Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 12: Problem 39

A source of sound waves of frequency \(1.0 \mathrm{kHz}\) is stationary. An observer is traveling at 0.50 times the speed of sound. (a) What is the observed frequency if the observer moves toward the source? (b) Repeat if the observer moves away from the source instead.

Short Answer

Expert verified
Answer: The observed frequencies are 1.5 kHz when the observer moves towards the source, and 0.5 kHz when the observer moves away from the source.

Step by step solution

01

Identify the Doppler Effect Formula

The Doppler effect formula for sound waves is given by: $$f_{obs} = f_{src} \frac{v \pm v_{obs}}{v \mp v_{src}},$$ where \(f_{obs}\) is the observed frequency, \(f_{src}\) is the source frequency, \(v\) is the speed of sound, \(v_{obs}\) is the observer's speed, and \(v_{src}\) is the source's speed. In our case, the source is stationary, so \(v_{src} = 0\). The observer's speed is 0.50 times the speed of sound, which can be represented as \(v_{obs} = 0.50v\). #Step 2: Calculate Observed Frequency when the Observer moves toward the source#
02

Calculate observed frequency when the observer moves toward the source

When the observer moves toward the source, we use the formula with the plus sign for the numerator and the minus sign for the denominator. Plugging in the given values, we get: $$f_{obs1} = (1.0 \mathrm{kHz}) \frac{v + 0.50v}{v - 0} = (1.0 \mathrm{kHz}) \frac{1.50v}{v}$$ We can simplify the expression: $$f_{obs1} = 1.5 \times (1.0 \mathrm{kHz}) = 1.5 \mathrm{kHz}$$ So, the observed frequency when the observer moves toward the source is 1.5 kHz. #Step 3: Calculate Observed Frequency when the Observer moves away from the source#
03

Calculate observed frequency when the observer moves away from the source

When the observer moves away from the source, we use the formula with the minus sign for the numerator and the plus sign for the denominator. Plugging in the given values, we get: $$f_{obs2} = (1.0 \mathrm{kHz}) \frac{v - 0.50v}{v + 0} = (1 \mathrm{kHz}) \frac{0.50v}{v}$$ We can simplify the expression: $$f_{obs2} = 0.5 \times (1.0 \mathrm{kHz}) = 0.5 \mathrm{kHz}$$ So, the observed frequency when the observer moves away from the source is 0.5 kHz. To summarize, the observed frequencies for the given situation are: (a) 1.5 kHz when the observer moves towards the source, and (b) 0.5 kHz when the observer moves away from the source.

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

An auditorium has organ pipes at the front and at the rear of the hall. Two identical pipes, one at the front and one at the back, have fundamental frequencies of \(264.0 \mathrm{Hz}\) at \(20.0^{\circ} \mathrm{C} .\) During a performance, the organ pipes at the back of the hall are at $25.0^{\circ} \mathrm{C},\( while those at the front are still at \)20.0^{\circ} \mathrm{C} .$ What is the beat frequency when the two pipes sound simultaneously?
A violin is tuned by adjusting the tension in the strings. Brian's A string is tuned to a slightly lower frequency than Jennifer's, which is correctly tuncd to \(440.0 \mathrm{Hz}\) (a) What is the frequency of Brian's string if beats of \(2.0 \mathrm{Hz}\) are heard when the two bow the strings together? (b) Does Brian need to tighten or loosen his A string to get in tune with Jennifer? Explain.
(a) Show that if \(I_{2}=10.01_{1},\) then $\beta_{2}=\beta_{1}+10.0 \mathrm{dB} .$ (A factor of 10 increase in intensity corresponds to a 10.0 dB increase in intensity level.) (b) Show that if \(I_{2}=2.0 I_{1}\) then \(\beta_{2}=\beta_{1}+3.0 \mathrm{dB} .\) (A factor of 2 increase in intensity corresponds to a 3.0 -dB increase in intensity level. tutorial: decibels)
A 30.0 -cm-long string has a mass of \(0.230 \mathrm{g}\) and is vibrating at its next-to-lowest natural frequency \(f_{2} .\) The tension in the string is \(7.00 \mathrm{N} .\) (a) What is \(f_{2} ?\) (b) What are the frequency and wavelength of the sound in the surrounding air if the speed of sound is $350 \mathrm{m} / \mathrm{s} ?$
(a) What should be the length of an organ pipe, closed at one end, if the fundamental frequency is to be \(261.5 \mathrm{Hz} ?\) (b) What is the fundamental frequency of the organ pipe of part (a) if the temperature drops to \(0.0^{\circ} \mathrm{C} ?\)
See all solutions

Recommended explanations on Physics Textbooks

Rotational Dynamics

Read Explanation

Torque and Rotational Motion

Read Explanation

Turning Points in Physics

Read Explanation

Scientific Method Physics

Read Explanation

Modern Physics

Read Explanation

Kinematics 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.

Sign-up for free