A bus is moving towards a huge wall with a velocity of 5 m/s. The driver sounds a horn of frequency

Q: Why does the passenger hear beats if they are moving with the source (the horn)?

The passenger is indeed stationary relative to the horn, so the direct sound they hear is unshifted at $200\text{ Hz}$. However, the reflected sound from the wall undergoes a double Doppler shift because the bus is moving relative to the wall. This creates a frequency difference between the direct and reflected sound, resulting in beats.

Q: Does the speed of sound in air affect the beat frequency?

Yes, the speed of sound $v$ is present in the final formula $f' = f_0 \left(\frac{v + v_b}{v - v_b}\right)$. A higher speed of sound would decrease the frequency shift, thereby reducing the beat frequency, while a lower speed of sound would increase it.

Q: What if there is wind blowing towards the wall? How does that change the calculation?

If wind is blowing with velocity $w$ towards the wall, the effective speed of sound becomes $v + w$ for the forward journey and $v - w$ for the return journey. This modifies the individual Doppler equations, and the cancellation of $v$ in the combined formula will no longer occur cleanly.

Q: Why did we round the final answer to 6 instead of keeping it as 5.93?

The question is from an older JEE paper where integer or numerical answers are expected to be rounded to the nearest whole number for standard grading, and $5.93$ rounds to $6$.

Q: Can we use the approximation formula $f_{\text{beat}} \approx \frac{2 f_0 v_b}{v}$ for all cases?

Only when the speed of the source/observer is much smaller than the speed of sound ($v_b \ll v$). If the bus were moving at a high speed (e.g., $50\text{ ms}^{-1}$), the binomial approximation would introduce significant error, and you must use the exact formula.

Question

A bus is moving towards a huge wall with a velocity of $5	ext{ ms}^{-1}$. The driver sounds a horn of frequency $200	ext{ Hz}$. The frequency of the beats heard by a passenger of the bus will be ...... Hz. (Speed of sound in air = $342	ext{ ms}^{-1}$)

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are sitting inside a bus that is cruising down a straight road directly toward a massive, solid concrete wall.

The bus is moving at a steady speed of $v_b = 5	ext{ ms}^{-1}$.

Suddenly, the driver presses the horn, emitting a continuous, sharp sound at a frequency of $f_0 = 200	ext{ Hz}$.

This sound wave travels through the air at a speed of $v = 342	ext{ ms}^{-1}$, hits the wall, and bounces right back toward the bus.

As a passenger, you hear two distinct sounds simultaneously: the direct sound from the horn right in front of you, and the reflected echo returning from the wall.

Because of the relative motion between the bus, the air, and the wall, these two sounds will not have the same frequency.

This frequency mismatch leads to a fascinating acoustic phenomenon known as **beats**.

Let us dive deep into the physics of how these frequencies shift and how to calculate the resulting beat frequency.

### The Two-Stage Doppler Effect

To find the frequency of the reflected sound heard by the passenger, we must break the physical process down into two distinct stages.

This is because the sound wave undergoes two separate frequency shifts due to the Doppler effect before it reaches the passenger's ears.

#### Stage 1: From the Bus to the Wall

In the first stage, the horn on the moving bus acts as a **moving source** of sound, propagating waves forward.

The stationary wall acts as a **stationary observer** receiving these waves.

Since the source is moving toward the observer, the sound waves in front of the bus are compressed.

This compression decreases the wavelength, which in turn increases the frequency perceived by the wall.

Using the standard Doppler effect formula for a moving source and a stationary observer, the frequency received by the wall ($f_{	ext{wall}}$) is:

$$f_{	ext{wall}} = f_0 \left(\frac{v}{v - v_b}ight)$$

Substituting the given values:
- $f_0 = 200	ext{ Hz}$
- $v = 342	ext{ ms}^{-1}$
- $v_b = 5	ext{ ms}^{-1}$

$$f_{	ext{wall}} = 200 \left(\frac{342}{342 - 5}ight) = 200 \left(\frac{342}{337}ight) \approx 202.97	ext{ Hz}$$

Thus, the wall "hears" a slightly higher frequency of approximately $202.97	ext{ Hz}$.

#### Stage 2: From the Wall back to the Passenger

In the second stage, the wall reflects the sound waves.

Since the wall is stationary, it acts as a **stationary source** emitting sound at the frequency it received, which is $f_{	ext{wall}}$.

Meanwhile, the passenger inside the bus is moving toward this stationary source at the speed of the bus, $v_b = 5	ext{ ms}^{-1}$.

Therefore, the passenger acts as a **moving observer** traveling toward a stationary source.

As the passenger moves toward the incoming wavefronts, they intercept more wave crests per second, causing the perceived frequency ($f'$) to increase even further.

The formula for a moving observer and a stationary source is:

$$f' = f_{	ext{wall}} \left(\frac{v + v_b}{v}ight)$$

### The Elegant Mathematical Simplification

Before we plug in the numbers, let us combine the equations from Stage 1 and Stage 2 to see if we can find a more direct relationship.

Substitute the expression for $f_{	ext{wall}}$ into the equation for $f'$:

$$f' = \left[ f_0 \left(\frac{v}{v - v_b}ight) ight] \left(\frac{v + v_b}{v}ight)$$

Notice how beautifully the speed of sound $v$ in the numerator of the first term cancels out with the $v$ in the denominator of the second term!

This leaves us with the master equation for a double Doppler shift with a stationary reflector:

$$f' = f_0 \left(\frac{v + v_b}{v - v_b}ight)$$

This formula is incredibly powerful because it bypasses the intermediate calculation of the frequency at the wall entirely.

Now, let us substitute our values into this simplified formula:

$$f' = 200 \left(\frac{342 + 5}{342 - 5}ight) = 200 \left(\frac{347}{337}ight) \approx 205.93	ext{ Hz}$$

### Calculating the Beat Frequency

Now we have the two frequencies heard by the passenger:
1. The direct sound from the horn: $f_0 = 200	ext{ Hz}$
2. The reflected sound from the wall: $f' \approx 205.93	ext{ Hz}$

When two sound waves of slightly different frequencies overlap, they interfere constructively and destructively over time.

This periodic variation in intensity is heard as beats.

The frequency of these beats ($f_{	ext{beat}}$) is simply the absolute difference between the two frequencies:

$$f_{	ext{beat}} = f' - f_0$$

$$f_{	ext{beat}} = 205.93 - 200 = 5.93	ext{ Hz}$$

Since the question asks for the frequency of the beats as an integer, we round $5.93	ext{ Hz}$ to the nearest whole number:

$$\mathbf{f_{	ext{beat}} \approx 6	ext{ Hz}}$$

### Alternative Approximation Method (The Physicist's Shortcut)

In many competitive exams like JEE, saving time is crucial.

Since the speed of the bus ($5	ext{ ms}^{-1}$) is extremely small compared to the speed of sound ($342	ext{ ms}^{-1}$), we can use a binomial approximation to find a quick estimate.

Let us rewrite our master equation:

$$f' = f_0 \left(1 + \frac{v_b}{v}ight) \left(1 - \frac{v_b}{v}ight)^{-1}$$

Using the binomial expansion $(1 - x)^{-1} \approx 1 + x$ for $x \ll 1$:

$$f' \approx f_0 \left(1 + \frac{v_b}{v}ight) \left(1 + \frac{v_b}{v}ight) \approx f_0 \left(1 + \frac{2v_b}{v}ight)$$

Now, the beat frequency is:

$$f_{	ext{beat}} = f' - f_0 \approx f_0 \left(\frac{2v_b}{v}ight)$$

Let us plug in the numbers to see how close this approximation is:

$$f_{	ext{beat}} \approx 200 \left(\frac{2 	imes 5}{342}ight) = \frac{2000}{342} \approx 5.85	ext{ Hz}$$

Rounding $5.85	ext{ Hz}$ to the nearest integer still gives us exactly **$6	ext{ Hz}$**!

This approximation is a fantastic tool to quickly verify your answer or solve the problem in under a minute.

Animated Solution for Physics - Waves: A bus is moving towards a huge wall with a velocity of $5 ms^{- 1}$ . The driver sounds a horn of frequency $200 Hz$ . The frequency of the beats heard by a passenger of the bus will be ...... Hz. (Speed of sound in air = $342 ms^{- 1}$ )

Enter Numerical Value:

Visualized Solution

Visualizing the Setup

The Two-Stage Doppler Effect

Stage 1: Frequency Received by the Wall

Stage 1: Substitution

Stage 1: Calculation

Stage 2: Frequency Heard by the Passenger

Combining the Equations

Calculating the Reflected Frequency

Understanding Beats

Final Calculation

Alternative Approximation Method

The Way Forward

The Sigma Insight: Doppler Effect

Analyzing the Setup

The Two-Stage Doppler Effect

# Stage 1

# Stage 2

The Elegant Mathematical Simplification

Calculating the Beat Frequency

Alternative Approximation Method (The Physicist's Shortcut)

Similar Questions

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Enter Numerical Value:

Visualized Solution

Visualizing the Setup

The Two-Stage Doppler Effect

Stage 1: Frequency Received by the Wall

Stage 1: Substitution

Stage 1: Calculation

Stage 2: Frequency Heard by the Passenger

Combining the Equations

Calculating the Reflected Frequency

Understanding Beats

Final Calculation

Alternative Approximation Method

The Way Forward

The Sigma Insight: Doppler Effect

Analyzing the Setup

The Two-Stage Doppler Effect

# Stage 1

# Stage 2

The Elegant Mathematical Simplification

Calculating the Beat Frequency

Alternative Approximation Method (The Physicist's Shortcut)

Similar Questions

The frequency of a car horn encountered a change from 400 Hz to 500 Hz, when the car approaches a vertical wall. If the speed of sound is 330 m/s, then the speed of car is …… km/h.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 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)

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 $= 330 m/s$ )