Answer

Imagine you are standing on a railway platform, and a train is approaching you from a distance. The driver blows the horn for a certain duration. Will you hear the horn for that exact same duration? The answer is no, and the reason lies in the fascinating physics of the Doppler effect.

### Analyzing the Setup

Let's break down the physical reality of the situation. We have a train moving at a constant speed $v = 90 	ext{ km/h}$, which is $25 	ext{ m/s}$. The speed of sound in still air is $c = 350 	ext{ m/s}$. The driver honks the horn for an actual duration of $	au = 44 	ext{ s}$.

The train starts honking when it is at a distance $x = 1.6 	ext{ km}$ ($1600 	ext{ m}$) from station P. But the train isn't stationary! During the $44 	ext{ s}$ that the horn is blowing, the train moves forward by a distance $v	au = 25 	imes 44 = 1100 	ext{ m}$.

We also have a village Q located at a perpendicular distance $y = 1.2 	ext{ km}$ ($1200 	ext{ m}$) from the station. We need to find the duration of the honk heard at both locations.

### The Master Equation for Heard Duration

Why does the heard duration differ from the actual duration? It's because the sound emitted at the beginning of the honk and the sound emitted at the end of the honk travel different distances to reach the listener.

The heard duration $\Delta t$ is simply the difference between the arrival time of the last sound pulse ($t_2$) and the arrival time of the first sound pulse ($t_1$):
$$ \Delta t = t_2 - t_1 $$

### Duration at Station P

Let's apply this to station P. The first sound pulse is emitted when the train is at distance $x$. It takes time $t_1 = \frac{x}{c}$ to reach P.

The last sound pulse is emitted $	au$ seconds later. By this time, the train has moved closer, and its new distance is $x - v	au$. This last pulse reaches P at time $t_2 = 	au + \frac{x - v	au}{c}$.

Subtracting $t_1$ from $t_2$, the $\frac{x}{c}$ terms cancel out beautifully:
$$ 	au_P = \left(	au + \frac{x - v	au}{c}ight) - \frac{x}{c} = 	au - \frac{v	au}{c} = 	au \left(1 - \frac{v}{c}ight) $$

Substituting the values:
$$ 	au_P = 44 \left(1 - \frac{25}{350}ight) = 44 \left(1 - \frac{1}{14}ight) = 44 	imes \frac{13}{14} \approx 40.86 	ext{ s} $$
The duration is compressed because the train is moving towards the station.

### Duration at Village Q

Now, let's shift our focus to village Q. The sound doesn't travel along the track; it travels along the hypotenuse of a right triangle.

The initial distance $d_1$ from the train to Q is found using the Pythagorean theorem:
$$ d_1 = \sqrt{x^2 + y^2} = \sqrt{1600^2 + 1200^2} = 2000 	ext{ m} $$
The first pulse reaches Q at $t_1 = \frac{d_1}{c}$.

The final distance $d_2$ from the train to Q is:
$$ d_2 = \sqrt{(x - v	au)^2 + y^2} = \sqrt{500^2 + 1200^2} = 1300 	ext{ m} $$
The last pulse reaches Q at $t_2 = 	au + \frac{d_2}{c}$.

### Final Calculation

The duration heard at Q is:
$$ 	au_Q = t_2 - t_1 = 	au + \frac{d_2 - d_1}{c} $$

Substituting our calculated distances:
$$ 	au_Q = 44 + \frac{1300 - 2000}{350} = 44 - \frac{700}{350} = 44 - 2 = 42 	ext{ s} $$

**The final durations are $40.86 	ext{ s}$ at station P and $42 	ext{ s}$ at village Q.** The duration at Q is also compressed, but less so than at P because the relative velocity of approach along the line of sight is smaller.

Visualized Solution

\text{Visualizing the Setup}

\text{The Honking Event}

\text{Concept of Heard Duration}

\text{Arrival Times at Station P}

\text{Duration at Station P}

\text{Calculating } \tau_P

\text{Final Answer for P}

\text{Arrival Times at Village Q}

\text{End of Honk for Q}

\text{Duration at Village Q}

\text{Calculating Distances}

\text{Final Answer for Q}

The Sigma Insight: Doppler Effect

Analyzing the Setup

The Master Equation for Heard Duration

Duration at Station P

Duration at Village Q

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 $300 m/s$ .)

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 = $320 ms^{- 1}$ )

Two sources of sound $S_{1}$ and $S_{2}$ produce sound waves of same frequency $660 Hz$ . A listener is moving from source $S_{1}$ towards $S_{2}$ with a constant speed $u m/s$ and he hears $10 beats/s$ . The velocity of sound is $330 m/s$ . Then, $u$ equal to