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A speedboat, starting from rest, moves along a straight line away from a dock. The boat has a constant acceleration of \(+3.00 \mathrm{m} / \mathrm{s}^{2}\) (see the figure). Attached to the dock is a siren that is producing a \(755-\mathrm{Hz}\) tone. If the air temperature is \(20^{\circ} \mathrm{C},\) what is the frequency of the sound heard by a person on the boat when the boat's displacement from the dock is \(+45.0 \mathrm{m} ?\)

Short Answer

Expert verified
The frequency heard by the person is approximately 791.04 Hz.

Step by step solution

01

Understand the Problem

The problem involves a moving source of sound (the siren on the dock) and a listener (person on the boat) moving away from the source with constant acceleration. We need to find the frequency of the sound heard by the person on the boat using the Doppler effect formula.
02

Calculate Final Velocity

Since the boat starts from rest, we can use the formula for constant acceleration to find the velocity: \[ v = u + at \]where \( v \) is the final velocity, \( u = 0 \) (initial velocity since it starts from rest), \( a = 3.00 \, \text{m/s}^2 \) (acceleration), and \( s = 45.0 \, \text{m} \) (displacement).Using the formula \[ v^2 = u^2 + 2as \]we find \[ v^2 = 0 + 2 \times 3.00 \, \text{m/s}^2 \times 45.0 \, \text{m} \]\[ v^2 = 270 \]\[ v = \sqrt{270} \approx 16.43 \, \text{m/s} \]
03

Calculate Speed of Sound in Air

The speed of sound at a given temperature can be calculated using \[ v_{\text{sound}} = 331.4 + 0.6 \times T \] where \( T = 20 \, ^\circ\text{C} \). Thus,\[ v_{\text{sound}} = 331.4 + 0.6 \times 20 \]\[ v_{\text{sound}} = 343.4 \, \text{m/s} \]
04

Apply the Doppler Effect Formula

The frequency perceived by a moving observer is given by the formula: \[ f' = \left( \frac{v_{\text{sound}} + v_{\text{observer}}}{v_{\text{sound}}} \right) f \]where\( f = 755 \, \text{Hz} \)(original frequency of the siren),\( v_{\text{observer}} = 16.43 \, \text{m/s} \), and\( v_{\text{sound}} = 343.4 \, \text{m/s} \).Plugging in these values:\[ f' = \left( \frac{343.4 + 16.43}{343.4} \right) \times 755 \]\[ f' = \left( \frac{359.83}{343.4} \right) \times 755 \]\[ f' \approx 791.04 \, \text{Hz} \]

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Key Concepts

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

Constant Acceleration
Constant acceleration refers to a situation where an object's rate of change of velocity remains the same over time. This is a key concept in this exercise. The speedboat starts from rest, meaning its initial velocity is zero. It then accelerates at a constant rate of \(3.00 \, \text{m/s}^2\). This uniform increase in velocity allows us to use the kinematic equations, specifically:
  • \(v = u + at\), which calculates the final velocity \(v\) based on initial velocity \(u\), acceleration \(a\), and time \(t\).
  • Alternatively, \(v^2 = u^2 + 2as\), used here to determine the velocity based on displacement \(s\).
In this problem, we utilize \(v^2 = u^2 + 2as\) because we have the displacement and need to find the velocity reached by that point. With constant acceleration, these equations become powerful tools for predicting future motion, aiding not just in academic exercises but also in real-world applications.
Frequency Change
The concept of frequency change is crucial in understanding how motion impacts sound perception. In this exercise, a siren emits a constant frequency of \(755 \, \text{Hz}\). However, a person on a boat moving away from the source perceives a different frequency due to the Doppler Effect.

The Doppler Effect causes the frequency of the sound waves to alter, depending on the relative motion of the source and observer. If the observer moves away from the source, the perceived frequency decreases and vice versa. Here, the boat accelerates away from the dock, causing an apparent increase in the wavelength received and thereby a shift in frequency.

Using the formula:\[f' = \left( \frac{v_{\text{sound}} + v_{\text{observer}}}{v_{\text{sound}}} \right) f\],we determine the new perceived frequency \(f'\). This shift is why moving objects and observers perceive notable changes in pitch—an intriguing real-world example of wave interactions!
Speed of Sound
Understanding the speed of sound is essential to compute changes in frequency due to motion. The speed at which sound travels through air depends on temperature and is calculated through the equation:\[v_{\text{sound}} = 331.4 + 0.6 \times T\],where \(T\) is the temperature in degrees Celsius.

For this problem, with an air temperature of \(20 \, ^\circ\text{C}\), the speed of sound is \(343.4 \, \text{m/s}\). This plays a vital role in how sound frequencies are perceived as it affects the propagation of sound waves. Calculating this accurately is crucial in practical applications, from acoustics to meteorology.

The fact that the speed of sound varies with temperature is why sound travels faster on a warm day compared to a cold day. Hence, both scientists and engineers must account for such environmental variables when working with sound in various fields.
Motion Along a Straight Line
Motion along a straight line refers to linear motion where an object moves in a single dimension. In this exercise, the speedboat's path directly impacts how sound is perceived by someone on the boat.

Starting from rest, the boat accelerates uniformly, allowing us to track its displacement, velocity, and time using straightforward kinematic equations. This helps simplify problems involving complex motion by reducing them to cases of linear path calculation.

Such linear motion ensures that the concepts like acceleration and displacement are easily quantifiable since only one spatial direction needs consideration. Understanding this type of motion is foundational in physics as it forms the basis for studying more complex trajectories later on.

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Most popular questions from this chapter

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 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?

The sound intensity level at a rock concert is \(115 \mathrm{dB},\) while that at a jazz fest is 95 dB. Determine the ratio of the sound intensity at the rock concert to the sound intensity at the jazz fest.

Consider the freight train in Figure \(16.6 .\) Suppose that 15 boxcars pass by in a time of 12.0 s and each has a length of \(14.0 \mathrm{m}\). (a) What is the frequency at which each boxcar passes? (b) What is the speed of the train?

Suppose that a public address system emits sound uniformly in all directions and that there are no reflections. The intensity at a location \(22 \mathrm{m}\) away from the sound source is \(3.0 \times 10^{-4} \mathrm{W} / \mathrm{m}^{2} .\) What is the intensity at a spot that is \(78 \mathrm{m}\) away?

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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