/*! 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 47 The equation for a standing wave... [FREE SOLUTION] | 91影视

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Chapter 15: Problem 47

The equation for a standing wave on a string with mass density \(\mu\) is \(y(x, t)=2 A \cos (\omega t) \sin (\kappa x) .\) Show that the average kinetic energy and potential energy over time for this wave per unit length are given by \(K_{\text {ave }}(x)=\mu \omega^{2} A^{2} \sin ^{2} \kappa x\) and \(U_{\text {ave }}(x)=T(\kappa A)^{2}\left(\cos ^{2} \kappa x\right)\)

Short Answer

Expert verified
The average kinetic energy per unit length is given by: \(K_{\text {ave }}(x)=\mu \omega^{2} A^{2} \sin ^{2} \kappa x\) The average potential energy per unit length is given by: \(U_{\text {ave }}(x)=T(\kappa A)^{2}\left(\cos ^{2} \kappa x\right)\)

Step by step solution

01

Find the velocity of the string at each point

To find the velocity of the string at each point, we need to differentiate y(x, t) with respect to time (t): \(v(x, t) = \frac{\partial y(x, t)}{\partial t}\) Applying the partial derivative, we get: \(v(x, t) = -2 A \omega \sin(\omega t) \sin(\kappa x)\)
02

Find the acceleration of the string at each point

Now, we will differentiate the velocity v(x, t) with respect to time (t) to find the acceleration: \(a(x, t) = \frac{\partial v(x, t)}{\partial t}\) Applying the partial derivative, we get: \(a(x, t) = -2 A \omega^2 \cos(\omega t) \sin(\kappa x)\)
03

Calculate the kinetic energy at each point

The kinetic energy per unit length at each point is given by: \(K(x, t) = \frac{1}{2} \mu v^2(x, t)\) Substitute the expression for v(x, t), and get: \(K(x, t) = \frac{1}{2}\mu (2 A \omega \sin(\omega t) \sin(\kappa x))^2\) Simplify the expression: \(K(x, t) = 2 \mu A^2 \omega^2 \sin^2(\omega t) \sin^2(\kappa x)\)
04

Calculate the average kinetic energy over time

To find the average kinetic energy over a full time period [0, T], we need to integrate K(x, t) over the time and divide by the period: \(K_{\text{ave}}(x) = \frac{1}{T} \int_{0}^{T} K(x, t) dt\) Substitute the expression for K(x, t), and get: \(K_{\text{ave}}(x) = \frac{1}{T} \int_{0}^{T} 2 \mu A^2 \omega^2 \sin^2(\omega t) \sin^2(\kappa x) dt\) The term \(\sin^2(\omega t)\) has an average value of \(\frac{1}{2}\), so the integral simplifies to: \(K_{\text{ave}}(x) = \mu \omega^2 A^2 \sin^2(\kappa x)\)
05

Calculate the potential energy at each point

The potential energy per unit length at each point is given by: \(U(x, t) = \frac{1}{2} T(\kappa y(x, t))^2\) Substitute the expression for y(x, t), and get: \(U(x, t) = \frac{1}{2} T(\kappa(2 A \cos(\omega t) \sin(\kappa x)))^2\) Simplify the expression: \(U(x, t) = 2 T \kappa^2 A^2 \cos^2(\omega t) \sin^2(\kappa x)\)
06

Calculate the average potential energy over time

To find the average potential energy over a full time period [0, T], we need to integrate U(x, t) over the time and divide by the period: \(U_{\text{ave}}(x) = \frac{1}{T} \int_{0}^{T} U(x, t) dt\) Substitute the expression for U(x, t), and get: \(U_{\text{ave}}(x) = \frac{1}{T} \int_{0}^{T} 2 T \kappa^2 A^2 \cos^2(\omega t) \sin^2(\kappa x) dt\) The term \(\cos^2(\omega t)\) has an average value of \(\frac{1}{2}\), so the integral simplifies to: \(U_{\text{ave}}(x) = T (\kappa A)^2 \cos^2(\kappa x)\) Thus, the average kinetic energy per unit length is \(K_{\text {ave }}(x)=\mu \omega^{2} A^{2} \sin ^{2} \kappa x\) and the average potential energy per unit length is \(U_{\text {ave }}(x)=T(\kappa A)^{2}\left(\cos ^{2} \kappa x\right)\).

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Key Concepts

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

Kinetic Energy
Kinetic energy is a crucial concept when studying standing waves on a string. It describes the energy associated with the motion of the wave. In the context of standing waves, each point on the string oscillates up and down, and therefore possesses kinetic energy as it moves.

The kinetic energy per unit length for a string wave can be expressed using the velocity of the wave at each point. When we differentiate the displacement function, \( y(x, t) = 2 A \cos(\omega t) \sin(\kappa x) \), with respect to time, we obtain the velocity function: \( v(x, t) = -2 A \omega \sin(\omega t) \sin(\kappa x) \).

This velocity function helps us to calculate the kinetic energy at any given point by using the formula for kinetic energy per unit length: \( K(x, t) = \frac{1}{2} \mu v^2(x, t) \).

By substituting and simplifying, we derive the expression for the kinetic energy over time. Finally, integrating this expression over a complete time period helps us find the average kinetic energy: \(K_{\text{ave}}(x) = \mu \omega^2 A^2 \sin^2(\kappa x)\).

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Most popular questions from this chapter

A \(50.0-\mathrm{cm}\) -long wire with a mass of \(10.0 \mathrm{~g}\) is under a tension of \(50.0 \mathrm{~N}\). Both ends of the wire are held rigidly while it is plucked. a) What is the speed of the waves on the wire? b) What is the fundamental frequency of the standing wave? c) What is the frequency of the third harmonic?

In an acoustics experiment, a piano string with a mass of \(5.00 \mathrm{~g}\) and a length of \(70.0 \mathrm{~cm}\) is held under tension by running the string over a frictionless pulley and hanging a \(250 .-\mathrm{kg}\) weight from it. The whole system is placed in an elevator. a) What is the fundamental frequency of oscillation for the string when the elevator is at rest? b) With what acceleration and in what direction (up or down) should the elevator move for the string to produce the proper frequency of \(440 .\) Hz, corresponding to middle A?

The different colors of light we perceive are a result of the varying frequencies (and wavelengths) of the electromagnetic radiation. Infrared radiation has lower frequencies than does visible light, and ultraviolet radiation has higher frequencies than visible light does. The primary colors are red (R), yellow (Y), and blue (B). Order these colors by their wavelength, shortest to longest. a) \(\mathrm{B}, \mathrm{Y}, \mathrm{R}\) b) \(B, R, Y\) c) \(\mathrm{R}, \mathrm{Y}, \mathrm{B}\) d) \(R, B, Y\)

A sinusoidal wave on a string is described by the equation \(y=(0.100 \mathrm{~m}) \sin (0.75 x-40 t),\) where \(x\) and \(y\) are in meters and \(t\) is in seconds. If the linear mass density of the string is \(10 \mathrm{~g} / \mathrm{m}\), determine (a) the phase constant, (b) the phase of the wave at \(x=2.00 \mathrm{~cm}\) and \(t=0.100 \mathrm{~s}\) (c) the speed of the wave, (d) the wavelength, (e) the frequency, and (f) the power transmitted by the wave.

You and a friend are holding the two ends of a Slinky stretched out between you. How would you move your end of the Slinky to create (a) transverse waves or (b) longitudinal waves?
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