Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

(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?

Short Answer

Expert verified
(a) Max velocity/acceleration depends on \(x\); (b) Amplitude is \(A\); (c) Time for displacement change is \(\frac{1}{2f}\).

Step by step solution

01

Understanding the Fundamental Mode of Vibration

In the fundamental mode of a string fixed at both ends, the string vibrates with a single antinode in the middle and nodes at both ends. The wavelength \( \lambda \) of the fundamental mode is twice the length of the string. The wave equation for a point \( x \) on the string is given by:\[ y(x, t) = A \sin\left(\frac{2\pi}{\lambda} x \right) \cos(2\pi f t) \]where \( y(x, t) \) is the displacement of the point at position \( x \) at time \( t \).
02

Calculating Maximum Transverse Velocity

The maximum transverse velocity \( v_{\text{max}} \) occurs when the time derivative of the displacement \( y(x, t) \) reaches its maximum. The velocity at point \( x \) is \[ v(x, t) = \frac{\partial y}{\partial t} = -2\pi f A \sin\left(\frac{2\pi}{\lambda} x \right) \sin(2\pi f t) \]Thus, the maximum transverse velocity at each point occurs when \( \sin(2\pi f t) = \pm 1 \), so:\[ v_{\text{max}}(x) = 2\pi f A \left| \sin\left(\frac{2\pi}{\lambda} x \right) \right| \]
03

Calculating Maximum Transverse Acceleration

The maximum transverse acceleration \( a_{\text{max}} \) occurs when the second time derivative of the displacement \( y(x, t) \) reaches its maximum. The acceleration is given by:\[ a(x, t) = \frac{\partial^2 y}{\partial t^2} = - (2\pi f)^2 A \sin\left(\frac{2\pi}{\lambda} x \right) \cos(2\pi f t) \]Thus, the maximum transverse acceleration at each point occurs when \( \cos(2\pi f t) = \pm 1 \), so:\[ a_{\text{max}}(x) = (2\pi f)^2 A \left| \sin\left(\frac{2\pi}{\lambda} x \right) \right| \]
04

Identifying the Amplitude for Each Point

The amplitude of motion at any point \( x \) on the string is the same and equal to \( A \) (the amplitude of the wave), because the entire wave moves uniformly across the string. So, the amplitude at all points \( x = \lambda/2, \lambda/4, \lambda/8 \) is the amplitude \( A \).
05

Determining Time for Largest Displacement Change

The time it takes for the string to go from its largest upward displacement to its largest downward displacement is half the period of the wave. The period \( T \) is the reciprocal of the frequency \( f \):\[ T = \frac{1}{f} \]The time taken for this change in displacement is:\[ \Delta t = \frac{T}{2} = \frac{1}{2f} \]

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!

Key Concepts

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

Transverse Velocity
In wave mechanics, transverse velocity refers to the speed at which a point on the string moves up and down as the wave travels. It's crucial to distinguish this from the wave speed, which is the speed at which the wave propagates along the string.

The formula for transverse velocity is derived from the time derivative of the displacement function, given as:
  • Velocity, \( v(x, t) = \frac{\partial y}{\partial t} = -2\pi f A \sin\left(\frac{2\pi}{\lambda} x \right) \sin(2\pi f t) \)
This expression tells us how fast a point on the string moves perpendicularly to the wave's travel direction. The maximum transverse velocity occurs when \( \sin(2\pi f t) = \pm 1 \), leading to the maximum expression:
  • \( v_{\text{max}}(x) = 2\pi f A \left| \sin\left(\frac{2\pi}{\lambda} x \right) \right| \)
This highlights that the transverse movement of the string depends on both the amplitude \( A \) of the vibration and the frequency \( f \) of the wave.
Transverse Acceleration
Transverse acceleration in the context of string vibrations measures how quickly the velocity of a point on the string changes. Like velocity, it acts perpendicular to the direction of wave propagation.

The maximum transverse acceleration is found by taking the second derivative of the displacement function:
  • Acceleration, \( a(x, t) = \frac{\partial^2 y}{\partial t^2} = - (2\pi f)^2 A \sin\left(\frac{2\pi}{\lambda} x \right) \cos(2\pi f t) \).
The maximum value occurs when \( \cos(2\pi f t) = \pm 1 \):
  • \( a_{\text{max}}(x) = (2\pi f)^2 A \left| \sin\left(\frac{2\pi}{\lambda} x \right) \right| \).
The factors affecting the transverse acceleration are similar to those affecting transverse velocity: the wave's frequency \( f \) and its amplitude \( A \). These tell us how energetic or vigorous the vibration is, influencing how quickly points on the string accelerate during their movement.
String Vibration
String vibration is a fundamental concept in wave mechanics where waves travel along a string fixed at both ends. This brings out standing waves, characterized by specific frequencies and patterns.

These waves are a result of the string's tension and its length, affecting the speed \( v \) at which waves move. In a vibrating string:
  • Nodes: Points that remain stationary, as seen at the ends of the string.
  • Antinodes: Points with maximum displacement, typically found in the middle sections between nodes.
The fundamental mode of vibration is the simplest possible wave on the string, with a single antinode in its center. It establishes the primary frequency of the string as it vibrates.
Fundamental Mode
The fundamental mode represents the simplest standing wave pattern a string can form when vibrated. It is the lowest possible frequency at which the string vibrates naturally.

In this mode:
  • There is only one antinode, lying midway between the two nodes at the ends of the string.
  • The wavelength \( \lambda \) of this mode is precisely twice the length of the string.
  • The frequency of the fundamental mode is the lowest resonance frequency the string can support.
The importance of the fundamental mode lies in its role as the basis for more complex vibrations, making it crucial for understanding harmonics and overtones in musical instruments.
Wavelength
Wavelength is a key parameter in wave mechanics that represents the spatial period of a wave—the distance over which the wave's shape repeats.

For a string vibrating in a standing wave pattern, the wavelength is determined by the number of nodes and antinodes:
  • In the fundamental mode, the wavelength is twice the length of the string.
  • Each subsequent mode of vibration (harmonics) divides the string into more segments, changing the effective wavelength.
Wavelength is associated with frequency \( f \) through the wave speed \( v \):
  • \( v = f\lambda \)
This relationship is crucial for calculating how waves travel on the string, important for both musical acoustics and general wave mechanics.

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 wave on a rope is given by $$y(x, t) = (0.750 \, \mathrm{cm}) \mathrm{cos} \space \pi[(10.400 \, \mathrm{cm}^{-1})x + (250 \mathrm s^{-1})t]$$ (a) Find the amplitude, period, frequency, wavelength, and speed of propagation. (b) Sketch the shape of the rope at these values of \(t:\) 0, 0.0005 s, 0.0010 s. (c) Is the wave traveling in the \(+x-\) or \(-x\)-direction? (d) The mass per unit length of the rope is 0.0500 kg/m. Find the tension. (e) Find the average power of this wave.

At a distance of \(7.00 \times 10^{12}\) m from a star, the intensity of the radiation from the star is 15.4 W/m\(^2\). Assuming that the star radiates uniformly in all directions, what is the total power output of the star?

A jet plane at takeoff can produce sound of intensity 10.0 W/m\(^2\) at 30.0 m away. But you prefer the tranquil sound of normal conversation, which is 1.0 \(\mu\)W/m\(^2\). Assume that the plane behaves like a point source of sound. (a) What is the closest distance you should live from the airport runway to preserve your peace of mind? (b) What intensity from the jet does your friend experience if she lives twice as far from the runway as you do? (c) What power of sound does the jet produce at takeoff?

A transverse sine wave with an amplitude of 2.50 mm and a wavelength of 1.80 m travels from left to right along a long, horizontal, stretched string with a speed of 36.0 m/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 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 transverse velocity of a particle 1.35 m to the right of the origin at time \(t = 0.0625\) s.

A wave on a string is described by \(y(x, t) = A \mathrm{cos}(kx - \omega t)\). (a) Graph \(y, v_y\), and \(a_y\) as functions of \(x\) for time \(t = 0\). (b) Consider the following points on the string: (i) \(x =\) 0; (ii) \(x = \pi/4k\); (iii) \(x = \pi/2k\); (iv) \(x = 3\pi/4k\); (v) \(x = \pi k\); (vi) \(x = 5\pi/4k\); (vii) \(x = 3\pi/2k\); (viii) \(x = 7\pi/4k\). For a particle at each of these points at \(t = 0\), describe in words whether the particle is moving and in what direction, and whether the particle is speeding up, slowing down, or instantaneously not accelerating.

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