/*! 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} Q62P A vertical, \(1.20\;{\rm{m}}\) l... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Q62P (page 501)

A vertical, \(1.20\;{\rm{m}}\) length of 18 gauge (diameter of 1.024 mm) copper wire has a 100.0 N ball hanging from it.

(a) What is the wavelength of the third harmonic for this wire?

(b) A 500.0 N ball now replaces the original ball. What is the change in the wavelength of the third harmonic caused by replacing the light ball with the heavy one? (Hint: Se Table 11.1 for Young鈥檚 modulus.)

Short Answer

Expert verified

Thus, (a) the wavelength of the third harmonic for the wire is 0.8 m

Step by step solution

01

(a) Given in the question

Tension in wire \(F = 100\;{\rm{N}}\).

Length of string \(L = 1.20\;m\)

Diameter of wire \(d = 1.024\;{\rm{mm}}\).

02

Use formula of wavelength

The formula of wavelength is:

\(\lambda = \frac{{2L}}{n}\).

Here,\(L\)length of string and\(n\)is harmonic number.

03

Calculate the wavelength

Solve the wavelength as follows:

\(\begin{array}{c}\lambda = \frac{{2L}}{n}\\ = \frac{2}{3} \times 1.2\\ = 0.8\;m\end{array}\)

Thus, the wavelength of the third harmonic for the wire is 0.8 m.

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

Two loudspeakers, A and B (Fig. E16.35), are driven by the same amplifier and emit sinusoidal waves in phase. Speaker B is 2.00 m to the right of speaker A. Consider point Q along the extension of the line connecting the speakers, 1.00 m to the right of speaker B. Both speakers emit sound waves that travel directly from the speaker to point Q. What is the lowest frequency for which (a) constructive interference occurs at point Q; (b) destructive interference occurs at point Q?

CP A police siren of frequency fsirenis attached to a vibrating platform. The platform and siren oscillate up and down in simple harmonic motion with amplitude APand frequency fp.(a) Find the maximum and minimum sound frequencies that you would hear at a position directly above the siren. (b) At what point in the motion of the platform is the maximum frequency heard? The minimum frequency? Explain.

Two strings of different mass per unit length 渭1and 渭2are tied together and stretched with a tension F. A wave travels along the string and passes the discontinuity in 渭. Which of the following wave properties will be the same on both sides of the discontinuity, and which will change: speed of the wave; frequency; wavelength? Explain the physical reasoning behind each answer.

Question:BIO Ultrasound and Infrasound. (a) Whale communication. Blue whales apparently communicate with each other using sound of frequency 17 Hz, which can be heard nearly 1000 km away in the ocean. What is the wavelength of such a sound in seawater, where the speed of sound is 1531 m/s? (b) Dolphin clicks. One type of sound that dolphins emit is a sharp click of wavelength 1.5 cm in the ocean. What is the frequency of such clicks? (c) Dog whistles. One brand of dog whistles claims a frequency of 25 kHz for its product. What is the wavelength of this sound? (d) Bats. While bats emit a wide variety of sounds, one type emits pulses of sound having a frequency between 39 kHz and 78 kHz. What is the range of wavelengths of this sound? (e) Sonograms. Ultrasound is used to view the interior of the body, much as x rays are utilized. For sharp imagery, the wavelength of the sound should be around one-fourth (or less) the size of the objects to be viewed. Approximately what frequency of sound is needed to produce a clear image of a tumor that is 1.0 mm across if the speed of sound in the tissue is 1550 m/s?

(a) A sound source producing 1.00-kHz waves moves toward a stationary listener at one-half the speed of sound. What frequency will the listener hear? (b) Suppose instead that the source is stationary and the listener moves toward the source at one-half the speed of sound. What frequency does the listener hear? How does your answer compare to that in part (a)? Explain on physical grounds why the two answers differ.
See all solutions

Recommended explanations on Physics Textbooks

Circular Motion and Gravitation

Read Explanation

Oscillations

Read Explanation

Energy Physics

Read Explanation

Atoms and Radioactivity

Read Explanation

Engineering Physics

Read Explanation

Radiation

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.