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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 with both ends held fixed is vibrating in its third harmonic. The waves have a speed of \(192 \mathrm{~m} / \mathrm{s}\) and a frequency of \(240 \mathrm{~Hz}\). The amplitude of the standing wave at an antinode is \(0.400 \mathrm{~cm}\). (a) Calculate the amplitude at points on the string a distance of (i) \(40.0 \mathrm{~cm}\) (ii) \(20.0 \mathrm{~cm} ;\) and (iii) \(10.0 \mathrm{~cm}\) from the left end of the string. (b) At each point in part (a), how much time does it take the string to go from its largest upward displacement to its largest downward displacement? (c) Calculate the maximum transverse velocity and the maximum transverse acceleration of the string at each of the points in part (a).

A \(1.50 \mathrm{~m}\) string of weight \(0.0125 \mathrm{~N}\) is tied to the ceiling at its upper end, and the lower end supports a weight \(W .\) Ignore 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 $$ y(x, t)=(8.50 \mathrm{~mm}) \cos (172 \mathrm{rad} / \mathrm{m} x-4830 \mathrm{rad} / \mathrm{s} t) $$ 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 waves traveling down the string?

For a string stretched between two supports, two successive standing-wave frequencies are \(525 \mathrm{~Hz}\) and \(630 \mathrm{~Hz}\). There are other standing-wave frequencies lower than \(525 \mathrm{~Hz}\) and higher than \(630 \mathrm{~Hz}\). If the speed of transverse waves on the string is \(384 \mathrm{~m} / \mathrm{s},\) what is the length of the string? Assume that the mass of the wire is small enough for its effect on the tension in the wire to be ignored.

A \(5.00 \mathrm{~m}, 0.732 \mathrm{~kg}\) wire is used to support two uniform \(235 \mathrm{~N}\) posts of equal length (Fig. P15.53). Assume that the wire is essentially horizontal and that the speed of sound is \(344 \mathrm{~m} / \mathrm{s}\) A strong wind is blowing, causing the wire to vibrate in its 5th overtone. What are the frequency and wavelength of the sound this wire produces?

A large rock that weighs \(164.0 \mathrm{~N}\) is suspended from the lower end of a thin wire that is \(3.00 \mathrm{~m}\) long. The density of the rock is \(3200 \mathrm{~kg} / \mathrm{m}^{3} .\) The mass of the wire is small enough that its effect on the tension in the wire can be ignored. The upper end of the wire is held fixed. When the rock is in air, the fundamental frequency for transverse standing waves on the wire is \(42.0 \mathrm{~Hz}\). When the rock is totally submerged in a liquid, with the top of the rock just below the surface, the fundamental frequency for the wire is \(28.0 \mathrm{~Hz}\). What is the density of the liquid?
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