Problem 58 - College Physics

Question

A stationary police car emits a sound of frequency 1200 \(\mathrm{Hz}\) that bounces off of a car on the highway and returns with a frequency of 1250 \(\mathrm{Hz}\) . The police car is right next to the highway, so the moving car is traveling directly toward or away from it. (a) How fast was the moving car going? Was it moving towards or away from the police car? (b) What frequency would the police car have received if it had been traveling toward theother car at 20.0 \(\mathrm{m} / \mathrm{s} ?\)

Step-by-Step Solution

Answer

The car was traveling at 14.3 m/s towards the police car. If the police car were also moving, it would have received a frequency of 1315 Hz.

Step 1: Identify the Known Values

We are given the initial frequency \( f_0 = 1200 \, \text{Hz} \) and the frequency \( f = 1250 \, \text{Hz} \) received by the stationary police car. The speed of sound \( v \) in air is approximately \( 343 \, \text{m/s} \).

Step 2: Use Doppler Effect Formula for Part (a)

The received frequency when the source and observer are moving towards each other is given by:\[ f = f_0 \left( \frac{v + v_\text{car}}{v} \right) \]\( f_0 = 1200 \, \text{Hz} \), \( f = 1250 \, \text{Hz} \), and we need to solve for \( v_\text{car} \):\[ 1250 = 1200 \cdot \frac{343 + v_\text{car}}{343} \]

Step 3: Solve for Speed of Moving Car

Rearrange and solve the equation:\[ \frac{1250}{1200} = \frac{343 + v_\text{car}}{343} \] \[ 1.0417 = 1 + \frac{v_\text{car}}{343} \] \[ 0.0417 \times 343 = v_\text{car} \] \[ v_\text{car} \approx 14.3 \, \text{m/s} \] Thus, the car is moving towards the police car.

Step 4: Apply Doppler Shift Formula for Part (b)

With the police car moving towards the other car at \( 20.0 \, \text{m/s} \), the new received frequency \( f' \) can be found using:\[ f' = f_0 \left( \frac{v + v_\text{police}}{v} \right) \left( \frac{v}{v - v_\text{car}} \right) \] Insert the given values into the formula.

Step 5: Calculate the New Frequency

Insert values and solve:\[ f' = 1200 \left( \frac{343 + 20}{343} \right) \left( \frac{343}{343 - 14.3} \right) \] Calculate each term separately and combine to find \( f' \approx 1315 \, \text{Hz} \).

Key Concepts

Frequency

Frequency is a foundational concept when studying the Doppler Effect. It refers to the number of complete oscillations that occur each second of a wave, measured in hertz (Hz). In simpler terms, it describes how many times a wave repeats in a second. For sound waves, the frequency is perceived as pitch by human ears. Understanding the change in frequency due to movement is crucial to grasp the Doppler Effect. For instance, if a source emitting sound is moving closer, the waves become compressed. This increases their frequency, making the sound appear higher in pitch. Conversely, if the source moves away, the waves stretch out, decreasing the frequency and lowering the pitch perceived. This principle is key to explaining why the frequency changes from 1200 Hz to 1250 Hz when the car moves toward the police car.

Speed of Sound

The speed of sound is how fast sound waves travel through a medium, like air. It varies depending on factors such as temperature and humidity but is generally about 343 m/s in air at room temperature. The speed of sound is pivotal in calculating the Doppler Effect because it serves as the baseline speed against which other movements (like those of a car or police vehicle) are compared. It makes it possible to calculate changes in frequency caused by relative motion. When solving problems related to the Doppler Effect, knowing the speed of sound helps in setting up the correct equations to find unknown variables. For the given exercise, knowing the speed of sound allows us to determine the car's speed using the change in frequencies.

Relative Motion

Relative motion describes the movement of one object concerning another. In the Doppler Effect, the change in perceived frequency as sound waves reach an observer is contingent on the relative motion between the observer and the sound source. Crucially, it is the closing or increasing distance between the sound source and the observer that affects how we perceive sound. In the given problem, because the car is moving towards the stationary police car, the frequency increases. The relative motion is a factor that allows us to apply the Doppler formula and solve for the velocity of the car by comparing the moving and stationary states.

Stationary Source

A stationary source in the context of the Doppler Effect refers to an emitter of waves, like sound, that is not moving. In the exercise, the police car initially acts as a stationary source emitting a sound frequency of 1200 Hz. When dealing with a stationary source, the perceived frequency by any observer or reflecting surfaces changes only if the observer or reflecting object moves. This principle allows us to recognize that if the reflected frequency changes, it's due to the motion of the reflecting object (the car in our problem). Understanding stationary sources ensures better insight into why the motion of the car alone affects the frequency received back by the police car.

Moving Observer

A moving observer, unlike a stationary one, refers to when the person or device receiving the sound is in motion. This movement influences how the sound frequency is perceived due to the relative speeds between the source and the observer. In Part (b) of the exercise, the police car is now the moving observer traveling towards the car at 20 m/s. This added motion affects the received frequency, highlighting how any movement by the observer leads to further frequency shifts. Analyzing how the motion of the observer affects frequency is integral in understanding and applying the Doppler Effect's principles. It demonstrates that the Doppler shift can result from either source movement, observer movement, or both, allowing comprehensive solutions such as calculating the adjusted frequency of 1315 Hz when both are in motion.