/*! 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 41 A thin, taut string tied at both... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 15: Problem 41

A thin, taut string tied at both ends and oscillating in its third harmonic has its shape described by the equation \(y(x, t)=\) \((5.60 \mathrm{cm}) \sin\) \([(0.0340 \mathrm{rad} / \mathrm{cm}) x]\) sin \([(50.0 \mathrm{rad} / \mathrm{s}) t]\) where the origin is at the left end of the string, the \(x\) -axis is along the string and the \(y\) -axis is perpendicular to the string. (a) Draw a sketch that shows the standing wave pattern. (b) Find the amplitude of the two traveling waves that make up this standing wave. (c) What is the length of the string? (d) Find the wavelength, frequency, period, and speed of the traveling waves. (e) Find the maximum transverse speed of a point on the string. (f) What would be the equation \(y(x, t)\) for this string if it were vibrating in its eighth harmonic?

Short Answer

Expert verified
(a) Sketch shows three antinodes. (b) 2.80 cm. (c) 276.15 cm. (d) 位 = 184.1 cm, f = 7.96 Hz, T = 0.126 s, v = 1466.6 cm/s. (e) 280 cm/s. (f) y(x,t) = 5.60 sin(0.091x) sin(400t).

Step by step solution

01

Identify Amplitude of Traveling Waves

A standing wave is formed by two traveling waves with equal amplitude moving in opposite directions. The total amplitude of the standing wave is twice the amplitude of one traveling wave. Given that the amplitude of the standing wave is 5.60 cm, the amplitude of each traveling wave is half of this, which is 2.80 cm.
02

Find the Length of the String

For a string vibrating in its third harmonic, the length of the string L must be \[ L = \frac{3\lambda}{2} \]where \( \lambda \) is the wavelength of the traveling wave. From the equation given, the wave number \( k = 0.0340 \text{ rad/cm} \), and the wavelength \( \lambda \) is \[ \lambda = \frac{2\pi}{k} = \frac{2\pi}{0.0340} \approx 184.1 \text{ cm}. \] Plugging \( \lambda \) back into the equation gives:\[ L = \frac{3 \times 184.1}{2} \approx 276.15 \text{ cm}. \]
03

Calculate Wavelength, Frequency, Period, and Speed

The wavelength \( \lambda \) is calculated previously as 184.1 cm. The frequency \( f \) of the wave is given by \[ f = \frac{\omega}{2\pi} = \frac{50}{2\pi} \approx 7.96 \text{ Hz}. \] The period \( T \) is the reciprocal of the frequency:\[ T = \frac{1}{f} \approx 0.126 \text{ s}. \] The speed of the wave \( v \) is given by the product of frequency and wavelength:\[ v = f \lambda = 7.96 \times 184.1 \approx 1466.6 \text{ cm/s}. \]
04

Determine Maximum Transverse Speed

The maximum transverse speed of a point on the string is given by \[ v_{max} = A\omega \] where \( A \) is the amplitude of the standing wave (5.60 cm) and \( \omega = 50 \text{ rad/s} \). Substituting these values gives:\[ v_{max} = 5.60 \times 50 = 280 \text{ cm/s}. \]
05

Derive Equation for Eighth Harmonic

For the eighth harmonic, the equation of the wave has the form \[ y(x, t) = A \sin(k x) \sin(\omega t) \] For the eighth harmonic, the wave number \( k \) is such that \[ k = \frac{8 \times \pi}{L}. \] With \( L \approx 276.15 \text{ cm} \), \[ k = \frac{8 \times \pi}{276.15} \approx 0.091 \text{ rad/cm}. \] The frequency \( f \) and angular frequency \( \omega \) would increase proportionally to the harmonic number:\[ \omega = 8 \times 50.0 \text{ rad/s} = 400.0 \text{ rad/s}. \] Therefore, the equation for the eighth harmonic is:\[ y(x,t) = 5.60 \sin(0.091x) \sin(400t).\]

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!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Harmonics
A harmonic is an integral multiple of a fundamental frequency, creating distinct standing wave patterns along a medium, like a string. Each harmonic corresponds to a specific mode of vibration. For example, the third harmonic involves three segments, or half-wavelengths, fitting along the string between its fixed ends.
In the given problem, the string vibrates in its third harmonic, which means the length of the string, denoted as \(L\), relates to the wavelength \(\lambda\) by \(L = \frac{3\lambda}{2}\). This configuration is crucial in determining the string's physical properties, as well as its potential shapes at higher harmonics, such as the eighth harmonic.
  • Higher harmonics involve more segments.
  • Different harmonics affect the frequency and wave speed.
Understanding harmonics is essential for analyzing how the string's length can accommodate various standing wave shapes. It shows the relationship between wavelength and frequency for different modes of vibration.
Amplitude
Amplitude measures the maximum displacement of points on the wave from the equilibrium position. In the context of standing waves, it is related to the height of the nodes and antinodes formed on the string.
In this exercise, the amplitude of the overarching standing wave on the string is given as 5.60 cm. This value represents the maximum height reached by the wave at the antinodes. To find the amplitude of each individual traveling wave contributing to this standing wave, divide the total amplitude by two. Thus, each traveling wave has an amplitude of 2.80 cm.
  • Amplitude is an indicator of wave intensity.
  • It helps determine wave energy distribution.
  • It plays a role in the maximum transverse speed calculation.
Understanding amplitude aids in comprehending the physics of standing waves and how different amplitudes influence wave characteristics.
Wavelength and Frequency
Wavelength \(\lambda\) is the spatial period of the wave, or the distance over which the wave's shape repeats. Frequency \(f\) describes how often these wave cycles occur per second and is measured in hertz (Hz).
In the context of the third harmonic of a standing wave on a string, the wavelength and frequency are tightly interwoven. The wavelength is calculated using the wave number \(k\), where \(\lambda = \frac{2\pi}{k}\). For this problem \(k = 0.0340 \, \text{rad/cm}\), leading to \(\lambda \approx 184.1 \, \text{cm}\).
The frequency is derived from the angular frequency \(\omega\), using \(f = \frac{\omega}{2\pi}\). With \(\omega = 50 \, \text{rad/s}\), the frequency comes to approximately 7.96 Hz. The interplay between wavelength and frequency helps set the foundation for understanding wave behavior. They determine the wave鈥檚 speed on the string, since wave speed \(v\) is the product of wavelength and frequency.
Transverse Wave Speed
The transverse wave speed denotes how quickly the wave's shape or disturbance travels through the medium, here the string. It's the rate which waves cycles propagate along the string.
Using calculated wavelength and frequency from previous sections, the wave speed \(v\) can be determined by the equation \(v = f\lambda\). For this instance, with \(f \approx 7.96\, \text{Hz}\) and \(\lambda \approx 184.1 \, \text{cm}\), the wave speed is calculated to be approximately 1466.6 cm/s.
Furthermore, maximum transverse speed \(v_{max}\), the maximum speed of oscillation of any point on the string, is calculated using the amplitude of the wave and the angular frequency \(\omega\), given by \(v_{max} = A\omega\). In this problem, \(A = 5.60\, \text{cm}\) and \(\omega = 50 \, \text{rad/s}\), leading to a maximum transverse speed of 280 cm/s.
  • Wave speed is vital for timing and synchronization in wave-related applications.
  • Transverse speed details energy transfer along the medium.
Understanding these aspects of wave speed can help students grasp the dynamics of wave motion both theoretically and practically.

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

Tuning an Instrument. A musician tunes the C.string of her instrument to a fundamental frequency of 65.4 \(\mathrm{Hz}\) . The vibrating portion of the string is 0.600 \(\mathrm{m}\) long and has a mass of 14.4 \(\mathrm{g}\) . (a) With what tension must the musician stretch it? (b) What percent increase in tension is needed to increase the frequency from 65.4 \(\mathrm{Hz}\) to 73.4 \(\mathrm{Hz}\) , corresponding to a rise in pitch from \(\mathrm{C}\) to \(\mathrm{D} ?\)

Guitar String. One of the 63.5 -cm-long strings of an ordinary guitar is tuned to produce the note \(\mathbf{B}_{3}\) (frequency 245 \(\mathrm{Hz} )\) when vibrating in its fundamental mode. (a) Find the speed of transverse waves on this string. (b) If the tension in this string is increased by \(1.0 \%,\) what will be the new fundamental frequency of the string? (c) If the speed of sound in the surrounding air is 344 \(\mathrm{m} / \mathrm{s}\) , find the frequency and wavelength of the sound wave produced in the air by the vibration of the \(\mathrm{B}_{3}\) string. How do these compare to the frequency and wavelength of the standing wave on the string?

Three pieces of string, each of length \(L,\) are joined together end to end, to make a combined string of length 3\(L\) . The first piece of string has mass per unit length \(\mu_{1}\) , the second piece has mass per unit length \(\mu_{2}=4 \mu_{1},\) and the third piece has mass per unit length \(\mu_{3}=\mu_{1} / 4\) . (a) If the combined string is under tension \(F\) ,how much time does it take a transverse wave to travel the entire length 3\(L ?\) Give your answer in terms of \(L, F,\) and \(\mu_{1}\) . (b) Does your answer to part (a) depend on the order in which the three pieces are joined together? Explain.

(a) A horizontal string tied at both ends is vibrating in its fundamental mode. The traveling waves have speed \(v,\) frequency\(f,\) amplitude \(A,\) and wavelength \(\lambda\) . Calculate the maximum transverse velocity and maximum transverse acceleration of points located at (i) \(x=\lambda / 2,\) (ii) \(x=\lambda / 4,\) and (iii) \(x=\lambda / 8\) from the left-hand end of the string. (b) At each of the points in part (a), what is the amplitude of the motion? (c) At each of the points in part (a), how much time does it take the string to go from its largest upward displacement to its largest downward displacement?

A sinusoidal transverse wave travels on a string. The string has length 8.00 \(\mathrm{m}\) and mass 6.00 \(\mathrm{g}\) . The wave speed is 30.0 \(\mathrm{m} / \mathrm{s}\) , and the wavelength is 0.200 \(\mathrm{m}\) (a) If the wave is to have an aver- age power of 50.0 \(\mathrm{W}\) , what must be the amplitude of the wave? (b) For this same string, if the amplitude and wavelength are the same as in part (a), what is the average power for the wave if the tension is increased such that the wave speed is doubled?
See all solutions

Recommended explanations on Physics Textbooks

Magnetism

Read Explanation

Fluids

Read Explanation

Wave Optics

Read Explanation

Oscillations

Read Explanation

Classical Mechanics

Read Explanation

Geometrical and Physical Optics

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.