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Chapter 16: Problem 77

From a vantage point very close to the track at a stock car race, you hear the sound emitted by a moving car. You detect a frequency that is 0.86 times as small as the frequency emitted by the car when it is stationary. The speed of sound is \(343 \mathrm{m} / \mathrm{s} .\) What is the speed of the car?

Short Answer

Expert verified
The speed of the car is approximately 55.37 m/s.

Step by step solution

01

Understand the Problem

We are given the observed frequency from a car moving away from us, which is 0.86 times the emitted frequency by the car when stationary. We need to find out the speed of the car, knowing that the speed of sound is 343 m/s.
02

Use the Doppler Effect Equation

The Doppler effect equation for a source moving away from the observer is \( f' = f \frac{v}{v+v_s} \), where \( f' \) is the observed frequency, \( f \) is the emitted frequency, \( v \) is the speed of sound, and \( v_s \) is the speed of the source (the car).
03

Plug in Known Values

We know that \( f' = 0.86f \). Substitute this and the speed of sound into the equation: \( 0.86f = f \frac{343}{343+v_s} \).
04

Simplify the Equation

Cancel out \( f \) from both sides of the equation since \( f eq 0 \): \( 0.86 = \frac{343}{343+v_s} \).
05

Solve for \( v_s \)

Rearrange the equation to solve for \( v_s \):1. Multiply both sides by \( 343+v_s \): \( 0.86(343 + v_s) = 343 \).2. Distribute and simplify: \( 295.38 + 0.86v_s = 343 \).3. Subtract 295.38 from both sides: \( 0.86v_s = 47.62 \).4. Divide by 0.86 to isolate \( v_s \): \( v_s \approx 55.37 \mathrm{m/s} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Speed of Sound
The speed of sound is a crucial factor when understanding how sound waves travel through the air. It's essentially the speed at which sound waves move from one point to another. In air, at room temperature, the speed of sound is approximately 343 m/s. This means that sound can cover a distance of 343 meters in one second. But what exactly influences the speed of sound?

  • **Medium:** Sound travels fastest in solids, slower in liquids, and slowest in gases due to the difference in particle density.
  • **Temperature:** Higher temperatures increase particle movement, allowing sound to travel faster.
  • **Humidity and Pressure:** Greater humidity and pressure can also increase sound speed, though their effects are less pronounced compared to temperature and medium type.
Understanding the speed at which sound travels is fundamental for calculating distance or speed as in the case of the Doppler Effect scenario presented in the exercise.
Frequency Change
The Doppler Effect describes how the frequency of a sound wave changes for an observer moving relative to the source of the sound. Imagine you're standing close to a race track and a car zooms past you. You might notice that the pitch of the car's engine sound is higher as it approaches and lower as it moves away. This change in frequency is because of the Doppler Effect.

When the sound source moves away from the observer, the sound waves stretch, resulting in a lower frequency and lower pitch. Conversely, if the source is coming towards the observer, the waves compress, leading to a higher frequency and higher pitch.
  • The formula provided in the exercise, which is used to calculate the changed frequency, is given by: \[ f' = f \frac{v}{v + v_s} \]
  • Where:
    • **\( f' \):** the observed frequency
    • **\( f \):** the original emitted frequency
    • **\( v \):** speed of sound
    • **\( v_s \):** speed of the source
Relative Motion in Physics
Relative motion plays a critical role in many physics problems, including those involving the Doppler Effect. It refers to analyzing the motion of objects from different viewpoints. In the exercise, the car's speed in relation to the observer creates a change in the frequency of sound waves emitted by the car. This is due to relative motion.

Some important points about relative motion include:
  • **Frame of reference:** Movement is always measured relative to a chosen frame of reference. For example, the ground can be a fixed reference point while analyzing the movement of a car.
  • **Relative velocity:** This describes how fast one object is moving compared to another. It involves understanding both speed and direction of the objects involved.
  • **Doppler Effect application:** When a source moves relative to the observer, the waves emitted are compressed or stretched depending on the relative direction of movement.
In physics, recognizing the impact of relative motion helps in accurately describing and predicting outcomes in various scenarios, such as frequency changes in sound as seen in this problem setting.

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Most popular questions from this chapter

Argon (molecular mass \(=39.9 \mathrm{u}\) ) is a monatomic gas. Assuming that it behaves like an ideal gas at \(298 \mathrm{K}(\gamma=1.67),\) find \((\mathrm{a})\) the rms speed of argon atoms and (b) the speed of sound in argon.

A jetskier is moving at \(8.4 \mathrm{m} / \mathrm{s}\) in the direction in which the waves on a lake are moving. Each time he passes over a crest, he feels a bump. The bumping frequency is \(1.2 \mathrm{Hz}\), and the crests are separated by \(5.8 \mathrm{m}\). What is the wave speed?

To navigate, a porpoise emits a sound wave that has a wavelength of \(1.5 \mathrm{cm} .\) The speed at which the wave travels in seawater is \(1522 \mathrm{m} / \mathrm{s} .\) Find the period of the wave.

To measure the acceleration due to gravity on a distant planet, an astronaut hangs a \(0.055-\mathrm{kg}\) ball from the end of a wire. The wire has a length of \(0.95 \mathrm{m}\) and a linear density of \(1.2 \times 10^{-4} \mathrm{kg} / \mathrm{m}\). Using electronic equipment, the astronaut measures the time for a transverse pulse to travel the length of the wire and obtains a value of 0.016 s. The mass of the wire is negligible compared to the mass of the ball. Determine the acceleration due to gravity.

A wireless transmitting microphone is mounted on a small platform that can roll down an incline, directly away from a loudspeaker that is mounted at the top of the incline. The loudspeaker broadcasts a tone that has a fixed frequency of \(1.000 \times 10^{4} \mathrm{Hz},\) and the speed of sound is \(343 \mathrm{m} / \mathrm{s} .\) At a time of \(1.5 \mathrm{s}\) following the release of the platform, the microphone detects a frequency of \(9939 \mathrm{Hz}\). At a time of \(3.5 \mathrm{s}\) following the release of the platform, the microphone detects a frequency of \(9857 \mathrm{Hz}\) What is the acceleration (assumed constant) of the platform?
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