A police car with a siren of frequency 8 kHz is moving with uniform velocity 36 km/h towards a tall

Q: Why does the building act as both an observer and a source?

The building first receives the sound waves from the moving car, acting as an observer. It then reflects these waves back into the medium, effectively acting as a stationary source emitting the frequency it just received.

Q: Can we use a single formula for reflection problems like this?

Yes! For a moving source/observer approaching a stationary reflector, the combined formula is $f' = f_0 \left(\frac{v + v_c}{v - v_c}\right)$. This represents a double Doppler shift.

Q: Why is the frequency higher when the car approaches the building?

As the car approaches the building, the sound waves in front of it are compressed (shorter wavelength), leading to a higher frequency. When the driver moves towards the reflected waves, they intercept more wave crests per second, increasing the perceived frequency even further.

Q: What if the car was moving away from the building?

If the car was moving away, the wave fronts would be stretched. The formula would become $f' = f_0 \left(\frac{v - v_c}{v + v_c}\right)$, resulting in a lower perceived frequency.

Q: Does the wind speed affect this calculation?

Yes, if wind were blowing, it would alter the effective speed of sound in the direction of propagation. However, in this standard problem, we assume still air.

Question

A police car with a siren of frequency $8	ext{ kHz}$ is moving with uniform velocity $36	ext{ km/h}$ towards a tall building which reflects the sound waves. The speed of sound in air is $320	ext{ m/s}$. The frequency of the siren heard by the car driver is

Options:

(A) $8.50	ext{ kHz}$

(B) $8.25	ext{ kHz}$

(C) $7.75	ext{ kHz}$

(D) $7.50	ext{ kHz}$

Sigma Percentile · Accepted Answer

### The Magic of Moving Sound

Imagine standing on a busy city street. A police car speeds past you, its siren wailing. You have undoubtedly noticed how the pitch of the siren suddenly drops as the car passes. This everyday phenomenon is the **Doppler Effect**, a cornerstone of wave physics.

But what happens when the sound doesn't just pass you by, but instead bounces off a massive obstacle and returns to the moving source? This is the thrilling scenario of acoustic reflection, where the Doppler shift occurs not once, but twice!

Let's dive deep into this problem and unravel the elegant mathematics that governs this double shift.

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### Breaking Down the Journey

To solve this problem without getting overwhelmed, we must divide the sound's journey into two distinct, sequential stages:

1. **Stage 1: Emission to Reflection** — The sound travels from the moving police car (the source) to the stationary building (the observer).
2. **Stage 2: Reflection to Reception** — The building now acts as a stationary source, reflecting the shifted sound back to the moving driver (the observer).

Before we begin our calculations, we must convert all given values into standard SI units. The velocity of the police car is given as:

$$v_c = 36	ext{ km/h}$$

Multiplying by the conversion factor $\frac{5}{18}$:

$$v_c = 36 	imes \frac{5}{18} = 10	ext{ m/s}$$

Now, we are ready to analyze each stage mathematically.

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### Stage 1: The Building as an Observer

In this first stage, the police car is a moving source approaching a stationary observer (the building). The formula for the apparent frequency $f_1$ received by the building is:

$$f_1 = f_0 \left(\frac{v}{v - v_s}ight)$$

Here, the source velocity $v_s = v_c = 10	ext{ m/s}$, the speed of sound $v = 320	ext{ m/s}$, and the original frequency $f_0 = 8	ext{ kHz}$. Substituting these values:

$$f_1 = 8 	imes \left(\frac{320}{320 - 10}ight) = 8 	imes \frac{320}{310} = 8 	imes \frac{32}{31}	ext{ kHz}$$

This is the frequency of the sound waves that strike the building's wall. Since the wall is stationary, it reflects this exact frequency back into the air without any further shift.

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### Stage 2: The Driver as an Observer

Now, the building acts as a stationary source emitting sound of frequency $f_1$. The driver in the police car is a moving observer approaching this source at speed $v_o = v_c = 10	ext{ m/s}$.

The formula for the final frequency $f'$ heard by the driver is:

$$f' = f_1 \left(\frac{v + v_o}{v}ight)$$

Substituting our expression for $f_1$ and the known velocities:

$$f' = \left(8 	imes \frac{32}{31}ight) 	imes \left(\frac{320 + 10}{320}ight)$$

$$f' = 8 	imes \frac{32}{31} 	imes \frac{330}{320}$$

Notice how beautifully the terms simplify. The fraction $\frac{330}{320}$ reduces to $\frac{33}{32}$:

$$f' = 8 	imes \frac{32}{31} 	imes \frac{33}{32}$$

The term $32$ in the numerator and denominator cancels out perfectly, leaving us with:

$$f' = 8 	imes \frac{33}{31} = \frac{264}{31} \approx 8.516	ext{ kHz}$$

Rounding to two decimal places, we get **$8.50	ext{ kHz}$**, which matches **Option (a)** perfectly.

Animated Solution for Physics - Waves: A police car with a siren of frequency $8 kHz$ is moving with uniform velocity $36 km/h$ towards a tall building which reflects the sound waves. The speed of sound in air is $320 m/s$ . The frequency of the siren heard by the car driver is

Select Answer:

Visualized Solution

Visualizing the Acoustic Setup

Converting Velocity to SI Units

Stage 1: Sound Reaching the Building

Substituting Values for Stage 1

Calculating Intermediate Frequency $f_{1}$

Stage 2: Reflected Sound Reaching the Driver

Substituting Values for Stage 2

Final Calculation of $f^{'}$

Selecting the Correct Option

The Sigma Insight: Doppler Effect

The Magic of Moving Sound

Breaking Down the Journey

Stage 1

Stage 2

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