A source of sound of frequency 256 Hz is moving rapidly towards a wall with a velocity of 5 m/s. How

Q: Why doesn't the direct sound undergo a Doppler shift?

The Doppler shift only occurs when there is relative motion between the source and the observer. Since the observer is on the source itself, their relative velocity is zero, meaning the direct sound is heard at its original frequency of $256\text{ Hz}$.

Q: Can we use the formula $f' = f\left(\frac{v+u}{v-u}\right)$ directly for any reflection problem?

Yes, when a source-observer moves directly towards a stationary reflector with speed $u$, the combined formula simplifies to $f' = f\left(\frac{v+u}{v-u}\right)$. This is highly useful for saving time in JEE!

Q: What happens to the beat frequency if the speed of the vehicle increases?

As the speed of the vehicle ($v_s = v_o$) increases, the term $\frac{v+u}{v-u}$ increases, which increases the reflected frequency $f'$. Consequently, the difference $f' - f$ increases, leading to a higher beat frequency.

Q: Does the wind speed affect the beat frequency in this scenario?

Yes, if wind is blowing, the effective speed of sound relative to the wall and the observer changes, which alters the Doppler shift equations and slightly changes the beat frequency.

Q: Why does the wall act as a stationary observer first and then a stationary source?

The wall receives the sound waves at a frequency determined by the moving source. It then reflects these waves back into the medium. Since the wall itself is not moving, it behaves exactly like a stationary source emitting the frequency it received.

Question

A source of sound of frequency $256	ext{ Hz}$ is moving rapidly towards a wall with a velocity of $5	ext{ m/s}$. How many beats per second will be heard by the observer on source itself if sound travels at a speed of $330	ext{ m/s}$?

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are riding on a vehicle that is equipped with a loud siren emitting a constant pitch of $256	ext{ Hz}$.

As you speed towards a massive, rigid concrete wall at $5	ext{ m/s}$, you notice something fascinating.

You don't just hear the steady hum of your own siren; you also hear a slightly higher-pitched sound reflecting back from the wall.

These two sound waves—the **direct wave** and the **reflected wave**—superimpose in space, creating a pulsating pattern of loudness known as **beats**.

Our goal is to determine exactly how many beats per second you, the observer on the source, will hear.

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### The Physics of Beats and Doppler Effect

To solve this, we must break down the frequencies reaching your ears.

First, the **direct sound** travels from the siren directly to your ears.

Since both you and the siren are moving together at the same speed, there is **no relative motion** between the source and the observer for this direct path.

Therefore, the frequency of the direct sound remains unchanged:

$$f_{	ext{direct}} = f = 256	ext{ Hz}$$

Second, we have the **reflected sound**.

This sound wave undergoes a two-step Doppler shift:
1. It travels from the moving source to the stationary wall.
2. It reflects off the wall and travels back to the moving observer.

Let's analyze these two steps mathematically.

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### Step 1: Sound Reaching the Wall

The wall acts as a stationary observer ($v_{	ext{wall}} = 0$).

The source is moving towards the wall with velocity $v_s = 5	ext{ m/s}$.

Using the standard Doppler shift formula, the frequency received by the wall ($f_{	ext{wall}}$) is:

$$f_{	ext{wall}} = f \left(\frac{v}{v - v_s}ight)$$

Substituting the values ($v = 330	ext{ m/s}$ and $v_s = 5	ext{ m/s}$):

$$f_{	ext{wall}} = 256 \left(\frac{330}{330 - 5}ight) = 256 \left(\frac{330}{325}ight)$$

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### Step 2: Sound Reflected to the Observer

Now, the wall acts as a stationary source emitting this new frequency $f_{	ext{wall}}$.

You, the observer, are moving towards this stationary source with velocity $v_o = 5	ext{ m/s}$.

The frequency you hear ($f'$) is shifted upwards:

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

Substituting our expression for $f_{	ext{wall}}$ into this equation:

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

Notice how beautifully the speed of sound $v$ in the numerator and denominator cancels out!

This leaves us with the master formula for reflection problems:

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

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### Final Calculation

Let's plug in our numbers into this elegant formula:

$$f' = 256 \left(\frac{330 + 5}{330 - 5}ight)$$

$$f' = 256 \left(\frac{335}{325}ight)$$

$$f' \approx 263.87	ext{ Hz}$$

Now, the beat frequency ($f_b$) is simply the absolute difference between the two frequencies heard by the observer:

$$f_b = f' - f_{	ext{direct}}$$

$$f_b = 263.87 - 256 = \mathbf{7.87	ext{ Hz}}$$

Thus, the observer will hear approximately **$7.87	ext{ beats per second}$**.

Animated Solution for Physics - Waves: A source of sound of frequency $256 Hz$ is moving rapidly towards a wall with a velocity of $5 m/s$ . How many beats per second will be heard by the observer on source itself if sound travels at a speed of $330 m/s$ ?

Enter Numerical Value:

Visualized Solution

Visualizing the Physical Setup

Understanding Beat Generation

Doppler Shift: Wall as the Observer

Doppler Shift: Wall as the Source

Doppler Shift: Observer Receiving Reflected Wave

Combining the Doppler Equations

Substituting the Values

Calculating the Reflected Frequency

Calculating the Beat Frequency

The Sigma Insight: Doppler Effect

Analyzing the Setup

The Physics of Beats and Doppler Effect

Step 1

Step 2

Final Calculation

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