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(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
Max velocity: 0, \( A \omega, \frac{A \omega \sqrt{2}}{2} \). Max accel: 0, \( A \omega^2, \frac{A \omega^2 \sqrt{2}}{2} \). Amplitude: 0, \( A, \frac{A \sqrt{2}}{2} \). Time: \( \frac{1}{2f} \).

Step by step solution

01

Understanding the waves on the string

The string vibrates in its fundamental mode, which means it forms a standing wave. A standing wave has nodes at both ends, antinodes in between, and can be described by \( y(x, t) = A \sin(kx) \cos(\omega t) \), where \( k = \frac{2\pi}{\lambda} \) and \( \omega = 2\pi f \). The amplitude at any point \( x \) is \( A \sin(kx) \).
02

Calculating maximum transverse velocity

The maximum transverse velocity occurs when the cosine function is maximum, which is 1. The transverse velocity is given by taking the time derivative of \( y(x, t) \), resulting in \( v_y = \frac{\partial y}{\partial t} = -A \omega \sin(kx) \sin(\omega t) \). The maximum value is \( A \omega \sin(kx) \) when \( \sin(\omega t) = 1 \).
03

Calculating maximum transverse acceleration

The maximum transverse acceleration is the maximum value of the time derivative of the transverse velocity, given by \( a_y = \frac{\partial^2 y}{\partial t^2} = -A \omega^2 \sin(kx) \cos(\omega t) \). The maximum value is \( A \omega^2 \sin(kx) \) when \( \cos(\omega t) = 1 \).
04

Calculating maximum transverse velocity and acceleration for given positions

For \( x = \frac{\lambda}{2} \), \( \sin(kx) = 0 \) so the maximum transverse velocity and acceleration are 0. For \( x = \frac{\lambda}{4} \), \( \sin(kx) = 1 \) so both maximum transverse velocity and acceleration are \( A \omega \) and \( A \omega^2 \) respectively. For \( x = \frac{\lambda}{8} \), \( \sin(kx) = \frac{\sqrt{2}}{2} \), resulting in \( \frac{A \omega \sqrt{2}}{2} \) for velocity and \( \frac{A \omega^2 \sqrt{2}}{2} \) for acceleration.
05

Determining amplitude of motion

The amplitude of motion at any point on the string is given by \( A \sin(kx) \): 0 at \( x = \frac{\lambda}{2} \), \( A \) at \( x = \frac{\lambda}{4} \), and \( \frac{A \sqrt{2}}{2} \) at \( x = \frac{\lambda}{8} \).
06

Calculating the time for largest displacement change

The time taken to move from the largest upward to the largest downward displacement is half of the wave's period, \( \frac{T}{2} = \frac{1}{2f} \), because this corresponds to half an oscillation.

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Key Concepts

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

Fundamental Mode
In wave motion, the fundamental mode refers to the simplest standing wave pattern that can be formed on a string. It's the lowest frequency at which a standing wave can exist, forming between the fixed ends of the string. In this mode, the string oscillates in a single segment with one antinode in the middle and nodes at both ends. The wavelength of the fundamental mode is actually twice the length of the string. This fundamental frequency establishes the basis frequency from which all harmonic modes are derived. When a string is vibrating in its fundamental mode, it is vibrating at its natural frequency, which ensures maximum energy transfer and minimal loss of energy.
Standing Waves
Standing waves occur when two waves traveling in opposite directions superpose to produce an apparent stationary wave. This is a common occurrence on strings tied at both ends. The points of zero amplitude, known as nodes, are where the waves cancel each other out. Meanwhile, points of maximum amplitude, known as antinodes, signify constructive interference. These waves do not transfer energy along the medium like a traveling wave but rather store energy within specific segments. This phenomenon is essential in musical instruments, where specific harmonics are desired. Standing waves are fundamental in shaping the sound and dynamics of stringed instruments, as the vibration modes determine the overtones and quality of the sound produced.
Transverse Velocity
The transverse velocity in wave motion refers to how fast a point on the medium, like a string, moves perpendicular to its rest position. It's an integral part of analyzing wave behavior because it describes the speed of the individual particles on the string moving up and down. The maximum transverse velocity occurs when the displacement of the string from its equilibrium position is zero, and it's the highest point-to-point speed the wave reaches. Mathematically, it's determined as the derivative of the wave equation with respect to time. Thus, it reaches its maximum when the temporal wave factor reaches its peak, i.e., when \(\sin(\omega t) = 1\).
  • At positions like \(x = \frac{\lambda}{2}\), the transverse velocity is zero.
  • Conversely, at \(x = \frac{\lambda}{4}\), the velocity reaches its maximum, \(A \omega\).
  • Intermediate positions, such as \(x = \frac{\lambda}{8}\), result in a reduced maximum velocity \(\frac{A \omega \sqrt{2}}{2}\).
Amplitude of Motion
Amplitude represents the maximum displacement of particles in the wave from their rest position. In standing waves, the amplitude differs along the length of the string, depending on the position between the nodes and antinodes. At the nodes, the amplitude is zero, meaning no up-and-down motion occurs. The maximum amplitude occurs at the antinodes, where the wave swings fully up and down.Understanding the amplitude helps in understanding the energy distribution along the wave, as energy in a wave is proportional to the square of the amplitude. For any point on the string, the amplitude \(A \sin(kx)\) indicates the maximum motion experienced by the point.
  • For example, at \(x = \frac{\lambda}{2}\), \(\sin(kx) = 0\), yielding zero amplitude.
  • At \(x = \frac{\lambda}{4}\), the amplitude is maximized to \(A\)
  • At \(x = \frac{\lambda}{8}\), it achieves a fraction, \(\frac{A \sqrt{2}}{2}\), indicating moderate motion.

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Most popular questions from this chapter

A uniform rope with length \(L\) and mass \(m\) is held at one end and whirled in a horizontal circle with angular velocity \(\omega\) . You can ignore the force of gravity on the rope. Find the time required for a transverse wave to travel from one end of the rope to the other.

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?

A wave on a string is described by \(y(x, t)=\) \(A \cos (k x-\omega t) \cdot(a)\) 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) \(x=\pi / 4 k ;\) (iii) \(x=\pi / 2 k ;\) (iv) \(x=3 \pi / 4 k ;\) (v) \(x=\pi / k\) (vi) \(x=5 \pi / 4 k ;\) (vii) \(x=3 \pi / 2 k ;\) (viii) \(x=7 \pi / 4 k\) . 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.

Energy in a Triangular Pulse. A triangular wave pulse on a taut string travels in the positive \(x\) -direction with speed \(v\) . The tension in the string is \(F,\) and the linear mass density of the string is \(\mu .\) At \(t=0,\) the shape of the pulse is given by $$ y(x, 0)=\left\\{\begin{array}{ll}{0} & {\text { if } x<-L} \\ {h(L+x) / L} & {\text { for }-LL}\end{array}\right. $$ (a) Draw the pulse at \(t=0 .\) (b) Determine the wave function \(y(x, t)\) at all times \(t\) (c) Find the instantancous power in the wave. Show that the power is zero except for \(-L<(x-v t)

One string of a certain musical instrument is 75.0 \(\mathrm{cm}\) long and has a mass of 8.75 \(\mathrm{g}\) . It is being played in a room where the speed of sound is 344 \(\mathrm{m} / \mathrm{s}\) . (a) To what tension must you adjust the string so that, when vibrating in its second overtone, it produces sound of wavelength 3.35 \(\mathrm{cm} ?\) (b) What frequency sound does this string produce in its fundamental mode of vibration?

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