\(1.50\;{\rm{m}}\)-long rope is stretched between two supports with a tension that makes the speed of transverse waves \(62.0\;{{\rm{m}} \mathord{\left/ {\vphantom {{\rm{m}} {\rm{s}}}} \right. \\} {\rm{s}}}\). What are the wavelength and frequency of

(a) the fundamental;

(b) the second overtone;

(c) the fourth harmonic?

Short Answer

Expert verified

(a) The wavelength and frequency of the fundamental are, \(3.0\;{\rm{m}}\) and \(20.67\;{\rm{Hz}}\).

Step by step solution

01

Identification of the given data

The given data can be listed below as,

  • The length of the rope is, \(L = 1.50\;{\rm{m}}\).
  • The value of \(n\) is, 1.
  • The speed of transverse wave is, \(v = 62.0\;{{\rm{m}} \mathord{\left/ {\vphantom {{\rm{m}} {\rm{s}}}} \right. \\} {\rm{s}}}\).
02

Concept

The relationship between frequency, wavelength, and speed as well as the condition for harmonics are the major concepts employed to address this problem. Use the fundamental tone condition to determine wavelength first, and then the relationship between frequency, wavelength, and speed to determine frequency.

The following expression is used to determine the wavelength of various tones:

\(\lambda=\frac{{2L}}{n}\) …(1)

Here\(\lambda \)is the wavelength,\(L\)is the length, and\(n\)is the number of tone.

03

Determination of the wavelength and frequency of the fundamental

The wavelength and frequency of the fundament is expressed as,

The wavelength of various tones is calculated using the expression as follows:

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

Substitute \(1.50\;{\rm{m}}\) for \(L\) and 1 for \(n\) in the above equation (1)

\(\begin{array}{c}\lambda = \frac{{2 \times 1.50\;{\rm{m}}}}{1}\\\lambda = 3.0\;{\rm{m}}\end{array}\)

Hence the wavelength is, \(3.0\;{\rm{m}}\).

The relation between frequency, wavelength, and speed is expressed as,

\(f=\frac{v}{\lambda}\) …(2)

Here \(f\) is the frequency, \(v\) is the speed of transverse wave and \(\lambda \) is wavelength.

Substitute \(62.0\;{{\rm{m}} \mathord{\left/ {\vphantom {{\rm{m}} {\rm{s}}}} \right. \\} {\rm{s}}}\) for \(v\) and \(3.0\;{\rm{m}}\) for \(\lambda \) in the above equation (2)

\(\begin{array}{c}f = \frac{{62.0\;{{\rm{m}} \mathord{\left/ {\vphantom {{\rm{m}} {\rm{s}}}} \right. \\} {\rm{s}}}}}{{3.0\;{\rm{m}}}}\\f = 20.67\;{\rm{Hz}}\end{array}\)

Hence the frequency is, \(20.67\;{\rm{Hz}}\).

Therefore, the wavelength and frequency of the fundamental are, \(3.0\;{\rm{m}}\) and \(20.67\;{\rm{Hz}}\).

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 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

Two small stereo speakers are driven in step by the same variable-frequency oscillator. Their sound is picked up by a microphone. For what frequencies does their sound at the speakers produce (a) constructive interference and (b) destructive interference?

The motors that drive airplane propellers are, in some cases, tuned by using beats. The whirring motor produces a sound wave having the same frequency as the propeller. (a) If one single-bladed propeller is turning at 575rpm and you hear 2Hz beats when you run the second propeller, what are the two possible frequencies (in rpm) of the second propeller? (b) Suppose you increase the speed of the second propeller slightly and find that the beat frequency changes to 2.1Hz. In part (a), which of the two answers was the correct one for the frequency of the second single-bladed propeller? How do you know?

A baby’s mouth is 30 cm from her father’s ear and 1.50 m from her mother’s ear. What is the difference between the sound intensity levels heard by the father and by the mother?

Two loudspeakers, Aand B, are driven by the same amplifier and emit sinusoidal waves in phase. The frequency of the waves emitted by each speaker is 172Hz. You are 8.00​ mfrom role="math" localid="1655809995061" A. What is the closest you can be to Band be at a point of destructive interference?

How fast (as a percentage of light speed) would a star have to be moving so that the frequency of the light we receive from it is 10.0% higher than the frequency of the light it is emitting? Would it be moving away from us or toward us? (Assume it is moving either directly away from us or directly toward us.)

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.

Sign-up for free