/*! 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 horizontal string tied at both... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 15: Problem 48

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 accelerationof points located at (i) \(x=\lambda / 2,\) (ii) \(x=\lambda / 4,\) and (ii) \(x=\lambda / 8\) 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?

Short Answer

Expert verified
At each point, the time to go from largest upward to downward displacement is \(\frac{1}{2f}\).

Step by step solution

01

Understanding the problem

We have a vibrating string in its fundamental mode, characterized by the wave speed \(v\), frequency \(f\), amplitude \(A\), and wavelength \(\lambda\). We need to calculate specific properties at three different points: \(x = \lambda/2\), \(x = \lambda/4\), and \(x = \lambda/8\) along the string.
02

Expression of the wave on the string

The displacement of a point on a string can be given by a wave function \( y(x, t) = A \sin(kx) \cos(\omega t) \), where \( k = \frac{2\pi}{\lambda} \) is the wave number, and \( \omega = 2\pi f \) is the angular frequency. To find the velocity and acceleration, we compute the derivatives with respect to time.
03

Compute maximum transverse velocity

The transverse velocity \( v_y \) is given by differentiating \( y(x, t) \) with respect to \( t \): \( v_y = \frac{\partial y}{\partial t} = -A \omega \sin(kx) \sin(\omega t) \). The maximum transverse velocity is \( A \omega |\sin(kx)| \).
04

Calculate maximum velocity at specific points

At \(x=\lambda/2\), \(\sin(kx) = 0\). At \(x=\lambda/4\), \(\sin(kx) = 1\). At \(x=\lambda/8\), \(\sin(kx) = \sin(\pi/4) = \frac{\sqrt{2}}{2}\). The maximum velocity expressions are:- \(x = \lambda/2\): \(0\)- \(x = \lambda/4\): \(A\omega\)- \(x = \lambda/8\): \(\frac{A\omega\sqrt{2}}{2}\)
05

Compute maximum transverse acceleration

The transverse acceleration \( a_y \) is given by the second derivative of \( y(x, t) \): \( a_y = \frac{\partial^2 y}{\partial t^2} = -A \omega^2 \sin(kx) \cos(\omega t) \). The maximum transverse acceleration is \( A \omega^2 |\sin(kx)| \).
06

Calculate maximum acceleration at specific points

Using the same \( \sin(kx) \) values:- \(x = \lambda/2\): \(0\)- \(x = \lambda/4\): \(A \omega^2\)- \(x = \lambda/8\): \(\frac{A \omega^2 \sqrt{2}}{2}\)
07

Determine amplitude of motion

The amplitude of transverse motion at each point (\(x\)) is given by \(A|\sin(kx)|\), which is:- \(x = \lambda/2\): \(0\)- \(x = \lambda/4\): \(A\)- \(x = \lambda/8\): \(\frac{A\sqrt{2}}{2}\)
08

Compute time for motion from largest up to down displacement

For all points, the time to move from largest upward displacement to largest downward displacement is half a period: \(T/2 = \frac{1}{2f}\) since it takes a full period \(T = \frac{1}{f}\) to complete a full up-and-down cycle.

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.

Fundamental Mode
When a string vibrates in its fundamental mode, it is oscillating in the simplest form. In this mode, the string exhibits a single antinode at the center and nodes at both ends. Imagine a jump rope moving up and down in one big arch鈥攖hat's the fundamental mode! In this case, the wavelength \( \lambda \) is twice the length of the string. The frequency \( f \) is the lowest possible frequency at which the string can vibrate. It's like playing the lowest possible note on a string instrument.
The fundamental mode is significant because it helps describe the base behavior of waves in strings, which can then be used to understand more complex modes involving higher frequencies.
Transverse Velocity
Transverse velocity refers to the speed at which a point on the string moves up and down as the wave passes. It is not the speed of the wave itself but rather the speed of the oscillation of the particles of the string. Think of it as how fast each point on the jump rope moves up and down.
To find the maximum transverse velocity, we use the derivative of the wave function with respect to time. This velocity reaches its peak when the sine component of the wave function, \( \sin(kx) \), is maximized.
  • At \( x=\lambda/2 \), transverse velocity is zero because the sine component is zero there.
  • At \( x=\lambda/4 \), it reaches a maximum value of \( A\omega \).
  • At \( x=\lambda/8 \), it is \( \frac{A\omega\sqrt{2}}{2} \).
Transverse Acceleration
Transverse acceleration is the rate of change of the transverse velocity. It tells us how quickly the velocity of the point is changing as the wave moves through. This is important in understanding how force is exerted on the points along the string.
Much like velocity, acceleration is time derivative of transverse velocity, and its maximum is determined by the sine term \( \sin(kx) \) in the second derivative.
  • At \( x=\lambda/2 \), there is no acceleration as the sine component is zero.
  • At \( x=\lambda/4 \), the acceleration is at its maximum, \( A\omega^2 \).
  • At \( x=\lambda/8 \), it is \( \frac{A\omega^2\sqrt{2}}{2} \).
Amplitude of Motion
The amplitude of motion at a particular point on the string indicates how far the point moves from its rest position as the wave passes. In essence, it describes the height of the wave at that point. Where the wave has a peak (antinode), the amplitude is at its greatest.
For a wave in its fundamental mode, the amplitude varies along the length of the string based on \( \sin(kx) \).
  • At \( x=\lambda/2 \), the amplitude is zero, meaning no movement from the rest position.
  • At \( x=\lambda/4 \), the full amplitude \( A \) is realized, showcasing peak displacement.
  • At \( x=\lambda/8 \), the amplitude is \( \frac{A\sqrt{2}}{2} \), indicating moderate displacement.

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 transverse sine wave with an amplitude of 2.50 \(\mathrm{mm}\) and a wavelength of 1.80 \(\mathrm{m}\) travels from left to right along a long, horizontal, stretched string with a speed of 36.0 \(\mathrm{m} / \mathrm{s} .\) Take the origin at the left end of the undisturbed string. At time \(t=0\) the left end of the string has its maximum upward displacement. (a) What are the frequency, angular frequency, and wave number of the wave? (b) What is the function \(y(x, t)\) that describes the wave? (c) What is \(y(t)\) for a particle at the left end of the string? (d) What is \(y(t)\) for a particle 1.35 \(\mathrm{m}\) to the right of the origin? (e) What is the maximum magnitude of transverse velocity of any particle of the string? (f) Find the transverse displacement and the transverss velocity of a particle 1.35 \(\mathrm{m}\) to the right of the origin at time \(t=0.0625 \mathrm{s}\)

Tuning an Instrument. A musician tunes the C-string of her instrument to a fundamental frequency of 65.4 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 \(\mathrm{D} ?\)

Combining Standing Waves. A guitar string of length \(L\) is plucked in such a way that the total wave produced is the sum of the fundamental and the second harmonic. That is, the standing wave is given by $$y(x, t)=y_{1}(x, t)+y_{2}(x, t)$$ where $$\begin{aligned} y_{1}(x, t) &=C \sin \omega_{1} t \sin k_{1} x \\ y_{2}(x, t) &=C \sin \omega_{2} t \sin k_{2} x \end{aligned}$$ with \(\omega_{1}=v k_{1}\) and \(\omega_{2}=v k_{2}\) . (a) At what values of \(x\) are the nodes of \(y_{1} ?\left(\) b) At what values of \(x\) are the nodes of \(y_{2} ?(\mathrm{c})\) Graph \right the total wave at \(t=0, t=\frac{1}{8} f_{1}, t=\frac{1}{4} f_{1}, t=\frac{3}{8} f_{1},\) and \(t=\frac{1}{2} f_{1}\) . (d) Does the sum of the two standing waves \(y_{1}\) and \(y_{2}\) produce a standing wave? Explain.

A water wave traveling in a straight line on a lake is described by the equation $$y(x, t)=(3.75 \mathrm{cm}) \cos \left(0.450 \mathrm{cm}^{-1} x+5.40 \mathrm{s}^{-1} t\right)$$ where \(y\) is the displacement perpendicular to the undisturbed surface of the lake. (a) How much time does it take for one complete wave pattem to go past a fisherman in a boat at anchor, and what horizontal distance does the wave crest travel in that time? (b) What are the wave number and the number of waves per second that pass the fisherman? (c) How fast does a wave crest travel past the fisherman, and what is the maximum speed of his cork floater as the wave causes it to bob up and down?

A 1.50 -m string of weight 0.0125 \(\mathrm{N}\) is tied to the ceiling at its upper end, and the lower end supports a weight \(W\) . Neglect the very small variation in tension along the length of the string that is produced by the weight of the string. When you pluck the string slightly, the waves traveling up the string obey the equation $$y(x, t)=(8.50 \mathrm{mm}) \cos \left(172 \mathrm{m}^{-1} x-4830 \mathrm{s}^{-1} t\right)$$ Assume that the tension of the string is constant and equal to \(W\) . (a) How much time does it take a pulse to travel the full length of the string? (b) What is the weight \(W ?\) (c) How many wavelengths are on the string at any instant of time? (d) What is the equation for wayes traveling down the string?
See all solutions

Recommended explanations on Physics Textbooks

Nuclear Physics

Read Explanation

Magnetism and Electromagnetic Induction

Read Explanation

Fields in Physics

Read Explanation

Energy Physics

Read Explanation

Waves Physics

Read Explanation

Atoms and Radioactivity

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.