/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 36 (II) In deriving Eq. \(2, \quad ... [FREE SOLUTION] | 91Ó°ÊÓ

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(II) In deriving Eq. \(2, \quad v=\sqrt{F_{\mathrm{T}} / \mu,}\) for the speed of a transverse wave on a string, it was assumed that the wave's amplitude \(A\) is much less than its wavelength \(\lambda\) . Assuming a sinusoidal wave shape \(D=A \sin (k x-\omega t),\) show via the partial derivative \(v^{\prime}=\partial D / \partial t \quad\) that the assumption \(A \ll \lambda\) implies that the maximum transverse speed \(v_{\text { max }}^{\prime}\) the string itself is much less than the wave velocity. If \(A=\lambda / 100\) determine the ratio \(v_{\max }^{\prime} / v.\) \(v=\sqrt{\frac{F_{\mathrm{T}}}{\mu}} \quad \quad \left[ \begin{array}{l}{\text { transverse }} \\ {\text { wave on a cord }}\end{array}\right]\) (2)

Short Answer

Expert verified
The ratio \(\frac{v'_{\text{max}}}{v} = \frac{\pi}{50}\), showing \(v'_{\text{max}} \ll v\).

Step by step solution

01

Identify the Aim

We want to show that the maximum transverse speed of the string, which is given by the partial derivative \(v' = \frac{\partial D}{\partial t}\), is much less than the speed of the wave \(v\). Further, if \(A = \frac{\lambda}{100}\), we are tasked with finding the ratio \(\frac{v_{\text{max}}'}{v}\).
02

Expression for Wave Displacement

The displacement of a sinusoidal wave is given by \(D = A \sin(kx - \omega t)\), where \(A\) is the amplitude, \(k\) is the wave number, and \(\omega\) is the angular frequency.
03

Compute Partial Derivative \( \partial D / \partial t \)

Take the partial derivative of \(D\) with respect to \(t\) to find the transverse speed of the string:\[v' = \frac{\partial}{\partial t}(A \sin(kx - \omega t)) = -A\omega \cos(kx - \omega t)\].
04

Find Maximum Transverse Speed

The maximum transverse speed \(v'_{\text{max}}\) occurs when \(\cos(kx - \omega t) = 1\), so \(v'_{\text{max}} = A\omega\).
05

Relate to Wave Velocity

The wave speed is given by \(v = \omega / k\). To relate the maximum transverse speed to wave velocity, we express \(\omega = vk\). Therefore, \(v'_{\text{max}} = A\omega = Avk\).
06

Compare Amplitude to Wavelength

The assumption \(A \ll \lambda\) implies that the amplitude is negligible compared to the wavelength. Since \(v = \omega/k\), \(v'_{\text{max}} = A\omega = Avk\), and the ratio \(\frac{v'_{\text{max}}}{v} = Ak\).
07

Calculate Ratio When \(A = \lambda / 100\)

If \(A = \lambda/100\), then \(k = 2\pi/\lambda\). Hence:\[ \frac{v'_{\text{max}}}{v} = A \cdot \frac{2\pi}{\lambda} = \frac{\lambda}{100} \cdot \frac{2\pi}{\lambda} = \frac{2\pi}{100} = \frac{\pi}{50} \].

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Key Concepts

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

Partial Derivatives in Physics
In physics, partial derivatives are essential for understanding how certain parameters affect a system when other parameters are held constant.
They help us analyze specific changes in multidimensional systems. For instance, when examining waves on a string, we often consider how the displacement of the wave changes over time while holding the spatial components constant.

In the given exercise, we use the partial derivative \( \frac{\partial D}{\partial t} \) to determine how the displacement \( D \) changes with time due to a sinusoidal wave. Given the wave equation \( D = A \sin(kx - \omega t) \), taking the partial derivative with respect to \( t \) yields \( -A\omega \cos(kx - \omega t) \).

This expresses the transverse velocity of a point on the string as the wave passes. It allows us to determine the maximum transverse speed, which is crucial in comparing it relative to the wave velocity \( v \). This calculation shows how the time-dependent component of wave motion is a pivotal concept in wave physics.
Wave Velocity
Wave velocity refers to the speed at which a wave propagates through a medium. It is given by the formula \( v = \sqrt{\frac{F_{\mathrm{T}}}{\mu}} \) for waves on a string, where \( F_{\mathrm{T}} \) is the tension in the string and \( \mu \) is the linear mass density.
This fundamental relationship shows that wave speed relies on both the tension applied and the mass per unit length of the string.

In our exercise, wave velocity is directly linked to angular frequency \( \omega \) and wave number \( k \), via the relationship \( v = \frac{\omega}{k} \). Understanding how angular frequency and the wave number interact helps explain the speed at which wave crests move.

It is important to contrast the wave velocity with the transverse speed of the particles in the medium, as illustrated in the solution when analyzing the maximum transverse speed of the string. This comparison underscores how the wave propagates energy through the medium, while individual particles oscillate around their equilibrium positions at much slower speeds.
Amplitude and Wavelength Relationship
The relationship between amplitude \( A \) and wavelength \( \lambda \) is a key aspect of wave behavior. Amplitude refers to the maximum displacement of points on a wave, while wavelength is the distance between successive wave peaks.
The assumption \( A \ll \lambda \) is crucial in wave analysis, particularly when simplifying mathematical models of wave behavior.

In the context of this exercise, this assumption implies that the transverse oscillations of points on the string are small compared to the wavelength.
When \( A = \lambda/100 \), it directly affects the maximum transverse speed \( v_{\text{max}}' \).

The exercise shows that, if amplitude is significantly smaller than wavelength, the transverse motion's speed ratio \( \frac{v_{\text{max}}'}{v} \) becomes \( \frac{\pi}{50} \).
This low ratio highlights that energy and information move through the medium much faster than the medium displacements, reaffirming the stability and efficiency of wave propagation.

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Most popular questions from this chapter

(II) Dimensional analysis. Waves on the surface of the ocean do not depend significantly on the properties of water such as density or surface tension. The primary "return force" for water piled up in the wave crests is due to the gravitational attraction of the Earth. Thus the speed \(v(\mathrm{m} / \mathrm{s})\) of ocean waves depends on the acceleration due to gravity \(g .\) It is reasonable to expect that \(v\) might also depend on water depth \(h\) and the wave's wavelength \(\lambda\) . Assume the wave speed is given by the functional form \(v=C g^{\alpha} h^{\beta} \lambda^{\gamma},\) where \(\alpha, \beta, \gamma,\) and \(C\) are numbers without dimension. \((a)\) In deep water, the water deep below does not affect the motion of waves at the surface. Thus \(v\) should be independent of depth \(h\) (i.e., \(\beta=0 ) .\) Using only dimensional analysis, determine the formula for the speed of surface waves in deep water. \((b)\) In shallow water, the speed of surface waves is found experimentally to be independent of the wavelength (i.e., \(\gamma=0 ) .\) Using only dimensional analysis, determine the formula for the speed of waves in shallow water.

(II) Suppose two linear waves of equal amplitude and frequency have a phase difference \(\phi\) as they travel in the same medium. They can be represented by $$ \begin{array}{l} D_{1}=A \sin (k x-\omega t) \\ D_{2}=A \sin (k x-\omega t+\phi) \end{array} $$ \(\begin{array}{llll}\text { (a) Use the trigonometric identity } & \sin \theta_{1}+\sin \theta_{2}=\end{array}\) \(2 \sin \frac{1}{2}\left(\theta_{1}+\theta_{2}\right) \cos \frac{1}{2}\left(\theta_{1}-\theta_{2}\right)\) to show that the resultant wave is given by $$ D=\left(2 A \cos \frac{\phi}{2}\right) \sin \left(k x-\omega t+\frac{\phi}{2}\right) $$ (b) What is the amplitude of this resultant wave? Is the wave purely sinusoidal, or not? (c) Show that constructive interference occurs if \(\phi=0,2 \pi, 4 \pi,\) and so on, and destructive interference occurs if \(\phi=\pi, 3 \pi, 5 \pi,\) etc. \((d)\) Describe the resultant wave, by equation and in words, if \(\phi=\pi / 2\).

(I) A transverse wave on a wire is given by \(D(x, t)=\) \(0.015 \sin (25 x-1200 t)\) where \(D\) and \(x\) are in meters and \(t\) is in seconds. (a) Write an expression for a wave with the same amplitude, wavelength, and frequency but traveling in the opposite direction. ( \(b\) ) What is the speed of either wave?

(I) An earthquake \(P\) wave traveling \(8.0 \mathrm{~km} / \mathrm{s}\) strikes a boundary within the Earth between two kinds of material. If it approaches the boundary at an incident angle of \(52^{\circ}\) and the angle of refraction is \(31^{\circ}\), what is the speed in the second medium?

A transverse wave pulse travels to the right along a string with a speed \(v=2.4 \mathrm{~m} / \mathrm{s}\). At \(t=0\) the shape of the pulse is given by the function $$ D=\frac{4.0 \mathrm{~m}^{3}}{x^{2}+2.0 \mathrm{~m}^{2}} $$ where \(D\) and \(x\) are in meters. \((a)\) Plot \(D\) vs. \(x\) at \(t=0\) from \(x=-10 \mathrm{~m}\) to \(x=+10 \mathrm{~m}\). ( \(b\) ) Determine a formula for the wave pulse at any time \(t\) assuming there are no frictional losses. (c) Plot \(D(x, t)\) vs. \(x\) at \(t=1.00 \mathrm{~s}\). (d) Repeat parts (b) and (c) assuming the pulse is traveling to the left.

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