91Ó°ÊÓ

(a) Show that for a wave on a string, the kinetic energy per unit length of string is $$ u_{k}(x, t)=\frac{1}{2} \mu v_{y}^{2}(x, t)=\frac{1}{2} \mu\left(\frac{\partial y(x, t)}{\partial t}\right)^{2} $$ where \(\mu\) is the mass per unit length. (b) Calculate \(u_{\mathrm{k}}(x, t)\) for a sinusoidal wave given by Eq. \((15.7) .\) (c) There is also elastic potential energy in the string, associated with the work required to deform and stretch the string. Consider a short segment of string at position \(x\) that has unstretched length \(\Delta x,\) as in \(\mathrm{Fg} .15 .13 .\) Ignoring the (small) curvature of the segment, its slope is \(\partial y(x, t) / \partial x\) Assume that the displacement of the string from equilibrium is small, so that \(\partial y / \partial x\) has a magnitude much less than unity. Show that the stretched length of the segment is approximately $$ \Delta x\left[1+\frac{1}{2}\left(\frac{\partial y(x, t)}{\partial x}\right)^{2}\right] $$ (Hint: Use the relationship \(\sqrt{1+u} \approx 1+\frac{1}{2} u,\) valid for \(|u| \ll 1 . )\) (d) The potential energy stored in the segment equalsthe work done by the string tension \(F\) (which acts along the string) to stretch the segment from its unstretched length \(\Delta x\) to the length calculated in part (c). Calculate this work and show that the potential energy per unit length of string is $$ u_{p}(x, t)=\frac{1}{2} F\left(\frac{\partial y(x, t)}{\partial x}\right)^{2} $$ (e) Calculate \(u_{\mathrm{o}}(x, t)\) for a sinusoidal wave given by Eq. \((15.7)\) (f) Show that \(u_{k}(x, t)=u_{p}(x, t)\) for all \(x\) and \(t .(g)\) Show \(y(x, t)\) \(u_{k}(x, t),\) and \(u_{p}(x, t)\) as functions of \(x\) for \(t=0\) in one graph with all three functions on the same axes. Explain why \(u_{k}\) and \(u_{p}\) are maximum where \(y\) is zero, and vice versa. (h) Show that the instantaneous power in the wave, given by Eq. \((15.22),\) is equal to the total energy per unit length multiplied by the wave speed \(v\) Explain why this result is reasonable.

Step by step solution

01

Understanding Kinetic Energy in a Wave

The kinetic energy per unit length of a wave on a string is given by:\[ u_{k}(x, t) = \frac{1}{2} \mu \left( \frac{\partial y(x, t)}{\partial t} \right)^2 \]where \( \mu \) is the mass per unit length. This expression comes from the definition of kinetic energy, which is \( KE = \frac{1}{2} mv^2 \). For small segments of the wave, the velocity \( v_y \) is the rate of change of displacement \( y \) with respect to time.

02

Calculate Kinetic Energy for Sinusoidal Wave

For a sinusoidal wave given by \( y(x,t) = A \sin(kx - \omega t) \), the partial derivative with respect to time is\[ \frac{\partial y}{\partial t} = -A\omega \cos(kx - \omega t) \]Substituting into kinetic energy equation gives:\[ u_k(x,t) = \frac{1}{2} \mu \left( -A\omega \cos(kx - \omega t) \right)^2 = \frac{1}{2} \mu A^2 \omega^2 \cos^2(kx - \omega t) \]

03

Approximate Stretched Length of String Segment

Assuming \( |\frac{\partial y}{\partial x}| \ll 1 \), the stretched length based on the small slope approximation and the hint is:\[ \Delta x \sqrt{1 + \left(\frac{\partial y}{\partial x}\right)^2} \approx \Delta x \left[ 1 + \frac{1}{2} \left(\frac{\partial y}{\partial x}\right)^2 \right] \]This uses the approximation \( \sqrt{1+u} \approx 1 + \frac{1}{2} u \).

04

Calculate Potential Energy per Unit Length

The work done to stretch the segment equals the potential energy, given the tension \( F \):\[ \text{Potential Energy} = F \Delta L = F \left( \Delta x \frac{1}{2} \left( \frac{\partial y}{\partial x} \right)^2 \right) \]Thus, the potential energy per unit length is:\[ u_p(x,t) = \frac{1}{2} F \left( \frac{\partial y}{\partial x} \right)^2 \]

05

Calculate Potential Energy for Sinusoidal Wave

For the sinusoidal wave, using \( y(x,t) = A \sin(kx - \omega t) \), the spatial derivative is:\[ \frac{\partial y}{\partial x} = A k \cos(kx - \omega t) \]Substituting gives:\[ u_p(x,t) = \frac{1}{2} F (A k \cos(kx - \omega t))^2 = \frac{1}{2} F A^2 k^2 \cos^2(kx - \omega t) \]

06

Show Equality of Kinetic and Potential Energies

The expressions for \( u_k(x,t) \) and \( u_p(x,t) \) are:\[ u_k(x,t) = \frac{1}{2} \mu A^2 \omega^2 \cos^2(kx - \omega t) \]\[ u_p(x,t) = \frac{1}{2} F A^2 k^2 \cos^2(kx - \omega t) \]Since \( \omega = vk \), we have \( F = \mu v^2 \) based on wave speed, showing \( u_k(x,t) = u_p(x,t) \).

07

Understanding Energy Distribution in the Wave

When graphing \( y(x, t) \), \( u_k(x, t) \), and \( u_p(x, t) \) for \( t = 0 \), \( y \) is zero when \( u_k \) and \( u_p \) are maximum due to the cosine-squared relationship, representing maximum energy exchange when displacement is zero.

08

Relate Power to Wave Energy

The instantaneous power is given by:\[ P = v( u_k + u_p ) \]This shows that the power is related to the total energy per unit length times the wave speed, illustrating energy flow through the wave which is conserved.

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Key Concepts

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

Kinetic Energy

Kinetic energy in the context of wave mechanics is the energy possessed by segments of a wave due to their motion. In a wave on a string, each small segment of the string has kinetic energy. This is because the segments are constantly moving as the wave propagates. The kinetic energy per unit length (\( u_{k}(x, t) \)) for a string wave can be expressed using the formula:

• \( u_{k}(x, t) = \frac{1}{2} \mu \left( \frac{\partial y(x, t)}{\partial t} \right)^2 \)

here \( \mu \) is the mass per unit length, and \( \frac{\partial y(x, t)}{\partial t} \) represents the rate of change of displacement with respect to time. This rate is essentially the velocity (\( v_{y} \)) of the wave segment as it oscillates back and forth.
This formula is derived from the basic kinetic energy equation \( KE = \frac{1}{2} mv^2 \). Think of it as each tiny piece of the string, stretching and compressing, possesses a little bit of kinetic energy as it moves. As a sinusoidal wave passes, each piece of string momentarily acquires velocity, and hence kinetic energy.

Potential Energy

Potential energy in a wave on a string is stored energy related to the position of the wave particles as they are displaced from equilibrium. As the wave vibrates, the string segments are stretched from their natural unstretched length. This stretching is where the potential energy comes from.
The potential energy per unit length (\( u_{p}(x, t) \)) can be calculated with the help of string tension \( F \). For a small string segment, this is given by:

• \( u_{p}(x, t) = \frac{1}{2} F \left( \frac{\partial y(x, t)}{\partial x} \right)^2 \)

The derivative \( \frac{\partial y(x, t)}{\partial x} \) describes how the wave shape (or slope) changes with position, which relates to the deformation of the string.
This formula arises from considering the work done to stretch the string segments based on their positions. Essentially, when the wave passes, the tension in the string does work due to displacement, storing energy in the process. In the case of a sinusoidal wave, this creates zones of high potential energy.

Sinusoidal Wave

A sinusoidal wave represents a smooth, repetitive oscillation most simply described by the sine and cosine functions. In the context of waves on a string, a typical equation for a sinusoidal wave is:

• \( y(x, t) = A \sin(kx - \omega t) \)

Where \( A \) is the amplitude, \( k \) is the wave number, and \( \omega \) is the angular frequency.
This waveform is crucial in wave mechanics due to its well-defined properties, like wavelength, frequency, and amplitude, which make mathematical manipulations straightforward.
For kinetic and potential energies of sinusoidal waves, both energies depend on trigonometric components (like \( \cos^2(kx - \omega t) \)) due to the derivatives these functions produce. These energy forms oscillate between kinetic and potential, showcasing how energy shifts back and forth in wave mechanics.

Energy Conservation

Energy conservation, in wave mechanics, highlights that the total energy within a wave system remains constant. This concept becomes evident when comparing kinetic and potential energy in waves. Both kinds of energy can transform into one another but the total energy stays unchanged.
For example, in a sinusoidal wave on a string, we see that:

• The maximum potential energy occurs when kinetic energy is at a minimum, and vice versa.
• \( u_{k}(x, t) = u_{p}(x, t) \) for all \( x \) and \( t \).

Such an interaction occurs under energy conservation principles, where energy is never lost but merely transferred.
This principle can also relate to the instantaneous power of the wave (\( P \)):

• \( P = v( u_{k} + u_{p} ) \)

Here, it's evident that energy conserved in the wave is continuously carried forward as the wave propagates. The total energy is effectively the sum of kinetic plus potential energy times the wave speed, illustrating constant energy flow through the medium.