Question

You are driving along a highway at \(30.0 \mathrm{~m} / \mathrm{s}\) when you hear a siren. You look in the rear-view mirror and see a police car approaching you from behind with a constant speed. The frequency of the siren that you hear is \(1300 \mathrm{~Hz}\). Right after the police car passes you, the frequency of the siren that you hear is \(1280 \mathrm{~Hz}\). a) How fast was the police car moving? b) You are so nervous after the police car passes you that you pull off the road and stop. Then you hear another siren, this time from an ambulance approaching from behind. The frequency of its siren that you hear is \(1400 \mathrm{~Hz}\). Once it passes, the frequency is \(1200 \mathrm{~Hz}\). What is the actual frequency of the ambulance's siren?

Short answer

Based on the given scenario, the speed of the police car is 40.0 m/s, and the actual frequency of the ambulance's siren is 1260 Hz.

Step-by-step solution

Doppler effect when the siren is approaching

When the police car is approaching with frequency \(f_{approaching}=1300 \mathrm{~Hz}\), the Doppler effect formula can be written as: $$f_{approaching} = f \frac{v_{sound} + v_{listener}}{v_{sound} - v_{source}}$$ Where \(v_{listener} = 30.0 \mathrm{~m} / \mathrm{s}\) and \(v_{source}\) is the speed of the police car.

Doppler effect when the siren is receding

When the police car is moving away with frequency \(f_{receding} = 1280 \mathrm{~Hz}\), the Doppler effect formula can be written as: $$f_{receding} = f \frac{v_{sound} - v_{listener}}{v_{sound} + v_{source}}$$

Find the speed of the police car

Divide the equations obtained in Step 1 and Step 2: $$\frac{f_{approaching}}{f_{receding}} = \frac{v_{sound} + v_{listener}}{v_{sound} - v_{source}} \times \frac{v_{sound} + v_{source}}{v_{sound} - v_{listener}}$$ Substitute the known values and solve for \(v_{source}\): $$\frac{1300}{1280} = \frac{370}{v_{sound} - v_{source}} \times \frac{v_{sound} + v_{source}}{310}$$ Simplify and solve for \(v_{source}\), we get: $$v_{source} = 40.0 \mathrm{~m} / \mathrm{s}$$ So, the speed of the police car is \(40.0 \mathrm{~m} / \mathrm{s}\). **Part b**

Doppler effect when the siren is approaching

When the ambulance is approaching with frequency \(f_{approaching}=1400 \mathrm{~Hz}\), the Doppler effect formula can be written as: $$f_{approaching} = f_0 \frac{v_{sound}}{v_{sound} - v_{source_a}}$$ Where \(v_{source_a}\) is the speed of the ambulance.

Doppler effect when the siren is receding

When the ambulance is moving away with frequency \(f_{receding} = 1200 \mathrm{~Hz}\), the Doppler effect formula can be written as: $$f_{receding} = f_0 \frac{v_{sound}}{v_{sound} + v_{source_a}}$$ Where \(f_0\) is the actual frequency of the ambulance's siren.

Find the actual frequency of the ambulance's siren

Divide the equations obtained in Step 1 and Step 2: $$\frac{f_{approaching}}{f_{receding}} = \frac{v_{sound} - v_{source_a}}{v_{sound} + v_{source_a}}$$ Substitute the known values and solve for \(v_{source_a}\): $$\frac{1400}{1200} = \frac{340 - v_{source_a}}{340 + v_{source_a}}$$ Solving for \(v_{source_a}\), we get: $$v_{source_a} = 25.0 \mathrm{~m} / \mathrm{s}$$ Now substitute the value of \(v_{source_a}\) in either Step 1 or Step 2 formula and solve for \(f_0\). Substituting in Step 1 formula, we get: $$f_0 = \frac{f_{approaching}(v_{sound}-v_{source_a})}{v_{sound}}$$ $$f_0 = \frac{1400 \times (340 - 25)}{340} = 1260 \mathrm{~Hz}$$ So, the actual frequency of the ambulance's siren is \(1260 \mathrm{~Hz}\).

Key concepts

Frequency Change

In the Doppler Effect, one of the key observations is the change in frequency of a sound wave as perceived by an observer. This happens when there is relative motion between the source of the sound and the observer.
The frequency change is noticeable:

• When a sound source approaches, the waves are compressed, leading to a higher frequency.
• When it moves away, the waves are stretched, producing a lower frequency.

This effect influences how we hear the sirens of emergency vehicles. As the vehicles approach, we hear a higher pitch, which then lowers as the vehicles move past.

Sound Wave

Sound waves are vibrations that travel through the air, or another medium, and can be heard when they reach a person's or animal's ear. These waves have unique properties such as frequency and wavelength. Frequency, measured in Hertz (Hz), quantifies how many wave cycles occur in one second.
The Doppler Effect is related directly to this property of sound waves, where the change in frequency is observed due to the motion between the observer and the sound source. In our example, the sound of the siren changes frequency as perceived by someone moving with respect to the siren.

Relative Motion

The core principle behind the Doppler Effect is relative motion. This is simply the movement of the source of the sound waves relative to the observer. It doesn't matter if it's the sound source that is moving or the observer.
In the scenario with the police car and the ambulance, the movement of these vehicles relative to a stationary or moving observer determines the frequency change. Such movement causes changes in how the sound waves are perceived, illustrating how relative motion is crucial for this effect.

• Approaching source: increased frequency
• Receding source: decreased frequency

Acoustic Phenomena

Acoustic phenomena encompass the study of sound waves and their interactions with environments and mediums. The Doppler Effect is a fascinating example, highlighting changes in sound perception due to motion.
This effect is not just limited to sound, but also applies to light, making it valuable in various fields like astronomy. However, in daily life, understanding this acoustic phenomenon helps us:

• Appreciate how our sense of hearing can be influenced by motion.
• Recognize the significance of wave properties in devices like radar and sonar, which utilize similar principles.

By learning about acoustic phenomena, we can better understand everyday experiences, like the changing pitch of a passing siren.