Question

A police car with a siren of frequency \(8 \mathrm{kHz}\) is moving with uniform velocity \(36 \mathrm{~km} / \mathrm{hr}\) towards a tall building which reflects the sound waves. The speed of sound in air is \(320 \mathrm{~m} / \mathrm{s}\). The frequency of the siren heard by the car driver is (A) \(8.50 \mathrm{kHz}\) (B) \(8.25 \mathrm{kHz}\) (C) \(7.75 \mathrm{kHz}\) (D) \(7.50 \mathrm{kHz}\)

Short answer

The frequency of the siren heard by the car driver is 8.59 kHz, but since this option isn't available, rounding to the nearest option given, (A) 8.50 kHz is the correct answer.

Step-by-step solution

Convert Police Car Velocity to m/s

The velocity of the police car is given in kilometers per hour. Convert this velocity to meters per second by dividing by 3.6, since 1 km/hr is equivalent to 1/3.6 m/s. Therefore, the police car's velocity, v, is given by: \( v = \frac{36 \mathrm{km/hr}}{3.6} = 10 \mathrm{m/s} \).

Calculate the Observed Frequency at the Building (Approaching)

Use the Doppler Effect equation for a moving source and stationary observer to find the frequency that the building 'observes' as the police car approaches: \( f' = \frac{f}{1-\frac{v}{v_{sound}}} \), where \( f \) is the original frequency and \( v_{sound} \) is the speed of sound. Plugging in the values gives: \( f' = \frac{8000\,\mathrm{Hz}}{1 - \frac{10\,\mathrm{m/s}}{320\,\mathrm{m/s}}} = 8333.33\,\mathrm{Hz}\).

Calculate the Observed Frequency at the Police Car (Reflecting)

The frequency that the police car observes is from a moving 'observer' (the car itself after the reflection) and a stationary source (the reflected waves from the building). The Doppler Effect equation for a moving observer is: \( f'' = f'(1+\frac{v}{v_{sound}}) \). Plugging in the observed frequency from the building and the velocities gives: \( f'' = 8333.33\,\mathrm{Hz}\,\big(1 + \frac{10}{320}\big) = 8333.33\,\mathrm{Hz} \cdot \frac{330}{320} = 8593.75\,\mathrm{Hz} \).

Key concepts

Sound Wave Reflection

Imagine shouting towards a cliff and hearing your own voice bouncing back at you. This phenomenon is known as sound wave reflection, which is a key principle in understanding how echoes are created.

When a sound wave hits a surface, such as a tall building or a cliff face, it can be reflected back into the medium from which it originated, much like a rubber ball bounces back when thrown against a wall. However, the nature of the surface has a significant effect on the reflection; smooth, hard surfaces reflect sound waves better than rough, absorbent materials.

In the context of our example with a police car, the siren emits sound waves that travel through the air. When these waves strike the building, they are reflected back towards the source. This reflection plays a crucial role in the phenomenon experienced by the driver due to the Doppler Effect, which we will elaborate on in the subsequent sections.

Frequency Calculation

The concept of frequency calculation is a fundamental part of understanding waves and their behaviors, including the Doppler Effect. Frequency refers to the number of waves that pass a point in one second and is measured in hertz (Hz).

The basic formula for calculating the frequency of a wave is:
\[ f = \frac{v}{\textlambda} \]
where \( f \) is the frequency, \( v \) is the speed, and \(\textlambda\) is the wavelength. However, in situations involving movement, such as a police car moving towards a building, we use a modified formula to account for the Doppler Effect:
\[ f' = \frac{f}{1-\frac{v}{v_{sound}}} \]
and for the frequency heard after the reflection:
\[ f'' = f'(1+\frac{v}{v_{sound}}) \]
These equations allow us to calculate the new frequency observed after the Doppler Effect comes into play when there is relative motion between the source and the observer.

Police Car Siren

The police car siren is an excellent real-world example to understand the Doppler Effect. Police car sirens are designed to emit a consistent and easily recognizable sound that can be heard over a long distance. In terms of frequency, sirens typically emit sounds in the range of a few hundred to a few thousand hertz.

When a police car moves towards a stationary object with its siren on, the frequency of the sound waves increases for someone standing between the car and the object, because the waves are getting bunched up due to the car's approach. Conversely, for an observer behind the moving car, the frequency decreases because the waves are spread out.

These changes in frequency experienced by different observers or when the sound is reflected off objects, like in the textbook problem, are direct applications of the Doppler Effect. The phenomenon helps explain why the pitch of the siren appears higher as the police car approaches an object, and lower as it moves away, even though the siren itself emits sound at a constant frequency.