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While sitting in your car by the side of a country road, you are approached by your friend, who happens to be in an identical car. You blow your car's horn, which has a frequency of 260 Hz. Your friend blows his car's horn, which is identical to yours, and you hear a beat frequency of 6.0 Hz. How fast is your friend approaching you?

Short Answer

Expert verified
Your friend is approaching you at approximately 8.4 m/s.

Step by step solution

01

Understand the Beat Frequency

The beat frequency is the difference between the frequencies of the two sound waves. In this scenario, the beat frequency you hear is 6 Hz, which means the frequency of your friend's horn as heard by you has shifted by 6 Hz.
02

Identify the Original Frequency

The original frequency of both car horns is 260 Hz when stationary. The observed frequency (either higher or lower) will depend on the direction of motion of your friend relative to you.
03

Use the Doppler Effect Formula for Frequency

The observed frequency \( f' \) can be calculated using the formula for the Doppler effect: \[ f' = f \left( \frac{v + v_O}{v + v_S} \right) \]where:- \( f \) is the original frequency (260 Hz), - \( v \) is the speed of sound in air (approximately 343 m/s), - \( v_O \) is the observer's velocity (0 m/s as you are stationary), - \( v_S \) is the source's velocity (your friend's speed).
04

Calculate the Observed Frequency

Since the beat frequency is 6 Hz and your friend's car is approaching, the frequency you hear is higher. Therefore:\[ f' = 260 + 6 = 266 ext{ Hz} \]
05

Solve for the Source Velocity

Substitute the known values into the Doppler effect equation:\[ 266 = 260 \left( \frac{343 + 0}{343 - v_S} \right) \]Solve for \( v_S \) by isolating it in the equation:1. Multiply both sides by \( 343 - v_S \): \[ 266 (343 - v_S) = 260 imes 343 \]2. Distribute and simplify the equation: \[ 266 imes 343 - 266 v_S = 260 imes 343 \]3. Rearrange to solve for \( v_S \): \[ 266 imes 343 - 260 imes 343 = 266 v_S \] \[ v_S = \frac{266 imes 343 - 260 imes 343}{266} \]
06

Perform the Calculation

Calculate the value of \( v_S \) using the given numbers:- Calculate \( 266 imes 343 = 91238 \\) and \( 260 imes 343 = 89180 \)- Subtract the results and divide by 266:\[ v_S = \frac{91238 - 89180}{266} \approx 8.4 ext{ m/s} \]

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Key Concepts

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

Beat Frequency
When two sound waves of slightly different frequencies meet, they interfere with each other and produce a phenomenon known as a beat frequency. This is heard as a periodic fluctuation in the volume or intensity of the sound. The beat frequency is simply the absolute difference between the frequencies of the two waves.
In our scenario, the car horns both originally have a frequency of 260 Hz. The beat frequency observed is 6 Hz. This indicates that there is a shift in the frequency due to the movement of your friend's car, altering what you hear to be either 254 Hz or 266 Hz. Since the car is approaching, the frequency increases to 266 Hz, leading to the beat frequency of 6 Hz.
Beat frequency is a practical application of sound wave interference and is useful in various fields, including music and engineering.
Sound Waves
Sound waves are longitudinal waves, moving through air by compressions and rarefactions of particles. These waves are characterized by their frequency (how often the particles vibrate), wavelength (distance between waves), and velocity (speed of wave propagation).
The frequency of sound waves is measured in Hertz (Hz), and it determines the pitch of the sound. Higher frequencies result in higher pitches, while lower frequencies correspond to lower pitches. Our situation involves a car horn with a known frequency of 260 Hz.
Sound wave interactions, such as the Doppler Effect and beat frequency phenomena, depend heavily on these properties, applying physics principles to understand how sounds are perceived when in motion.
Velocity Calculation
Calculating velocity, especially in sound-related problems, often involves understanding the speed of sound and the relative motion of the source and listener. In our exercise, we applied these principles through the Doppler Effect to find your friend's velocity.
  • Start by recognizing the observed frequency, which was 266 Hz due to your friend's approach.
  • The original frequency was 260 Hz, and since you're stationary, your velocity as an observer is 0 m/s.
To find the velocity of your friend's car, or the source, we rearrange the Doppler Effect formula:\[ f' = f \left( \frac{v + v_O}{v + v_S} \right) \] Substituting the known values, solve for the unknown velocity. This allows for precise calculation of how fast the car is moving, which in this case turned out to be approximately 8.4 m/s.
Frequency Shift
The frequency shift is a change in the apparent frequency of a wave as perceived by an observer relative to the source of the wave. This shift is caused by the Doppler Effect, a common wave phenomenon.
The Doppler Effect formula describes how the frequency changes when there is relative motion between an observer and the source. If the source moves towards the observer, the frequency increases and shifts higher; if it moves away, the frequency decreases.
For two identical car horns, the shift is exactly what creates the beat frequency. Because the friend's car is moving towards you, the frequency rises from 260 Hz to 266 Hz, resulting in the beat frequency of 6 Hz. This frequency shift makes it possible to calculate the velocity at which your friend is approaching, using principles of wave dynamics.

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Most popular questions from this chapter

You blow across the open mouth of an empty test tube and produce the fundamental standing wave of the air column inside the test tube. The speed of sound in air is 344 m/s and the test tube acts as a stopped pipe. (a) If the length of the air column in the test tube is 14.0 cm, what is the frequency of this standing wave? (b) What is the frequency of the fundamental standing wave in the air column if the test tube is half filled with water?

A long tube contains air at a pressure of 1.00 atm and a temperature of 77.0\(^\circ\)C. The tube is open at one end and closed at the other by a movable piston. A tuning fork that vibrates with a frequency of 500 Hz is placed near the open end. Resonance is produced when the piston is at distances 18.0 cm, 55.5 cm, and 93.0 cm from the open end. (a) From these values, what is the speed of sound in air at 77.0\(^\circ\)C? (b) From the result of part (a), what is the value of \(\gamma\)? (c) These results show that a displacement antinode is slightly outside the open end of the tube. How far outside is it?

An organ pipe has two successive harmonics with frequencies 1372 and 1764 Hz. (a) Is this an open or a stopped pipe? Explain. (b) What two harmonics are these? (c) What is the length of the pipe?

(a) By what factor must the sound intensity be increased to raise the sound intensity level by 13.0 dB? (b) Explain why you don't need to know the original sound intensity.

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