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Two waves are generated on a string of length \(3.0 \mathrm{~m}\) 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 cquation 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 \(\omega ?\)

Short Answer

Expert verified
(a) \( y_m = 1.0 \mathrm{~cm} \); (b) \( k = \pi \mathrm{~m}^{-1} \); (c) \( \omega = 100\pi \mathrm{~rad/s} \); (d) Negative sign.

Step by step solution

01

Identify the Parameters of the Standing Wave

A standing wave with three loops in a string of length 3.0 m implies that the wavelength is twice the length of one loop. Therefore, the wavelength \( \lambda \) is \( \lambda = \frac{2 \times 3.0}{3} = 2.0 \mathrm{~m} \). Since the amplitude of the standing wave is the same as the amplitude of the individual waves, \( y_m = 1.0 \mathrm{~cm} \).
02

Calculate the Wavenumber \( k \)

The wavenumber \( k \) is defined as \( k = \frac{2\pi}{\lambda} \). Substituting the wavelength from the previous step, \( k = \frac{2\pi}{2.0 \mathrm{~m}} = \pi \mathrm{~m}^{-1} \).
03

Determine the Angular Frequency \( \omega \)

The angular frequency \( \omega \) can be found using the wave speed formula: \( v = \lambda f \), where \( v \) is the speed, and \( f \) is the frequency. Then \( f = \frac{v}{\lambda} = \frac{100 \mathrm{~m/s}}{2.0 \mathrm{~m}} = 50 \mathrm{~Hz} \). The angular frequency \( \omega \) is given by \( \omega = 2\pi f \), so \( \omega = 2\pi \times 50 \mathrm{~Hz} = 100\pi \mathrm{~rad/s} \).
04

Determine the Sign in Front of \( \omega \) in the Second Wave Equation

The second wave, which interferes with the first to produce the standing wave, must have a phase difference that is typically \( \Delta\omega = \pm \omega \). For a standing wave, if the first wave is represented as \( \sin(kx + \omega t) \), the second wave is represented as \( \sin(kx - \omega t) \) to account for the opposite direction of propagation. Therefore, the sign in front of \( \omega \) is negative.

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Key Concepts

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

Wavelength
The concept of wavelength is crucial in understanding wave behavior. Wavelength, often denoted by the symbol \( \lambda \), represents the distance between two consecutive points that are in phase on a wave, such as from crest to crest or trough to trough. In this exercise, we are working with a three-loop standing wave on a string.
To calculate the wavelength, we consider one loop to be half of a wavelength. Since there are three loops in a standing wave on a 3.0 m string, the full wavelength \( \lambda \) becomes two meters: \( \lambda = \frac{2 \times 3.0}{3} = 2.0 \, \mathrm{m} \).
Knowing the wavelength helps us derive other important properties such as wavenumber and frequency. Recognizing the relationship between different wave parameters can simplify solving wave problems and deepen understanding.
Wavenumber
Wavenumber is a measure related to the spatial frequency of a wave. It indicates the number of wave cycles in a given distance and is represented by \( k \).
The formula to calculate wavenumber is \( k = \frac{2\pi}{\lambda} \). This means that wavenumber is inversely proportional to wavelength, with a larger wavelength resulting in a smaller wavenumber, and vice versa. From the calculated wavelength \( \lambda = 2.0 \, \mathrm{m} \), we can determine the wavenumber as follows: \( k = \frac{2\pi}{2.0 \, \mathrm{m}} = \pi \, \mathrm{m^{-1}} \).
The wavenumber is an essential concept in wave physics because it links the physical distance between repeating features in the wave to the wave’s measurable frequency.
Angular Frequency
Angular frequency, denoted \( \omega \), describes how quickly a wave oscillates in time. It’s related to the regular frequency \( f \) of the wave but expressed in radians per second.
Angular frequency is calculated using the formula \( \omega = 2\pi f \), where \( f \) is the wave's frequency. The frequency of a wave can be derived from its wave speed and wavelength with \( f = \frac{v}{\lambda} \). Given the wave speed \( v = 100 \, \mathrm{m/s} \) and wavelength \( \lambda = 2.0 \, \mathrm{m} \), the frequency \( f \) is \( 50 \, \mathrm{Hz} \), yielding an angular frequency of \( \omega = 100\pi \, \mathrm{rad/s} \).
Understanding angular frequency provides insights into how energetic a wave is and is crucial for analyzing wave behavior in physics.
Wave Interference
Wave interference is a phenomenon that occurs when two or more waves overlap and combine to form a new wave pattern. In the case of standing waves, interference is what allows the formation of fixed nodes and antinodes along a medium.
For standing waves, such as in this exercise, two waves of the same amplitude and frequency travel in opposite directions. The mathematical representation of these waves often involves a phase difference. If one wave is expressed as \( \sin(kx + \omega t) \), the other will be \( \sin(kx - \omega t) \). This setup creates a perfectly destructive and constructive interference pattern leading to a standing wave.
Comprehending wave interference helps explain many natural and technological phenomena, including the behavior of musical instruments and electromagnetic waves.

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Most popular questions from this chapter

A \(125 \mathrm{~cm}\) length of string has mass \(2.00 \mathrm{~g}\) and tension \(7.00 \mathrm{~N}\). (a) What is the wave speed for this string? (b) What is the lowest resonant frequency of this string?

A stretched string has a mass per unit length of \(5.00 \mathrm{~g} / \mathrm{cm}\) and a tension of \(10.0 \mathrm{~N}\). A sinusoidal wave on this string has an amplitude of \(0.12 \mathrm{~mm}\) and a frequency of \(100 \mathrm{~Hz}\) and is traveling in the negative direction of an \(x\) axis. If the wave equation is of the form \(y(x, t)=y_{m} \sin (k x \pm \omega t),\) what are (a) \(y_{n n}\), (b) \(k,\) (c) \(\omega\), and (d) the correct choice of sign in front of \(\omega ?\) (a) \(y_{y=}\) (b) \(k,\) (c) \(\omega\), and (d) the correct choice of sign in front of \(\omega ?\)

A sand scorpion can detect the motion of a nearby beetle (its prey) by the waves the motion sends along the sand surface (Fig. \(16-30\) ). The waves are of two types: transverse waves traveling at \(v_{i}=50 \mathrm{~m} / \mathrm{s}\) and longitudinal waves traveling at \(v_{l}=150 \mathrm{~m} / \mathrm{s} .\) If a sudden motion sends out such waves, a scorpion can tell the distance of the bectle from the difference \(\Delta t\) in the arrival times of the waves at its leg nearest the beetle. If \(\Delta t=4.0 \mathrm{~ms}\) what is the beetle's distance?

A \(100 \mathrm{~g}\) wire is held under a tension of \(250 \mathrm{~N}\) with one end at \(x=0\) and the other at \(x=10.0 \mathrm{~m}\). At time \(t=0\), pulse 1 is sent along the wire from the end at \(x=10.0 \mathrm{~m}\). At time \(t=30.0 \mathrm{~ms}\), pulse 2 is sent along the wire from the end at \(x=0 .\) At what position \(x\) do the pulses begin to meet?

These two waves travel along the same string: \(y_{1}(x, t)=(4.60 \mathrm{~mm}) \sin (2 \pi x-400 \pi t)\) \(y_{2}(x, t)=(5.60 \mathrm{~mm}) \sin (2 \pi x-400 \pi t+0.80 \pi \mathrm{rad})\) What are (a) the amplitude and (b) the phase angle (relative to wave 1 ) of the resultant wave? (c) If a third wave of amplitude \(5.00 \mathrm{~mm}\) is also to be sent along the string in the same direction as the first two waves, what should be its phase angle in order to maximize the amplitude of the new resultant wave?

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