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Chapter 16: Problem 59

A string oscillates according to the equation $$ y^{\prime}=(0.80 \mathrm{~cm}) \sin \left[\left(\frac{\pi}{3} \mathrm{~cm}^{-1}\right) x\right] \cos \left[\left(40 \pi \mathrm{s}^{-1}\right) t\right] $$ What are the (a) amplitude and (b) speed of the two waves (identical except for direction of travel) whose superposition gives this oscillation? (c) What is the distance between nodes? (d) What is the transverse speed of a particle of the string at the position \(x=2.1 \mathrm{~cm}\) when \(t=0.50 \mathrm{~s} ?\)

Short Answer

Expert verified
(a) 0.40 cm, (b) 120 cm/s, (c) 3 cm, (d) 0 cm/s.

Step by step solution

01

Identify the wave function parameters

The given equation for the string oscillation is\[y' = (0.80 \, \text{cm}) \sin\left(\frac{\pi}{3} \, \text{cm}^{-1} x\right) \cos\left(40 \pi \, \text{s}^{-1} t\right)\] This is a form of a standing wave, which can be expressed as the superposition of two waves traveling in opposite directions. We identify the parameters as:- Amplitude, \(A = 0.80 \, \text{cm}\)- Wave number, \(k = \frac{\pi}{3} \, \text{cm}^{-1}\)- Angular frequency, \(\omega = 40\pi \, \text{s}^{-1}\).
02

Determine the amplitude of the individual waves

The amplitude of each traveling wave that forms the standing wave is half of the standing wave's amplitude. So,\[A_{wave} = \frac{1}{2} \times 0.80 \, \text{cm} = 0.40 \, \text{cm}.\]
03

Calculate the speed of the individual waves

The speed \(v\) of a wave is given by the relation \(v = \frac{\omega}{k}\). Using the identified parameters:\[\omega = 40\pi \, \text{s}^{-1}, \, k = \frac{\pi}{3} \, \text{cm}^{-1}\]\[v = \frac{40\pi}{\pi/3}\ = 120 \, \text{cm/s}.\]
04

Calculate the distance between nodes

The distance between nodes in a standing wave is half of the wavelength, given by \(\lambda/2\). The wavelength \(\lambda\) can be calculated as:\[\lambda = \frac{2\pi}{k} = \frac{2\pi}{\pi/3} = 6 \, \text{cm}\]Hence, the distance between nodes is:\[\frac{\lambda}{2} = \frac{6}{2} = 3 \, \text{cm}.\]
05

Determine the transverse speed of a particle at specific position and time

The transverse speed of a particle on the string is given by the derivative of the displacement with respect to time, \(\frac{dy'}{dt}\). Starting from:\[y' = 0.80 \, \sin\left(\frac{\pi}{3} x\right) \cos(40 \pi t)\]Taking the derivative, we get:\[\frac{dy'}{dt} = -0.80 \cdot \sin\left(\frac{\pi}{3} x\right) \cdot 40\pi \cdot \sin(40 \pi t) = -32\pi \, \sin\left(\frac{\pi}{3} x\right) \sin(40 \pi t)\]Substitute \(x = 2.1 \, \text{cm}\) and \(t = 0.50 \, \text{s}\):\[\sin\left(\frac{\pi}{3} \times 2.1\right) = \sin\left(\frac{2.1\pi}{3}\right)\]\[\sin(40\pi \times 0.50) = \sin(20 \pi) = 0\]Therefore, the transverse speed is \(0 \, \text{cm/s}\) at this position and time.

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Key Concepts

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

Wave amplitude
The concept of wave amplitude revolves around understanding the extent of displacement from the central axis of oscillation. In a standing wave, like the one described in the string equation, the amplitude signifies the maximum peak height the wave reaches from its equilibrium position. Here, it is given as an amplitude of 0.80 cm. However, for individual traveling waves that superimpose to form this standing wave, the amplitude is reduced by half.
This reduction is because each traveling wave contributes equally to the standing wave鈥檚 structure, resulting in an individual amplitude of 0.40 cm.
  • Standing Wave Amplitude: 0.80 cm
  • Individual Wave Amplitude: 0.40 cm
Understanding the amplitude is crucial, as it impacts the energy carried by the wave and its visible "height." The concept is vital in fields such as acoustics, optics, and various engineering applications.
Wave speed
Wave speed is a fundamental concept in wave mechanics, indicating how fast a wave propagates through a medium. In our exercise, the speed of individual traveling waves can be determined using their angular frequency and wavenumber.
The formula to find the speed of a wave (\( v \)) is:\[ v = \frac{\omega}{k} \]where \( \omega \) is the angular frequency, and \( k \) is the wavenumber.
  • Angular Frequency: 40蟺 s-1
  • Wavenumber: \( \frac{\pi}{3} \text{cm}^{-1} \)
  • Wave Speed: 120 cm/s
Substituting these values in, we calculate the wave speed as 120 cm/s. Knowing the wave speed helps determine how quickly information or energy travels through the medium, and it is essential in various physical sciences like meteorology and electromagnetic theory.
Standing wave
Standing waves occur when two identical waves traveling in opposite directions superimpose. This superposition creates a pattern that appears stationary, hence the term 'standing wave.' The exercise explores a standing wave formed on a string with certain parameters.
Standing waves are unique because they exhibit distinct nodes and antinodes. Nodes are points of no displacement, whereas antinodes are points of maximum displacement.
  • Node: Point of zero amplitude
  • Antinode: Point of maximum amplitude
The particular standing wave in this scenario is characterized by its wave equation and features such as amplitude and wavenumber, illustrating the interplay between dynamic wave properties and equilibrium patterns. Understanding standing waves is key in musical instruments acoustics, microwave technology, and many engineering fields.
Wave node distance
Wave node distance relates to the distance between consecutive nodes in a standing wave, where there is no movement or "zero" amplitude. This node spacing is essentially half the wavelength of the wave. To find this, we first calculate the wavelength ( \( \lambda \)) using the wavenumber, where\[ \lambda = \frac{2\pi}{k} \]Subsequently, the node distance becomes half of this wavelength:\[\text{Node Distance} = \frac{\lambda}{2} = \frac{6 \text{cm}}{2} = 3 \text{cm}\]
  • Wavelength (\(\lambda\)): 6 cm
  • Distance Between Nodes: 3 cm
Knowing the distance between nodes is vital for practical applications like determining resonance conditions in physics or optimizing sound production in acoustic instruments. This concept reflects how wave properties translate into spatial characteristics in physical systems.

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Most popular questions from this chapter

Two sinusoidal waves of the same frequency are to be sent in the same direction along a taut string. One wave has an amplitude of \(5.50 \mathrm{~mm}\), the other \(12.0 \mathrm{~mm}\). (a) What phase difference \(\phi_{1}\) between the two waves results in the smallest amplitude of the resultant wave? (b) What is that smallest amplitude? (c) What phase difference \(\phi_{2}\) results in the largest amplitude of the resultant wave? (d) What is that largest amplitude? (e) What is the resultant amplitude if the phase angle is \(\left(\phi_{1}-\phi_{2}\right) / 2 ?\)

A standing wave pattern on a string is described by $$ y(x, t)=0.040(\sin 4 \pi x)(\cos 40 \pi t), $$ where \(x\) and \(y\) are in meters and \(t\) is in seconds. For \(x \geq 0\), what is the location of the node with the (a) smallest, (b) second smallest, and (c) third smallest value of \(x ?\) (d) What is the period of the oscillatory motion of any (nonnode) point? What are the (e) speed and (f) amplitude of the two traveling waves that interfere to produce this wave? For \(t \geq 0\), what are the (g) first, (h) second, and (i) third time that all points on the string have zero transverse velocity?

A sinusoidal wave of frequency \(500 \mathrm{~Hz}\) has a speed of \(320 \mathrm{~m} / \mathrm{s}\). (a) How far apart are two points that differ in phase by \(\pi / 3 \mathrm{rad}\) ? (b) What is the phase difference between two displacements at a certain point at times \(1.00 \mathrm{~ms}\) apart?

Two waves are generated on a string of length \(4.0 \mathrm{~m}\) to produce a three-loop standing wave with an amplitude of \(1.0 \mathrm{~cm}\). The wave speed is \(100 \mathrm{~m} / \mathrm{s}\). Let the equation for one of the waves be of the form \(y(x, t)=y_{m} \sin (k x+\omega t)\). In the equation for the other wave, what are (a) \(y_{m}\), (b) \(k\), (c) \(\omega\), and (d) the sign in front of \(\omega\) ?

Two sinusoidal waves of the same frequency travel in the same direction along a string. If \(y_{m 1}=2.0 \mathrm{~cm}, y_{m 2}=\) \(4.0 \mathrm{~cm}, \phi_{1}=0\), and \(\phi_{2}=\pi / 2 \mathrm{rad}\), what is the amplitude of the resultant wave?
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