Question

A policeman with a very good ear and a good understanding of the Doppler effect stands on the shoulder of a freeway assisting a crew in a 40 -mph work zone. He notices a car approaching that is honking its horn. As the car gets closer, the policeman hears the sound of the horn as a distinct \(\mathrm{B} 4\) tone \((494 \mathrm{~Hz}) .\) The instant the car passes by, he hears the sound as a distinct \(\mathrm{A} 4\) tone \((440 \mathrm{~Hz}) .\) He immediately jumps on his motorcycle, stops the car, and gives the motorist a speeding ticket. Explain his reasoning.

Short answer

Answer: Yes, the motorist was speeding at 126.62 mph, which is significantly higher than the 40 mph speed limit in the work zone.

Step-by-step solution

Understand the Doppler effect formula

The Doppler effect formula is given by: \(f' = f \frac{v + v_{0}}{v + v_{s}}\) where \(f'\) is the frequency observed by the listener (policeman), \(f\) is the source frequency (car's honk), \(v\) is the speed of sound, \(v_{0}\) is the speed of the listener relative to the air (usually 0 for a stationary listener), and \(v_{s}\) is the speed of the source relative to the air.

Set up the equations for approaching and passing frequencies

Now, we can set up two separate equations for when the car is approaching and when it passes the policeman. As the car approaches, the observed frequency is higher (494 Hz), and as it passes, the observed frequency is lower (440 Hz). For approaching: \(f' = f \frac{v}{v - v_{s}}\) For passing: \(f' = f \frac{v}{v + v_{s}}\)

Solve for the car's speed, \(v_s\)

Now, we can solve for \(v_s\). First, we will need to find the ratio of approaching to passing frequencies, which is \(\frac{494}{440}\). Then, we can write the following equation: \(\frac{v}{v - v_{s}} = \frac{494}{440}\frac{v}{v + v_{s}}\) Next, we will solve this equation for \(v_s\). First, multiply both sides by \((v-v_s)(v+v_s)\) to get: \(v(v + v_{s})(v - v_{s}) = \frac{494}{440}(v + v_{s})(v - v_{s})(v)\) Now we can simplify this equation to: \(v_s^2 = \frac{494}{440}v^2 - v^2\) Let \(v\) be the typical speed of sound, about 343 m/s: \(v_s^2 = \frac{494}{440}(343^2) - 343^2\) Solving for \(v_s\): \(v_s = \sqrt{(\frac{494}{440})(343^2) - 343^2} = 56.62~\mathrm{m/s}\) Convert this to mph: \(v_s = 56.62 \cdot \frac{3600}{1609.34} = 126.62~\mathrm{mph}\)

Determine if the car was speeding

Now that we've calculated the speed of the car 126.62 mph, we can compare it to the speed limit in the work zone, which is 40 mph. Since the car's speed is much higher than the speed limit, the policeman's decision to give the motorist a speeding ticket is justified.

Key concepts

Sound Frequency

Sound frequency is an important concept in understanding how we hear sounds. Frequencies are measured in hertz (Hz), which indicates how many cycles or vibrations occur per second.
High frequency sounds have more vibrations per second and typically sound higher in pitch, like a whistle.
Low frequency sounds have fewer vibrations per second and sound deeper, like a drum. In the Doppler Effect context, as a sound source moves towards you, the frequency of the sound waves increases, making the pitch seem higher.
When the sound source moves away, the frequency decreases, resulting in a lower pitch. This is why the policeman heard different tones ( B4 and A4) as the car approached and then passed by.

• Approaching sound source: Frequency seems higher.
• Receding sound source: Frequency seems lower.

Understanding sound frequency helps explain why we perceive changes in sound as objects move past us. This is a practical demonstration of how the movement affects sound perception.

Wave Motion

Wave motion is how sound travels through the air. Sound waves are a type of mechanical wave that requires a medium, like air, to travel through.
They spread out in a series of compressions and rarefactions, much like ripples in a pond. These waves carry the frequency information from the sound source to our ears. The motion of these waves is key to understanding the Doppler Effect.
When the sound source moves towards an observer, the waves are compressed, increasing their frequency.
Conversely, when moving away, they are spread apart, decreasing frequency.

• Sound waves travel through air as compressions and rarefactions.
• Wave motion explains frequency changes due to relative motion.

Comprehending wave motion not only helps in understanding sound frequency changes but also in seeing how sounds move through different environments.

Speed of Sound

The speed of sound is a critical factor in how we experience sound in our daily lives. It defines how fast sound travels through different mediums, such as air, water, or solid objects.
In air, under normal conditions, the speed of sound is about 343 meters per second (m/s). In the exercise, knowing the speed of sound allows us to apply the Doppler Effect formula accurately. It is used to calculate the change in frequency heard by the policeman when the car approaches and passes by.
This speed can vary slightly with temperature, humidity, and air pressure, but 343 m/s is a good approximation for most practical calculations.

• The speed of sound in air is approximately 343 m/s.
• Knowing this speed is essential for calculations involving moving sound sources.

Understanding the speed of sound helps in solving problems involving sound propagation and perception in various environments.