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Chapter 15: Problem 74

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).

Short Answer

Expert verified
(a)(i) 0 cm, (ii) 0.400 cm, (iii) 0.282 cm; (b) 2.08 ms for all; (c) 6.03 m/s, scales with amplitude.

Step by step solution

01

Understanding the Third Harmonic

For a string in its third harmonic, it has three antinodes and two nodes (not including the nodes at the ends). The wavelength \( \lambda \) of the wave can be expressed as \( \frac{2L}{3} \), where \( L \) is the length of the string. We use the wave speed and frequency to find the wavelength as:\[ \lambda = \frac{v}{f} = \frac{192}{240} = 0.8 \text{ m} = 80 \text{ cm} \].
02

Node and Antinode Positions

In the third harmonic, nodes divide the string into segments of \( \frac{\lambda}{2} \). Since \( \lambda = 80 \) cm, nodes are at positions 0 cm, 40 cm, 80 cm, etc. Antinodes are midway between the nodes at 20 cm and 60 cm.
03

Calculate Amplitude at Specific Points

The amplitude at a point depends on its distance from the nearest node. It can be given as a function: \( A(x) = A_0 \sin(kx) \), where \( k = \frac{2\pi}{\lambda} \). So the calculations for amplitude at (i) 40 cm, (ii) 20 cm, (iii) 10 cm are:- (i) At 40 cm (node): \( A(x) = A_0 \sin(\frac{2\pi}{80} \times 40) = 0 \).- (ii) At 20 cm (antinode): \( A(x) = 0.400 \sin(\frac{2\pi}{80} \times 20) = 0.400 \). - (iii) At 10 cm: \( A(x) = 0.400 \sin(\frac{2\pi}{80} \times 10) \approxeq 0.282 \).
04

Calculate Time for Maximum Displacement Changes

The time for the string to move from maximum upward to downward displacement is half the period of oscillation. The period \( T \) is given by \( T = \frac{1}{f} = \frac{1}{240} \text{ s} \). Therefore, \( \frac{T}{2} \approx 2.08 \times 10^{-3} \text{ s} \) for all points.
05

Calculate Maximum Transverse Velocity at Each Point

The maximum velocity \( v_{max} \) at each point is \( v_{max} = A_0 \omega \), where \( \omega = 2\pi f \). For antinodes, \( v_{max} = 0.004 \times 2\pi \times 240 \approx 6.03 \text{ m/s} \). For an intermediate point, \( v_{max} \) scales with the local amplitude.
06

Calculate Maximum Transverse Acceleration

The maximum acceleration \( a_{max} \) is \( a_{max} = A_0 \omega^2 \). With \( \omega = 2\pi \times 240 \), \( a_{max} = 0.004 \times (2\pi \times 240)^2 \) for antinodes. This scales down for other points based on local amplitude.

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Key Concepts

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

Understanding Harmonics
When a string vibrates while fixed at both ends, it produces standing waves that can resonate at specific frequencies called harmonics. Each harmonic corresponds to a different mode of vibration. The first harmonic is the fundamental frequency, which has one antinode in the center and nodes at each end. The third harmonic, explored in this problem, contains three antinodes and two nodes in addition to the ends.

Harmonics can be visualized by imagining waves of distinct shapes fitting perfectly along the length of the string. Each harmonic involves a different pattern of nodes and antinodes, which are places where the wave does not move (node) or it has maximum displacement (antinodes).
  • The nodes are points on the string that do not move at all.
  • The antinodes are where the motion of the string is maximal.
The position of these nodes and antinodes is reliant on the wavelength of the standing wave. The wavelength, in turn, depends on the speed of the wave and its frequency. This interplay of speed, frequency, and resulting wavelength helps to determine the specific node and antinode patterns of each harmonic.
Exploring Wave Speed
Wave speed is a critical concept in understanding how waves propagate through different mediums. In this exercise, the wave speed given is 192 m/s. Wave speed is a measure of how fast a wave travels from one point to another and is calculated using the formula: \[ v = f \lambda \]where \( v \) is the wave speed, \( f \) is the frequency, and \( \lambda \) is the wavelength.

The speed of a wave on a string is influenced by the tension in the string and its mass per unit length. For a standing wave, like the one in the problem, wave speed helps in determining the wavelength when coupled with the known frequency. Plugging our known values into this formula gives us the wavelength used in calculating the positions of nodes and antinodes.
  • Higher tension or lower mass per length increases wave speed.
  • Wave speed is constant for a given string and tension but varies with different materials and settings.
Understanding wave speed allows us to predict how quickly energy is transferred along the string, and how the frequency and wavelength interplay to form different harmonics.
Deciphering Frequency
Frequency refers to how many complete cycles or oscillations a wave undergoes in a period of one second. In this problem, the frequency is given as 240 Hz, meaning the wave oscillates 240 times per second. Frequency is pivotal because it dictates the pitch of the sound produced and is related to the wave's energy.

To link frequency with other wave properties, the formula:\[ f = \frac{v}{\lambda} \]is used, interconnecting frequency \( f \), wave speed \( v \), and wavelength \( \lambda \). This relationship is fundamental in understanding how changes to the length or tension of a string affect its resonant frequency and the subsequent harmonic patterns.
  • Frequency remains constant for a harmonic unless the length or tension is altered.
  • A high frequency results in closely spaced antinodes and a high-pitched sound.
By manipulating frequency, the same string can produce various sounds and vibrational patterns, offering insight into the versatility of musical and physical systems that rely on wave behavior.
Concept of Amplitude
Amplitude is essentially the height of a wave and reflects the energy carried by the wave. Higher amplitudes correspond to waves that have more energy or are more intense. In the context of standing waves on a string, amplitude varies along the string's length, with maximum values at antinodes and zero at nodes.

In the given problem, the amplitude is 0.400 cm at an antinode. This is the maximum displacement found on the string. Generally, the amplitude at any point \( x \) on the string can be determined by: \[ A(x) = A_0 \sin(kx) \]where \( A_0 \) is the maximum amplitude, and \( k \) is the wave number \( \frac{2\pi}{\lambda} \). This equation shows how amplitude decreases as you move closer to a node and increases towards antinodes.
  • The amplitude of a wave affects its intensity and the loudness perceived in a string instrument.
  • Maximum transverse velocity and acceleration of the string rely on amplitude, affecting how quickly and forcefully the string can move.
Amplitude is key in determining the physical power of the wave, impacting not just scientific calculations but also practical applications like sound production in musical instruments.

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Most popular questions from this chapter

A fellow student with a mathematical bent tells you that the wave function of a traveling wave on a thin rope is \(y(x, t)=\) 2.30 \(\mathrm{mm} \cos [(6.98 \mathrm{rad} / \mathrm{m}) x+(742 \mathrm{rad} / \mathrm{s}) t] .\) Being more practical, you measure the rope to have a length of 1.35 \(\mathrm{m}\) and a mass of 0.00338 \(\mathrm{kg}\) . You are then asked to determine the following: (a) amphtude; (b) frequency; (c) wavelength; (d) wave speed; (e) direction the wave is traveling; (f) tension in the rope; (g) aver- age power transmitted by the wave.

Energy Output. By measurement you determine that sound waves are spreading out equally in all directions from a point source and that the intensity is 0.026 \(\mathrm{W} / \mathrm{m}^{2}\) at a distance of 4.3 \(\mathrm{m}\) from the source. (a) What is the intensity at a distance of 3.1 \(\mathrm{m}\) from the source? (b) How much sound energy does the source emit in one hour if its power output remains constant?

(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?

Waves on a Stick. A flexible stick 2.0 \(\mathrm{m}\) long is not fixed in any way and is free to vibrate. Make clear drawings of this stick vibrating in its first three harmonics, and then use your drawings to find the wavelengths of each of these harmonics. (Hint: Should the ends be nodes or antinodes?)

A jet plane at take-off can produce sound of intensity 10.0 \(\mathrm{W} / \mathrm{m}^{2}\) at 30.0 \(\mathrm{m}\) away. But you prefer the tranguil sound of normal conversation, which is 1.0\(\mu \mathrm{W} / \mathrm{m}^{2}\) . 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 take-off?
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