A band playing music at a frequency f is moving towards a wall at a speed vb. A motorist is followin

Q: Why is the reflected frequency higher than the direct frequency?

For the reflected sound, the source (band) is moving towards the reflector (wall), which shifts the frequency up. For the direct sound, the source is moving away from the observer (motorist), which shifts the frequency down. Thus, the reflected sound always has a higher frequency in this setup.

Q: Can we use the mirror image method for the reflected sound?

Yes! You can imagine an 'image' of the band behind the wall, moving towards the wall (and thus towards you) with speed $v_b$, emitting sound of frequency $f$. Applying the Doppler formula for an observer moving towards an approaching source gives the exact same result: $f_1 = f \frac{v + v_m}{v - v_b}$.

Q: What happens to the beat frequency if the band stops moving?

If the band stops moving ($v_b = 0$), the numerator of our beat frequency expression becomes zero, meaning $f_b = 0$. This makes physical sense because if the source is stationary, both the direct and reflected waves will reach the motorist with the exact same Doppler-shifted frequency $f' = f \frac{v + v_m}{v}$.

Q: Why did the speed of sound $v$ cancel out in the reflected wave calculation?

It cancels out because the wall acts as a perfect intermediate relay. The factor of $v$ in the denominator of the first step (source to wall) is exactly balanced by the factor of $v$ in the numerator of the second step (wall to motorist) when the two shifts are multiplied.

Q: Is this formula valid if the motorist is moving faster than the speed of sound?

No, the standard Doppler formulas assume subsonic speeds ($v_m, v_b < v$). If any speed exceeds the speed of sound, shockwaves (sonic booms) are created, and these linear wave equations are no longer valid.

Question

A band playing music at a frequency $f$ is moving towards a wall at a speed $v_b$. A motorist is following the band with a speed $v_m$. If $v$ is the speed of sound, obtain an expression for the beat frequency heard by the motorist.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine standing on a straight highway.

Ahead of you is a musical band playing on a moving truck, and further ahead is a massive, solid concrete wall.

Both you (the motorist) and the band are moving towards this wall.

Because of this motion, the sound waves reaching your ears don't just take one path; they take **two distinct paths**.

First, there is the **direct sound** that travels backward from the band directly to your ears.

Second, there is the **reflected sound** that travels forward from the band, hits the wall, and bounces back to you.

Because both the source (the band) and the observer (you) are moving relative to the air, both of these sound paths will experience a **Doppler shift**.

Since the paths are different, the frequencies of these two sounds will be slightly different when they reach you.

This difference in frequency is what produces the phenomenon of **beats**.

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### The Master Tool: Doppler Effect

To calculate these shifted frequencies, we must use the general formula for the Doppler effect in sound:

$$f' = f \left( \frac{v \pm v_o}{v \mp v_s} ight)$$

Here, $v$ is the speed of sound in air, $v_o$ is the speed of the observer, and $v_s$ is the speed of the source.

Let's establish our sign conventions carefully:
- If the observer moves **towards** the source, it tends to increase the frequency (use $+$ in the numerator).
- If the source moves **away** from the observer, it tends to decrease the frequency (use $+$ in the denominator).
- If the source moves **towards** the observer, it tends to increase the frequency (use $-$ in the denominator).

---

### Step 1: Calculating the Direct Frequency ($f_2$)

Let's first analyze the direct sound path.

The band (source) is moving towards the wall with speed $v_b$.

Since you are following the band, the band is moving **away** from you.

Therefore, the source speed is $v_s = v_b$ (receding), which means we use a plus sign in the denominator:

$$	ext{Denominator} = v + v_b$$

You (the observer) are moving towards the band with speed $v_m$.

Since you are approaching the source, we use a plus sign in the numerator:

$$	ext{Numerator} = v + v_m$$

Putting these together, the frequency of the direct sound $f_2$ heard by you is:

$$f_2 = f \left( \frac{v + v_m}{v + v_b} ight)$$

---

### Step 2: Calculating the Reflected Frequency ($f_1$)

Now, let's analyze the reflected sound path. This is a two-step process.

#### Part A: Sound Reaching the Wall

The sound first travels from the band to the wall.

The wall is stationary, so the observer speed is $v_o = 0$.

The band (source) is moving **towards** the wall with speed $v_b$.

Since the source is approaching, the frequency reaching the wall ($f_{	ext{wall}}$) is shifted upwards:

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

#### Part B: Sound Bouncing Back to the Motorist

The wall now acts as a stationary source emitting sound at this new frequency $f_{	ext{wall}}$.

You (the observer) are moving **towards** the wall with speed $v_m$.

Since you are approaching this stationary source, the frequency $f_1$ you hear is:

$$f_1 = f_{	ext{wall}} \left( \frac{v + v_m}{v} ight)$$

Now, substitute the expression for $f_{	ext{wall}}$ into this equation:

$$f_1 = \left[ f \left( \frac{v}{v - v_b} ight) ight] \left( \frac{v + v_m}{v} ight)$$

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

This leaves us with:

$$f_1 = f \left( \frac{v + v_m}{v - v_b} ight)$$

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### Step 3: Finding the Beat Frequency ($f_b$)

The beat frequency is simply the difference between the two frequencies heard by the motorist:

$$f_b = f_1 - f_2$$

Since $v - v_b < v + v_b$, the reflected frequency $f_1$ is larger than the direct frequency $f_2$.

Let's substitute our expressions:

$$f_b = f \left( \frac{v + v_m}{v - v_b} ight) - f \left( \frac{v + v_m}{v + v_b} ight)$$

We can factor out the common term $f(v + v_m)$:

$$f_b = f(v + v_m) \left[ \frac{1}{v - v_b} - \frac{1}{v + v_b} ight]$$

Now, let's find a common denominator for the terms inside the bracket:

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

Simplify the numerator and denominator:

$$f_b = f(v + v_m) \left[ \frac{2v_b}{v^2 - v_b^2} ight]$$

Rearranging this gives our final, elegant expression:

$$f_b = \frac{2v_b(v + v_m)f}{v^2 - v_b^2}$$

This is the final expression for the beat frequency heard by the motorist.

Animated Solution for Physics - Waves: A band playing music at a frequency $f$ is moving towards a wall at a speed $v_{b}$ . A motorist is following the band with a speed $v_{m}$ . If $v$ is the speed of sound, obtain an expression for the beat frequency heard by the motorist.

Visualized Solution

Visualizing the Dual-Path Sound Propagation

The Doppler Effect Principle

Calculating Direct Frequency $f_{2}$

Reflected Sound - Step 1: Frequency Reaching the Wall

Reflected Sound - Step 2: Reflected Frequency Heard by Motorist $f_{1}$

Setting up the Beat Frequency Equation

Algebraic Simplification

The Final Beat Frequency Expression

The Sigma Insight: Doppler Effect

Analyzing the Setup

The Master Tool

Step 1

Step 2

# Part A

# Part B

Step 3

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