/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 55 A whistle of frequency \(540 \ma... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A whistle of frequency \(540 \mathrm{~Hz}\) moves in a circle of radius \(60.0 \mathrm{~cm}\) at an angular speed of \(15.0 \mathrm{rad} / \mathrm{s}\). What are the (a) lowest and (b) highest frequencies heard by a listener a long distance away, at rest with respect to the center of the circle?

Short Answer

Expert verified
Lowest frequency: 526.2 Hz, Highest frequency: 554.5 Hz.

Step by step solution

01

Understanding the Problem

We are asked to find the lowest and highest frequencies heard by a stationary observer from a whistle moving in circular motion. The Doppler Effect will cause the frequency observed to change based on the motion of the whistle source relative to the listener.
02

Analyze the Doppler Effect

The frequency observed by a listener due to circular motion of a source is affected by the Doppler Effect. The frequency at the observer is given by the formula \[ f' = \frac{f}{1 \pm \frac{v_s}{v}} \]where \( f \) is the source frequency, \( v_s \) is the speed of the source, and \( v \) is the speed of sound.
03

Calculate the Speed of the Source

The speed of the source \( v_s \) moving in a circular path is computed using the formula for linear velocity: \[ v_s = r \cdot \omega \]where \( r = 0.6 \text{ m} \) is the radius, and \( \omega = 15 \text{ rad/s} \) is the angular speed. Substituting the values gives:\[ v_s = 0.6 \cdot 15 = 9 \text{ m/s} \]
04

Determine Speed of Sound

The speed of sound in air at room temperature (about 20°C) is approximately \( v = 343 \text{ m/s} \).
05

Calculate Lowest Observed Frequency

The lowest frequency occurs when the source is moving directly away from the observer, using \[ f'_{min} = \frac{f}{1 + \frac{v_s}{v}} \]Substitute in the known values:\[ f'_{min} = \frac{540}{1 + \frac{9}{343}} \approx \frac{540}{1.026} \approx 526.2 \text{ Hz} \]
06

Calculate Highest Observed Frequency

The highest frequency occurs when the source is moving directly towards the observer, using \[ f'_{max} = \frac{f}{1 - \frac{v_s}{v}} \]Substitute in the known values:\[ f'_{max} = \frac{540}{1 - \frac{9}{343}} \approx \frac{540}{0.974} \approx 554.5 \text{ Hz} \]
07

Final Results

Based on the calculations, the lowest frequency heard by the listener is approximately \(526.2 \text{ Hz}\), and the highest frequency is approximately \(554.5 \text{ Hz}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Circular Motion
Circular motion occurs when an object moves in a path where its distance from a certain point remains constant. This point is known as the center of the circle. Imagine a tetherball tied to a pole and swinging around the stick. It keeps moving around the pole in a circle. For the movement to stay circular, the ball experiences a force directed towards the center of the circle, often called the centripetal force.
Understanding circular motion requires familiarizing oneself with certain key terms:
  • **Radius ( )**: The distance from the center of the circle to any point on its circumference. In our exercise, the whistle is moving in a circle with a radius of 0.6 meters.
  • **Angular Speed ( ext{ω})**: This is how fast the object travels around the circle. It's measured in radians per second, which describes the angle covered over time as the object travels along the circle. In the exercise, the given angular speed is 15 rad/s.
  • **Linear Speed ( ext{v}_s)**: This is the speed that depicts how fast the object moves along the actual path of the circle. It can be found by multiplying the radius by the angular speed: ext{v}_s = r imes ext{ω}.
By calculating with these terms, we find that the whistle has a linear speed of 9 m/s while undergoing circular motion.
Frequency Modulation
Frequency modulation (FM) occurs when the frequency of a wave changes in correspondence to certain stimulus or changes. In our context, the stimulus is the changing position of the moving whistle relative to the listener. The Doppler Effect is responsible for this kind of modulation.
The Doppler Effect describes how the frequency of a wave (like sound) changes for an observer moving relative to the source of the wave:
  • **When the source approaches the observer:** The sound waves get "squished" together, making the observer perceive a higher frequency.
  • **When the source moves away from the observer:** The waves get "stretched" out, and the observer perceives a lower frequency.
The key formula capturing this change is given by: \[ f' = \frac{f}{1 \pm \frac{v_s}{v}} \]where \( f \) represents the source's original frequency, \( v_s \) is the speed of the source (whistle, in our case), and \( v \) is the speed of sound.
By plugging in these values, we determine the frequencies that can be heard by a distant observer depending on the whistle's movement. The change in frequency we've calculated occurs because of the varying distances between the moving whistle and the listener due to circular motion.
Speed of Sound
The speed of sound is how fast sound waves travel through a medium, such as air, water, or solids. It varies based on the environment and its conditions, such as temperature and pressure. For our calculations, the sound moves through typical air conditions, at approximately 20°C, where it travels roughly at 343 m/s.
This constant speed is crucial for figuring out the effect of any change in the frequency of sound waves due to movement. When dealing with problems involving the Doppler Effect, it is essential to know the speed of sound since it acts as the basis for calculating perceived frequency changes.
The formula for perceived frequency further depends on the ratio of the source's speed to the speed of sound: \[ f'_{min} = \frac{f}{1 + \frac{v_s}{v}} \] \[ f'_{max} = \frac{f}{1 - \frac{v_s}{v}} \] where \( f \) is the original frequency, and \( v_s \) is the speed of the source. These equations allow for calculating the highest and lowest observed frequencies, considering the movement of the sound source.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Two sound waves, from two different sources with the same frequency, \(540 \mathrm{~Hz}\), travel in the same direction at \(330 \mathrm{~m} / \mathrm{s}\). The sources are in phase. What is the phase difference of the waves at a point that is \(4.40 \mathrm{~m}\) from one source and \(4.00 \mathrm{~m}\) from the other?

A point source emits sound waves isotropically. The intensity of the waves \(2.50 \mathrm{~m}\) from the source is \(1.91 \times 10^{-4} \mathrm{~W} / \mathrm{m}^{2}\). Assuming that the energy of the waves is conserved, find the power of the source.

You have five tuning forks that oscillate at close but different resonant frequencies. What are the (a) maximum and (b) minimum number of different beat frequencies you can produce by sounding the forks two at a time, depending on how the resonant frequencies differ?

A girl is sitting near the open window of a train that is moving at a velocity of \(10.00 \mathrm{~m} / \mathrm{s}\) to the east. The girl's uncle stands near the tracks and watches the train move away.The locomotive whistle emits sound at frequency \(500.0 \mathrm{~Hz}\). The air is still. (a) What frequency does the uncle hear? (b) What frequency does the girl hear? A wind begins to blow from the east at \(10.00 \mathrm{~m} / \mathrm{s} .\) (c) What frequency does the uncle now hear? (d) What frequency does the girl now hear?

In pipe \(A\), the ratio of a particular harmonic frequency to the next lower harmonic frequency is \(1.2 .\) In pipe \(B,\) the ratio of a particular harmonic frequency to the next lower harmonic frequency is 1.4. How many open ends are in (a) pipe \(A\) and (b) pipe \(B ?\)

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.