/*! 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 72 A physics professor demonstrates... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 20: Problem 72

A physics professor demonstrates the Doppler effect by tying a \(600 \mathrm{Hz}\) sound generator to a 1.0 -m-long rope and whirling it around her head in a horizontal circle at 100 rpm. What are the highest and lowest frequencies heard by a student in the classroom?

Short Answer

Expert verified
So, the highest frequency heard by the student will be \(635\,Hz\) and the lowest frequency will be \(567\,Hz\).

Step by step solution

01

Calculation of Tangential Velocity

The first step is to calculate the tangential velocity (\(v\)) of the sound source. The tangential velocity can be calculated through the formula \(v = 2\pi r f_s\), where \(r\) is the radius and \(f_s\) is the frequency of rotation. Converting the given frequency of revolution to Hz, \(f_s = 100\,rpm \times (1\,min/60\,s) = 1.67\,Hz\). Substituting these values into the formula, we get \(v = 2\pi \times 1.0\,m \times 1.67\,Hz = 10.47\,m/s\).
02

Calculating Highest and Lowest Frequencies

Then, the highest and lowest frequencies heard by the observer can be calculated using the formula \(f_o = f_s [ (v + v_o)/(v - v_s) ]\) for highest frequency and \(f_o = f_s [ (v - v_o)/(v + v_s) ]\) for lowest frequency, where \(v\) is the velocity of sound, \(v_o\) is the velocity of the observer and \(v_s\) is the velocity of the source. Since the observer is stationary in the classroom, \(v_o = 0\). The velocity of sound in air is \(343\,m/s\). Substituting these values into the formula, we get: for the highest frequency, \(f_o = 600\, Hz [ 343\,m/s / ( 343\,ms^-1 -10.47\,ms^-1 ) ] = 635\,Hz\); for the lowest frequency, \(f_o = 600\,Hz [ 343\,m/s / ( 343\,ms^-1 + 10.47\,ms^-1 ) ] = 567\,Hz\).

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 91影视!

Key Concepts

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

Understanding the Sound Frequency
Sound frequency refers to the number of sound wave cycles that pass a point per second, measured in hertz (Hz). In the exercise given, the source of sound is a sound generator that emits 600 Hz. This is the baseline frequency when there is no relative motion between the source and the observer.
However, due to the Doppler effect, when the source moves, the frequency heard by the observer changes. The movement alters the compression and rarefaction of sound waves, leading to a perceived change in frequency.
- If the source approaches the observer, the frequency increases, resulting in a higher pitch known as the blueshift. - If the source moves away, the frequency decreases, resulting in a lower pitch known as the redshift.
The change in frequency is directly related to the speed at which the source moves relative to the speed of sound.
Decoding Tangential Velocity
Tangential velocity is the linear speed of something moving along a circular path. In our context, it defines the speed at which the sound source moves around the observer. This speed impacts the Doppler shift of the sound frequency.
To calculate the tangential velocity, you use the formula:\[ v = 2\pi r f_s \]where:- \( r \) is the radius of the circle.- \( f_s \) is the frequency of the source's rotation in hertz.
For the given problem, substituting the values \( r = 1.0\,m \) and \( f_s = 1.67\,Hz \) (converted from 100 revolutions per minute), results in a tangential velocity of 10.47 m/s.

This velocity then helps calculate the Doppler effect by adjusting the observed frequency due to the movement of the sound source.
The Role of Velocity of Sound
The velocity of sound is the speed at which sound waves propagate through a medium, usually air. For standard calculations in air, the velocity of sound is approximately 343 m/s. This constant plays a crucial role when applying the Doppler effect equations to find perceived frequency changes.
In a physics problem involving relative motion, like this one, it provides the baseline speed used to determine shifts in sound frequency.
- When the sound source is moving towards an observer, the effective speed of sound waves increases, raising the observed frequency. - Conversely, when it moves away, the speed of the sound waves appearing to the observer decreases, lowering the observed frequency.
This steady value allows us to predict and calculate how sound frequency changes with motion.
Approach to Physics Problem-Solving
Physics problem-solving is about breaking down complex situations into understandable parts and applying known principles. For the Doppler effect problem, you can follow a systematic approach:
  • Identify known quantities, such as the frequency of the sound source and its speed of rotation.
  • Determine relevant physical constants like the velocity of sound in the medium.
  • Break the problem into manageable steps, like calculating tangential velocity first.
  • Apply formulas correctly by substituting in known values to find unknowns, such as observed frequencies.

In this problem, calculating tangential velocity and using it with the Doppler formula provides the solution for the frequency observed by a stationary listener. Problem-solving often involves checking units and ensuring logical consistency.

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

A spherical wave with a wavelength of \(2.0 \mathrm{m}\) is emitted from the origin. At one instant of time, the phase at \(r=4.0 \mathrm{m}\) is \(\pi\) rad. At that instant, what is the phase at \(r=3.5 \mathrm{m}\) and at \(r=4.5 \mathrm{m} ?\)

a. An FM radio station broadcasts at a frequency of 101.3 MHz. What is the wavelength? b. What is the frequency of a sound source that produces the same wavelength in \(20^{\circ} \mathrm{C}\) air?

The wave speed on a string is \(150 \mathrm{m} / \mathrm{s}\) when the tension is 75 N. What tension will give a speed of 180 m/s?

You have just been pulled over for running a red light, and the police officer has informed you that the fine will be \(\$ 250 .\) In desperation, you suddenly recall an idea that your physics professor recently discussed in class. In your calmest voice, you tell the officer that the laws of physics prevented you from knowing that the light was red. In fact, as you drove toward it. the light was Doppler shifted to where it appeared green to you. "OK," says the officer, "Then T'll ticket you for speeding. The fine is \(\$ 1\) for every \(1 \mathrm{km} / \mathrm{hr}\) over the posted speed limit of \(50 \mathrm{km} / \mathrm{hr} .^{\prime \prime}\) How big is your fine? Use 650 \(nm\) as the wavelength of red light and \(540 \mathrm{nm}\) as the wavelength of green light.

Lasers can be used to drill or cut material. One such laser generates a series of high-intensity pulses rather than a continuous beam of light. Each pulse contains \(500 \mathrm{mJ}\) of energy and lasts 10 \(ns\). The laser fires 10 such pulses per second. a. What is the peak power of the laser light? The peak power is the power output during one of the 10 ns pulses. b. What is the average power output of the laser? The average power is the total energy delivered per second. c. Alens focuses the laser beam to a \(10-\mu\) m-diameter circle on the target. During a pulse, what is the light intensity on the target? d. The intensity of sunlight at midday is about \(1100 \mathrm{W} / \mathrm{m}^{2} .\) What is the ratio of the laser intensity on the target to the intensity of the midday sun?
See all solutions

Recommended explanations on Physics Textbooks

Modern Physics

Read Explanation

Fundamentals of Physics

Read Explanation

Force

Read Explanation

Kinematics Physics

Read Explanation

Fluids

Read Explanation

Fields in Physics

Read Explanation
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.