/*! 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

Consider a linear array of \(n\) masses, each equal to \(m,\) connected by \(n+1\) springs, all massless and having spring constant \(k\), with the outer ends of the first and last springs fixed. The masses can move without friction in the linear dimension of the array. a) Write the equations of motion for the masses. b) Configurations of motion for which all parts of a system oscillate with the same angular frequency are called normal modes of the system; the corresponding angular frequencies are the system's normal-mode angular frequencies. Find the normal-mode angular frequencies of this array.

One of the main things allowing humans to determine whether a sound is coming from the left or the right is the fact that the sound will reach one ear before the other. Given that the speed of sound in air is \(343 \mathrm{~m} / \mathrm{s}\) and that human ears are typically \(20.0 \mathrm{~cm}\) apart, what is the maximum time resolution for human hearing that allows sounds coming from the left to be distinguished from sounds coming from the right? Why is it impossible for a diver to be able to tell from which direction the sound of a motor boat is coming? The speed of sound in water is \(1.50 \cdot 10^{3} \mathrm{~m} / \mathrm{s}\).

Students in a lab produce standing waves on stretched strings connected to vibration generators. One such wave is described by the wave function \(y(x, t)=(2.00 \mathrm{~cm}) \sin \left[\left(20.0 \mathrm{~m}^{-1}\right) x\right] \cos \left[\left(150 . \mathrm{s}^{-1}\right) t\right],\) where \(y\) is the transverse displacement of the string, \(x\) is the position along the string, and \(t\) is time. Rewrite this wave function in the form for a right- moving and a left-moving wave: \(y(x, t)=\) \(f(x-v t)+g(x+v t)\); that is, find the functions \(f\) and \(g\) and the speed, \(v\)

A ping-pong ball is floating in the middle of a lake and waves begin to propagate on the surface. Can you think of a situation in which the ball remains stationary? Can you think of a situation involving a single wave on the lake in which the ball remains stationary?

Why do circular water waves on the surface of a pond decrease in amplitude as they travel away from the source?
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