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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
(a) Calculate max velocities as \( A \omega \sin\left(\frac{n\pi}{4}\right) \) for \( n=2,4,8 \). (b) Amplitude at each point is \( A \sin\left(\frac{n\pi}{4}\right) \). (c) Time is \( \frac{1}{2f} \) for each point.

Step by step solution

01

Understanding the Fundamental Mode

When a string is vibrating in its fundamental mode, it has only one antinode and two nodes at the endpoints. The wavelength, \( \lambda \), for the fundamental frequency is twice the length of the string. In this mode, the string can be described by the wave function: \\[ y(x, t) = A \sin\left(\frac{\pi x}{L}\right) \cos(\omega t) \] \where \( L \) is the length of the string, \( A \) is the amplitude, and \( \omega = 2\pi f \) is the angular frequency.
02

Maximum Transverse Velocity

The transverse velocity \( v_y \) of the string at any point \( x \) is given by the partial derivative of the displacement with respect to time: \\[ v_y = \frac{\partial y}{\partial t} = -A \omega \sin\left(\frac{\pi x}{L}\right) \sin(\omega t) \] \The maximum transverse velocity occurs when \( \sin(\omega t) = \pm 1 \): \\[ v_y^{\text{max}} = \pm A \omega \left| \sin\left(\frac{\pi x}{L}\right) \right| \] \Substitute \( \omega = 2\pi f \) for each point.
03

Maximum Transverse Acceleration

The transverse acceleration \( a_y \) is the second partial derivative of the displacement with respect to time: \\[ a_y = \frac{\partial^2 y}{\partial t^2} = -A \omega^2 \sin\left(\frac{\pi x}{L}\right) \cos(\omega t) \] \The maximum transverse acceleration occurs when \( \cos(\omega t) = \pm 1 \): \\[ a_y^{\text{max}} = \pm A \omega^2 \left| \sin\left(\frac{\pi x}{L}\right) \right| \] \Substitute \( \omega = 2\pi f \) for each point.
04

Amplitude of Motion

The amplitude of motion at any point \( x \) on the string in the fundamental mode is given by \( A \left| \sin\left(\frac{\pi x}{L}\right) \right| \). Evaluate this expression at each specific point to determine the amplitude of the motion.
05

Time from Largest Upward to Downward Displacement

The time to go from the largest upward to the largest downward displacement is half a period of the wave. The period \( T \) is given by \( T = \frac{1}{f} \). Thus, the time taken is \( \frac{T}{2} = \frac{1}{2f} \). This value will be the same for all points because it only depends on the frequency.

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Key Concepts

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

Fundamental Mode
In the context of vibrating string waves, the fundamental mode is a key concept. When a string vibrates in this mode, it maintains the simplest pattern of oscillation with only one antinode (point of maximum amplitude) centered and nodes (points of zero amplitude) at both ends.
The string’s vibration can be represented by a wave function: \[ y(x, t) = A \sin\left(\frac{\pi x}{L}\right) \cos(\omega t) \] where:
  • \( A \) is the amplitude.
  • \( L \) is the length of the string.
  • \( \omega = 2\pi f \), with \( f \) being the frequency.
In this mode, the entire length of the string vibrates in phase as a single segment. The wavelength is twice the length of the string \( \lambda = 2L \). This setup creates the most efficient and simple vibration mode, which is why it is termed the 'fundamental'. Understanding this basic mode helps in analyzing more complex oscillations.
Transverse Velocity
The transverse velocity of a vibrating string refers to the speed at which a point on the string moves up and down as the wave passes through. It is important because it reveals how energetic the wave is.
Mathematically, the transverse velocity \( v_y \) is defined as the time derivative of the wave function: \[ v_y = \frac{\partial y}{\partial t} = -A \omega \sin\left(\frac{\pi x}{L}\right) \sin(\omega t) \] The maximum transverse velocity is attained when \( \sin(\omega t) = \pm 1 \): \[ v_y^{\text{max}} = \pm A \omega \left| \sin\left(\frac{\pi x}{L}\right) \right| \] This shows that the maximum transverse velocity is proportional to both the wave's amplitude \( A \) and its angular frequency \( \omega \). Calculating it at different points along the string helps in understanding how energy is distributed across the string.
Transverse Acceleration
Transverse acceleration describes how quickly the velocity of a point on the vibrating string changes. Rapid changes in velocity signify high acceleration and energy transfers.
The transverse acceleration \( a_y \) can be derived as the second time derivative of the wave function: \[ a_y = \frac{\partial^2 y}{\partial t^2} = -A \omega^2 \sin\left(\frac{\pi x}{L}\right) \cos(\omega t) \] The maximum occurs when \( \cos(\omega t) = \pm 1 \): \[ a_y^{\text{max}} = \pm A \omega^2 \left| \sin\left(\frac{\pi x}{L}\right) \right| \] This formula indicates that the maximum acceleration is directly proportional to the amplitude \( A \) and the square of the angular frequency \( \omega^2 \). It highlights how different points along the string experience varying accelerations though rooted in the same fundamental mode.
Wave Function
The wave function is a fundamental concept in this discussion. It mathematically describes the displacement of points along the string over time due to the wave. Structured as: \[ y(x, t) = A \sin\left(\frac{\pi x}{L}\right) \cos(\omega t) \] This equation shows:
  • The amplitude \( A \) affects how high the wave peaks.
  • The factor \( \sin\left(\frac{\pi x}{L}\right) \) determines the structure of the wave across the string length, indicating nodes and antinodes.
  • \( \cos(\omega t) \) governs the wave's time-dependent oscillation.
The wave function is vital because it encapsulates all the information about the vibrating string, letting us predict the motion at any point \( x \) and time \( t \). It serves as a blueprint for the vibrational behavior.
Amplitude of Motion
Amplitude of motion shows the vertical displacement a point on the string achieves during its vibration. It reflects the energy of the wave and determines the maximum extent to which the string moves from its rest position.
Specifically, the amplitude of motion at point \( x \) in the fundamental mode is given by: \[ A \left| \sin\left(\frac{\pi x}{L}\right) \right| \] This equation reveals:
  • How amplitude varies with position \( x \) along the string.
  • It is maximum at the antinode (center of the string) and zero at the nodes (ends of the string).
Understanding the amplitude of motion is crucial as it conveys the oscillation range available and sheds light on the distribution of energy along the string.

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Most popular questions from this chapter

A continuous succession of sinusoidal wave pulses are produced at one end of a very long string and travel along the length of the string. The wave has frequency 70.0 Hz, amplitude 5.00 mm, and wavelength 0.600 m. (a) How long does it take the wave to travel a distance of 8.00 m along the length of the string? (b) How long does it take a point on the string to travel a distance of 8.00 m, once the wave train has reached the point and set it into motion? (c) In parts (a) and (b), how does the time change if the amplitude is doubled?

In your physics lab, an oscillator is attached to one end of a horizontal string. The other end of the string passes over a frictionless pulley. You suspend a mass \(M\) from the free end of the string, producing tension \(Mg\) in the string. The oscillator produces transverse waves of frequency \(f\) on the string. You don't vary this frequency during the experiment, but you try strings with three different linear mass densities \(\mu\). You also keep a fixed distance between the end of the string where the oscillator is attached and the point where the string is in contact with the pulley's rim. To produce standing waves on the string, you vary \(M\); then you measure the node-to-node distance \(d\) for each standingwave pattern and obtain the following data: (a) Explain why you obtain only certain values of \(d\). (b) Graph \(\mu d^2\) (in kg \(\cdot\) m) versus \(M\) (in kg). Explain why the data plotted this way should fall close to a straight line. (c) Use the slope of the best straight-line fit to the data to determine the frequency \(f\) of the waves produced on the string by the oscillator. Take \(g = 9.80 \, \mathrm{m/s}^2\). (d) For string A (\(\mu = 0.0260\) g/cm), what value of \(M\) (in grams) would be required to produce a standing wave with a node-to-node distance of 24.0 cm? Use the value of \(f\) that you calculated in part (c).

A piano wire with mass 3.00 g and length 80.0 cm is stretched with a tension of 25.0 N. A wave with frequency 120.0 Hz and amplitude 1.6 mm travels along the wire. (a) Calculate the average power carried by the wave. (b) What happens to the average power if the wave amplitude is halved?

You are exploring a newly discovered planet. The radius of the planet is \(7.20 \times 10^7\) m. You suspend a lead weight from the lower end of a light string that is 4.00 m long and has mass 0.0280 kg. You measure that it takes 0.0685 s for a transverse pulse to travel from the lower end to the upper end of the string. On the earth, for the same string and lead weight, it takes 0.0390 s for a transverse pulse to travel the length of the string. The weight of the string is small enough that you ignore its effect on the tension in the string. Assuming that the mass of the planet is distributed with spherical symmetry, what is its mass?

A simple harmonic oscillator at the point \(x = 0\) generates a wave on a rope. The oscillator operates at a frequency of 40.0 Hz and with an amplitude of 3.00 cm. The rope has a linear mass density of 50.0 g/m and is stretched with a tension of 5.00 N. (a) Determine the speed of the wave. (b) Find the wavelength. (c) Write the wave function \(y(x, t)\) for the wave. Assume that the oscillator has its maximum upward displacement at time \(t = 0\). (d) Find the maximum transverse acceleration of points on the rope. (e) In the discussion of transverse waves in this chapter, the force of gravity was ignored. Is that a reasonable approximation for this wave? Explain.

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