/*! 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 21 A source \(x\) of unknown freque... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 9: Problem 21

A source \(x\) of unknown frequency produces 8 beats with a source of \(250 \mathrm{~Hz}\) and 12 beats with a source of \(270 \mathrm{~Hz}\). The frequency of source \(x\) is (A) \(258 \mathrm{~Hz}\) (B) \(242 \mathrm{~Hz}\) (C) \(262 \mathrm{~Hz}\) (D) \(282 \mathrm{~Hz}\)

Short Answer

Expert verified
The frequency of source $x$ is 258 Hz, which corresponds to option (A).

Step by step solution

01

Write down the given information and the beat frequency formula

Two sources, x of unknown frequency and a known frequency source (250 Hz and 270 Hz), produce 8 and 12 beats respectively. The beat frequency formula is given by: Beat frequency = |f1 - f2| where f1 and f2 are the frequencies of the two sources that are combined.
02

Set up two equations for the unknown frequency

Using the beat frequency formula for each pair of sources we have: 1) |x - 250 Hz| = 8 Hz 2) |x - 270 Hz| = 12 Hz
03

Solve the system of equations

We have two cases to solve for both equations. Case 1: x > 250 Hz for equation 1, and x > 270 Hz for equation 2. 1a) x - 250 Hz = 8 Hz -> x = 258 Hz 2a) x - 270 Hz = 12 Hz -> x = 282 Hz Since we got two different answers in this case, this isn't the right solution. Case 2: x < 250 Hz for equation 1, and x < 270 Hz for equation 2. 1b) 250 Hz - x = 8 Hz -> x = 242 Hz 2b) 270 Hz - x = 12 Hz -> x = 258 Hz In this case, we still have two different answers, but we need only one of these frequencies to satisfy both equations. Since there is no unique solution in which x > 250 Hz and x > 270 Hz or x < 250 Hz and x < 270 Hz, let's try a mixed case. Case 3: x > 250 Hz for equation 1, and x < 270 Hz for equation 2. 1c) x - 250 Hz = 8 Hz -> x = 258 Hz 2c) 270 Hz - x = 12 Hz -> x = 258 Hz Now we have a unique solution, where both equations give us the same frequency value for x. We can conclude that source x has a frequency of 258 Hz, which corresponds to option (A).

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!

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

The displacement of a particle varies according to the relation. \(x=4(\cos \pi t+\sin \pi t)\) The amplitude of the particle is (A) \(-4\) (B) 4 (C) \(4 \sqrt{2}\) (D) 8

Change in temperature of the medium changes (A) frequency of sound waves. (B) amplitude of sound waves. (C) wavelength of sound waves. (D) loudness of sound waves.

A string of mass \(2.5 \mathrm{~kg}\) is under tension of \(200 \mathrm{~N}\). The length of the stretched string is \(20.0 \mathrm{~m}\). If the transverse jerk is struck at one end of the string, the disturbance will reach the other end in (A) \(1 \mathrm{~s}\) (B) \(0.5 \mathrm{~s}\) (C) \(2 \mathrm{~s}\) (D) Data given is insufficient

A \(3.6 \mathrm{~m}\) long vertical pipe is filled completely with a liquid. A small hole is drilled at the base of the pipe due to which liquids starts leaking out. This pipe resonates with a tuning fork. The first two resonances occur when height of water column is \(3.22 \mathrm{~m}\) and \(2.34 \mathrm{~m}\), respectively. The area of cross-section of pipe is (A) \(25 \pi \mathrm{cm}^{2}\) (B) \(100 \pi \mathrm{cm}^{2}\) (C) \(200 \pi \mathrm{cm}^{2}\) (D) \(400 \pi \mathrm{cm}^{2}\)

Two sources of sound are moving in opposite directions with velocities \(v_{1}\) and \(v_{2}\left(v_{1}>v_{2}\right)\). Both are moving away from a stationary observer. The frequency of both the source is \(1700 \mathrm{~Hz}\). What is the value of \(\left(v_{1}-v_{2}\right)\) so that the beat frequency observed by the observer is \(10 \mathrm{~Hz} v_{\text {sound }}=340 \mathrm{~m} / \mathrm{s}\) and assume that \(v_{1}\) and \(v_{2}\) both are very much less than \(v_{\text {sound }}\) (A) \(1 \mathrm{~m} / \mathrm{s}\) (B) \(2 \mathrm{~m} / \mathrm{s}\) (C) \(3 \mathrm{~m} / \mathrm{s}\) (D) \(4 \mathrm{~m} / \mathrm{s}\)
See all solutions

Recommended explanations on Physics Textbooks

Fluids

Read Explanation

Circular Motion and Gravitation

Read Explanation

Magnetism

Read Explanation

Geometrical and Physical Optics

Read Explanation

Physical Quantities And Units

Read Explanation

Oscillations

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.