/*! 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 48 A sound source \(A\) and a refle... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A sound source \(A\) and a reflecting surface \(B\) move directly toward each other. Relative to the air, the speed of source \(A\) is \(29.9 \mathrm{~m} / \mathrm{s}\), the speed of surface \(B\) is \(65.8 \mathrm{~m} / \mathrm{s}\), and the speed of sound is \(329 \mathrm{~m} / \mathrm{s}\). The source emits waves at frequency \(1200 \mathrm{~Hz}\) as measured in the source frame. In the reflector frame, what are (a) the frequency and (b) the wavelength of the arriving sound waves? In the source frame, what are (c) the frequency and (d) the wavelength of the sound waves reflected back to the source?

Short Answer

Expert verified
The frequency at the reflector is approximately 1385.62 Hz, wavelength at the reflector is 0.237 m, the frequency of reflected sound in the source frame is 1562.75 Hz, and the wavelength of reflected sound is 0.210 m.

Step by step solution

01

- Understanding the Problem and Given Data

Identify the givens: speed of source A, speed of reflective surface B, speed of sound, and frequency emitted by the source.
02

- Determine the Apparent Frequency at the Reflective Surface

Use the Doppler effect formula to calculate the frequency heard by the reflective surface. This involves relative speeds of the source and the reflector.
03

- Doppler Effect Formula

The Doppler effect formula for the scenario where both the source and observer are moving is:\[ f_{b} = f_{s} \frac{v + v_b}{v - v_a} \]where \( f_b \) is the frequency at the reflector, \( f_s \) is the emitted frequency (1200 Hz), \( v \) is the speed of sound (329 m/s), \( v_a \) is the speed of the source (29.9 m/s), and \( v_b \) is the speed of the reflector (65.8 m/s).
04

- Calculation of Apparent Frequency

Substitute the given values into the Doppler effect formula:\[ f_{b} = 1200 \cdot \frac{329 + 65.8}{329 - 29.9} \]Calculate the result to find the frequency at the reflector.
05

- Calculate the Wavelength at the Reflective Surface

Use the formula for wavelength: \( \lambda = \frac{v}{f_{b}} \), where \( v \) is the speed of sound and \( f_{b} \) is the frequency at the reflector calculated in Step 4.
06

- Determine the Frequency of Reflected Sound in the Source Frame

Again use the Doppler effect formula for the reflected sound wave. This time, the source is moving towards the reflector but it will act as the receiver of the reflected frequency.
07

- Doppler Effect Formula for Reflected Sound

The formula now is:\[ f_{r} = f_{b} \frac{v + v_a}{v - v_b} \]where \( f_r \) is the frequency heard by the source, \( f_b \) is the frequency at the reflector, and \( v, v_a, v_b \) are the same as before.
08

- Calculation of Reflected Frequency in Source Frame

Substitute the known values into the formula:\[ f_{r} = f_{b} \cdot \frac{329 + 29.9}{329 - 65.8} \]Calculate the result to find the reflected frequency in the source frame.
09

- Calculate Wavelength of Reflected Sound in Source Frame

Use the formula for wavelength: \( \lambda = \frac{v}{f_r} \) to find the wavelength of reflected sound as heard by 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.

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.

Sound Wave Frequency
Sound wave frequency is the number of sound waves that pass a certain point in one second. It is measured in Hertz (Hz). In this exercise, the sound source emits waves at a frequency of 1200 Hz. This frequency will change depending on the relative motion of the source and the observer due to the Doppler Effect. Understanding the initial frequency is crucial as it will help us determine the apparent frequency at the reflecting surface. To summarize: frequency measures how often the sound waves occur, and in this problem, it starts at 1200 Hz.
Relative Motion
Relative motion is the motion of an object as seen from another object. In the context of the Doppler Effect, relative motion between the sound source and the observer (or reflector) affects the frequency and wavelength of the sound waves detected. In this problem, the source A is moving towards the reflecting surface B at 29.9 m/s, and the reflecting surface B is moving towards the source at 65.8 m/s. The relative speed between these two objects is significant because it determines how the sound waves are perceived. Essentially, the effective speed is the sum of their individual speeds when moving towards each other.
Wavelength Calculation
Wavelength is the distance between successive crests of a wave. It is inversely proportional to frequency, meaning that as frequency increases, wavelength decreases and vice versa. To calculate the wavelength of the sound waves as heard by the reflecting surface, we use the formula: \[ \lambda = \frac{v}{f} \] where \( v \) is the speed of sound, and \( f \) is the frequency at any given point. For instance, if the apparent frequency at the reflecting surface has been calculated, we can substitute it in place of \( f \) and use \( 329 \ \text{m/s} \) for the speed of sound to find the wavelength at that point.
Doppler Effect Formula
The Doppler Effect describes how the frequency of a wave changes for an observer moving relative to the source of the wave. The general formula used for sound waves when both the source and the observer are moving is: \[ f' = f \frac{v + v_o}{v - v_s} \] where \( f' \) is the observed frequency, \( f \) is the emitted frequency, \( v \) is the velocity of sound, \( v_o \) is the velocity of the observer, and \( v_s \) is the velocity of the source. This formula is key to solving problems involving the Doppler Effect as it allows us to calculate the frequency changes due to relative motion. By carefully identifying the velocities of the source and observer/reflector, we can apply this formula to determine the apparent frequency at different points during the motion.
Velocity of Sound
The velocity of sound in a given medium (like air) is the speed at which sound waves travel through that medium. In this exercise, the speed of sound is given as 329 m/s. This value is crucial in both the Doppler Effect formula and the wavelength calculation. The velocity of sound is affected by factors like temperature, humidity, and the medium itself. In most physics problems, a standard value is used unless otherwise specified. For our problem, knowing the velocity of sound allows us to determine how the relative motion between the source and the reflector impacts the frequency and wavelength of the sound waves detected.

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 certain violin string is \(30 \mathrm{~cm}\) long between its fixed ends and has a mass of \(2.0 \mathrm{~g}\). The "open" string (no applied finger) sounds an A note \((440 \mathrm{~Hz})\). (a) To play a \(\mathrm{C}\) note \((523\) \(\mathrm{Hz}\) ), how far down the string must one place a finger? (b) What is the ratio of the wavelength of the string waves required for an A note to that required for a C note? (c) What is the ratio of the wavelength of the sound wave for an A note to that for a \(\mathrm{C}\) note?

The source of a sound wave has a power of \(1.00 \mu \mathrm{W}\). If it is a point source, (a) what is the intensity \(3.00 \mathrm{~m}\) away and (b) what is the sound level in decibels at that distance?

A tuning fork of unknown frequency makes three beats per second with a standard fork of frequency \(384 \mathrm{~Hz}\). The beat frequency decreases when a small piece of wax is put on a prong of the first fork. What is the freauency of this fork?

A tube \(1.20 \mathrm{~m}\) long is closed at one end. A stretched wire is placed near the open end. The wire is \(0.330 \mathrm{~m}\) long and has a mass of \(9.60 \mathrm{~g} .\) It is fixed at both ends and oscillates in its fundamental mode. By resonance, it sets the air column in the tube into oscillation at that column's fundamental frequency. Find (a) that frequency and (b) the tension in the wire.

A well with vertical sides and water at the bottom resonates at \(7.00 \mathrm{~Hz}\) and at no lower frequency. (The air-filled portion of the well acts as a tube with one closed end and one open end.) The air in the well has a density of \(1.10 \mathrm{~kg} / \mathrm{m}^{3}\) and a bulk modulus of \(1.33 \times 10^{5} \mathrm{~Pa}\). How far down in the well is the water surface?

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.