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Two trucks travel at the same speed. They are far apart on adjacent lanes and approach each other essentially head-on. One driver hears the horn of the other truck at a frequency that is 1.14 times the frequency he hears when the trucks are stationary. The speed of sound is \(343 \mathrm{m} / \mathrm{s} .\) At what speed is each truck moving?

Short Answer

Expert verified
Each truck is moving at approximately 22.43 m/s.

Step by step solution

01

Understand the Doppler Effect

The Doppler Effect describes the change in frequency of a wave in relation to an observer who is moving relative to the wave source. For sound waves, if the source and observer are moving towards each other, the observed frequency increases.
02

Set up the Doppler Effect Formula

The formula for the Doppler Effect when both the source and observer are moving towards each other is given by:\[ f' = \frac{f(v + v_o)}{v - v_s} \]where \(f'\) is the observed frequency, \(f\) is the source frequency, \(v\) is the speed of sound, \(v_o\) is the speed of the observer, and \(v_s\) is the speed of the source.
03

Define Known Values and Equation Setup

We know that \( f' = 1.14f \) and the speed of sound \( v = 343 \text{ m/s} \). Since the trucks travel at the same speed, let that speed be \( v_t \). Both \(v_o\) and \(v_s\) equal \(v_t\). Substitute these into the Doppler equation.
04

Substitute and Rearrange the Formula

Substitute the known values into the formula:\[ 1.14f = \frac{f(343 + v_t)}{343 - v_t} \]Cancel \(f\) from both sides to simplify:\[ 1.14 = \frac{343 + v_t}{343 - v_t} \]
05

Solve for truck speed \(v_t\)

Rearrange to solve for \(v_t\):\[ 1.14(343 - v_t) = 343 + v_t \]\[ 391.02 - 1.14v_t = 343 + v_t \]\[ 391.02 - 343 = 1.14v_t + v_t \]\[ 48.02 = 2.14v_t \]\[ v_t = \frac{48.02}{2.14} \approx 22.43 \text{ m/s} \]

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Key Concepts

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

Frequency Change
The concept of frequency change is central to understanding the Doppler Effect. When two objects, like the trucks in our problem, are in relative motion toward each other, the frequency of the sound waves emitted by one truck's horn appears to increase for the driver of the other truck. This is because the waves are "bunched up" as they are encountered more frequently, leading to a higher observed frequency. In contrast, if the objects were moving apart, the sound waves would spread out, and the frequency would decrease.
In the context of this exercise, the observed frequency is 1.14 times the frequency when stationary, indicating that the sound waves are being compressed due to the trucks approaching each other.
This change in frequency does not mean the true frequency of the horn has altered; it results purely from the relative motion between the source and observer. This phenomenon allows us to calculate how fast the trucks are traveling by using the changes in sound wave frequency.
Sound Waves
Sound waves are vibrations that propagate through air, or another medium, as acoustic waves. These waves have characteristics such as wavelength, speed, and frequency.
When discussing the Doppler Effect, understanding sound wave properties is crucial. The speed of sound is given as 343 m/s in the given problem. This speed contributes to how quickly the sound waves move from the source truck to the observer truck.
Sound frequency is the rate at which the waves pass a point and is measured in Hertz (Hz). In this problem, while the actual frequency of the horn is never directly stated, we know the observed frequency (the frequency heard by the other truck's driver) is amplified by a factor of 1.14. This increase is a result of sound waves compressing as both trucks move toward each other.
Relative Motion
Relative motion describes how an observer's motion regarding the motion of the source affects the perception of various physical phenomena, like sound waves in our exercise.
For our trucks, relative motion is key because both trucks are moving towards each other, which influences how the driver perceives the horn's frequency. If the trucks were stationary or moving apart, the frequency experienced by the driver would differ significantly.
In this problem, applying the relative motion concept allows us to use the Doppler Effect formula to establish a relationship between observed frequency, speed of sound, and trucks' speeds. By rearranging the formula \[ f' = \frac{f(v + v_o)}{v - v_s} \] we capture the interaction of these factors. With the given frequency change and known speed of sound, we solve for the trucks' speed, illustrating the powerful role of relative motion in modifying perceived sound.

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Most popular questions from this chapter

Dolphins emit clicks of sound for communication and echolocation. A marine biologist is monitoring a dolphin swimming in seawater where the speed of sound is \(1522 \mathrm{m} / \mathrm{s}\). When the dolphin is swimming directly away at \(8.0 \mathrm{m} / \mathrm{s},\) the marine biologist measures the number of clicks occurring per second to be at a frequency of \(2500 \mathrm{Hz}\). What is the difference (in Hz) between this frequency and the number of clicks per second actually emitted by the dolphin?

(a) A uniform rope of mass \(m\) and length \(L\) is hanging straight down from the ceiling. A small-amplitude transverse wave is sent up the rope from the bottom end. Derive an expression that gives the speed \(v\) of the wave on the rope in terms of the distance \(y\) above the bottom end of the rope and the magnitude \(g\) of the acceleration due to gravity. (b) Use the expression that you have derived to calculate the speeds at distances of \(0.50 \mathrm{m}\) and \(2.0 \mathrm{m}\) above the bottom end of the rope.

A microphone is attached to a spring that is suspended from the ceiling, as the drawing indicates. Directly below on the floor is a stationary \(440-\mathrm{Hz}\) source of sound. The microphone vibrates up and down in simple harmonic motion with a period of 2.0 s. The difference between the maximum and minimum sound frequencies detected by the microphone is 2.1 Hz. Ignoring any reflections of sound in the room and using \(343 \mathrm{m} / \mathrm{s}\) for the speed of sound, determine the amplitude of the simple harmonic motion.

A man stands at the midpoint between two speakers that are broadcasting an amplified static hiss uniformly in all directions. The speakers are \(30.0 \mathrm{m}\) apart and the total power of the sound coming from each speaker is \(0.500 \mathrm{W}\). Find the total sound intensity that the man hears (a) when he is at his initial position halfway between the speakers, and (b) after he has walked \(4.0 \mathrm{m}\) directly toward one of the speakers.

Have you ever listened for an approaching train by kneeling next to a railroad track and putting your ear to the rail? Young's modulus for steel is \(Y=2.0 \times 10^{11} \mathrm{N} / \mathrm{m}^{2},\) and the density of steel is \(\rho=7860 \mathrm{kg} / \mathrm{m}^{3} .\) On a day when the temperature is \(20^{\circ} \mathrm{C},\) how many times greater is the speed of sound in the rail than in the air?

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