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Chapter 15: Problem 44

(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
The maximum transverse velocity and acceleration for points \(\lambda / 2\), \(\lambda / 4\), and \(\lambda / 8\) from the left-hand end of the string are \(A\omega\) and \(A\omega^2\) respectively. The amplitude of the motion at these positions is \(A\). The time it takes for the string to move from the maximum upward displacement to the maximum downward displacement is \(T/2=1/2f\) regardless of position on the string.

Step by step solution

01

Setup

The wave equation for the string vibrating in its fundamental mode is given by:\(y(x,t) = A \sin(kx - \omega t)\), where \(k = 2\pi / \lambda\) is the wave number, \(\omega = 2\pi f\) is the angular frequency and \(A\) is the amplitude of the wave.
02

Maximum velocity and acceleration

The transverse velocity and acceleration are given by the first and second time-derivatives of \(y\). This gives \(v_y = \frac{dy}{dt} = -A \omega \cos(kx - \omega t)\) and \( a_y = \frac{d^2y}{dt^2} = -A \omega^2 \sin(kx - \omega t)\). The maximum values occur when \(\cos\) or \(\sin\) in these equations is equal to 1 or -1. Thus, the maximum velocity and acceleration are \(v_{max} = A\omega\) and \(a_{max} = A \omega^2\) respectively.
03

Calculate for different locations

Now these results should be applied to the individual parts of the question, for points at \(x=\lambda / 2\), \(x=\lambda / 4\) and \(x=\lambda / 8.\)
04

Amplitude of Motion

The amplitude of the motion part of the wave does not depend on the position of \(x\), so it is equal to the given amplitude \(A\) at all positions.
05

Time Interval Calculation

The string takes a time \(T/2\) to travel from the maximum upward displacement to the maximum downward displacement, where \(T = 1/f\) is the period of the wave. So, regardless of the location on the wave, the string takes a time \(T/2\) to go from maximum upward to maximum downward displacement.

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Key Concepts

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

Transverse Velocity
In wave motion, transverse velocity represents the speed at which a point on the wave moves up and down as the wave passes by. It's crucial in understanding how energy is propagated along the wave. The transverse velocity is derived by taking the first derivative of the wave equation with respect to time.

The equation for transverse velocity in a wave is given by:
  • \( v_y = \frac{dy}{dt} = -A \omega \cos(kx - \omega t) \)
Here:
  • \( A \) is the amplitude of the wave.
  • \( \omega = 2\pi f \) is the angular frequency.
  • \( k = 2\pi / \lambda \) is the wave number.
  • The maximum transverse velocity occurs when the \( \cos(\cdot) \) is equal to 1 or -1, giving \( v_{max} = A \omega \).
Understanding how fast points on a wave can move helps in applications ranging from music to broadcasting and more.
Transverse Acceleration
Transverse acceleration is the rate of change of the transverse velocity over time, essentially showing how quickly a part of the wave is speeding up or slowing down. You determine it by taking the second time-derivative of the wave equation.

The equation looks like this:
  • \( a_y = \frac{d^2y}{dt^2} = -A \omega^2 \sin(kx - \omega t) \)
A few points to consider:
  • \( A \) is still the amplitude of the wave.
  • \( \omega^2 \) significantly impacts the acceleration.
  • Maximum transverse acceleration (\( a_{max} \)) happens when \(\sin(\cdot)\) equals 1 or -1, resulting in \( a_{max} = A \omega^2 \).
The concept of transverse acceleration is important as it gives insight into the forces acting on the particles of the medium through which the wave propagates.
Wave Equation
The wave equation encapsulates how waves move through a medium, and it's vital for understanding wave mechanics. For a string vibrating in its fundamental mode, the wave equation is expressed as:
  • \( y(x,t) = A \sin(kx - \omega t) \)
Key elements include:
  • \( A \) - the amplitude, indicating the maximum displacement from equilibrium.
  • \( k = 2\pi / \lambda \) - the wave number, indicating how many wave cycles exist over a unit distance.
  • \( \omega = 2\pi f \) - the angular frequency in radians per second.
  • \( y(x,t) \) - the displacement as a function of position \(x\) and time \(t\).
This equation helps us understand how wave phenomena like sound, light, and water waves move through various environments.
Amplitude of Motion
Amplitude is one of the fundamental characteristics of waves, indicating how much a medium is displaced by a wave. In the context of motion on a vibrating string, amplitude refers to the maximum extent of a vibration or oscillation, measured from the position of equilibrium.

For standing waves, like those in a vibrating string, amplitude remains constant regardless of position along the wave. Thus, no matter where you measure on the string, it will always be equal to \( A \), the given amplitude.
  • Amplitude is crucial as it determines the energy carried by the wave. Higher amplitude means more energy.
  • Understanding amplitude helps in practical applications such as calibrating instruments and engineering machinery susceptible to oscillations.
By knowing the amplitude, we can predict wave behavior without having to measure every single point.
Time Interval Calculation
Understanding the time interval needed for a string to move from its maximum upward displacement to its maximum downward displacement is essential for analyzing wave timing.

This time interval can be tied to the wave's period \(T\), which is the time for a complete cycle of the wave. For a string under harmonic motion, this journey takes half of its period:
  • The formula is: \( T/2 \)
  • Where \( T = 1/f \)
Regardless of where you look at the string along its length, the transition time remains the same.
  • This uniformity helps in creating synchronized systems where timing precision is mandatory.
  • In practice, knowing \( T/2 \) ensures efficient designing and testing of equipment responding to wave-like motion.

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Most popular questions from this chapter

A string or rope will break apart if it is placed under too much tensile stress [see Eq. (11.8)]. Thicker ropes can withstand more tension without breaking because the thicker the rope, the greater the cross-sectional area and the smaller the stress. One type of steel has density \(7800 \mathrm{~kg} / \mathrm{m}^{3}\) and will break if the tensile stress exceeds \(7.0 \times 10^{8} \mathrm{~N} / \mathrm{m}^{2}\). You want to make a guitar string from \(4.0 \mathrm{~g}\) of this type of steel. In use, the guitar string must be able to withstand a tension of \(900 \mathrm{~N}\) without breaking. Your job is to determine (a) the maximum length and minimum radius the string can have; (b) the highest possible fundamental frequency of standing waves on this string, if the entire length of the string is free to vibrate.

A strong string of mass \(3.00 \mathrm{~g}\) and length \(2.20 \mathrm{~m}\) is tied to supports at each end and is vibrating in its fundamental mode. The maximum transverse speed of a point at the middle of the string is \(9.00 \mathrm{~m} / \mathrm{s}\) The tension in the string is \(330 \mathrm{~N}\). (a) What is the amplitude of the standing wave at its antinode? (b) What is the magnitude of the maximum transverse acceleration of a point at the antinode?

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 transverse velocity of a particle \(1.35 \mathrm{~m}\) to the right of the origin at time \(t=0.0625 \mathrm{~s}\)

A water wave traveling in a straight line on a lake is described by the equation $$ y(x, t)=(2.75 \mathrm{~cm}) \cos (0.410 \mathrm{rad} / \mathrm{cm} x+6.20 \mathrm{rad} / \mathrm{s} t) $$ where \(y\) is the displacement perpendicular to the undisturbed surface of the lake. (a) How much time does it take for one complete wave pattern 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 horizontal wire is tied to supports at each end and vibrates in its second- overtone standing wave. The tension in the wire is \(5.00 \mathrm{~N}\), and the node-to-node distance in the standing wave is \(6.28 \mathrm{~cm}\). (a) What is the length of the wire? (b) A point at an antinode of the standing wave on the wire travels from its maximum upward displacement to its maximum downward displacement in \(8.40 \mathrm{~ms}\). What is the wire's mass?
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