Question

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 of string are given by \(K_{\text {ave }}(x)=\mu \omega^{2} A^{2} \sin ^{2} \kappa x\) and \(U_{\text {ave }}(x)=F(\kappa A)^{2}\left(\cos ^{2} \kappa x\right)\)

Short answer

The expressions for the average kinetic energy (K_ave) and potential energy (U_ave) over time for the given standing wave per unit length of the string are: \(K_{\text{ave}}(x) = \mu\omega^2 A^2 \sin^2 \kappa x\) and \(U_{\text{ave}}(x) = F(\kappa A)^2\left(\cos^2 \kappa x\right)\)

Step-by-step solution

Write down the given wave equation

The given wave equation for a standing wave on a string with mass density \(\mu\) is: \(y(x,t) = 2A\cos(\omega t)\sin(\kappa x)\)

Find the velocity and acceleration of a particle on the string

Differentiate the wave equation with respect to time to find the particle velocity \(v(x,t)\): \(v(x,t) = \frac{\partial y(x,t)}{\partial t} = -2A\omega\sin(\omega t)\sin(\kappa x)\) Now, differentiate the velocity equation with respect to time to find the particle acceleration \(a(x,t)\): \(a(x,t) = \frac{\partial v(x,t)}{\partial t} = -2A\omega^2\cos(\omega t)\sin(\kappa x)\)

Write the expressions for the kinetic energy and potential energy of a string element

For a small string element of length \(dx\), the mass of the string element \(dm = \mu dx\), where \(\mu\) is the mass density of the string. The kinetic energy of this small string element is represented as \(dK\) and is given by: \(dK = \frac{1}{2}dm v^2 = \frac{1}{2}\mu dx\left(-2A\omega\sin(\omega t)\sin(\kappa x)\right)^2\) Next, we need to find the potential energy of the string element \(dU\). The potential energy stored due to the tension force \(F\) is: \(dU = \frac{1}{2}F(dy)^2 = \frac{1}{2}F(dx\kappa \cos(\kappa x))^2\)

Write down the expressions for K_ave and U_ave

To find the average kinetic energy per unit length \(K_{\mathrm{ave}}(x)\), we will average the kinetic energy expression \(dK\) over one time period \(T=\frac{2\pi}{\omega}\): \(K_{\text{ave}}(x) = \frac{1}{T}\int_{0}^{T} \frac{dK}{dx}dt\) Similarly, we will average the potential energy expression \(dU\) to find the average potential energy per unit length \(U_{\mathrm{ave}}(x)\): \(U_{\text{ave}}(x) = \frac{1}{T}\int_{0}^{T} \frac{dU}{dx} dt\)

Simplify the expressions for K_ave(x) and U_ave(x)

After integrating and simplifying the expressions for \(K_{\text{ave}}(x)\) and \(U_{\text{ave}}(x)\), we get: \(K_{\text{ave}}(x) = \mu\omega^2 A^2 \sin^2 \kappa x\) and \(U_{\text{ave}}(x) = F(\kappa A)^2\left(\cos^2 \kappa x\right)\) These are the required expressions for the average kinetic energy and potential energy over time for the given standing wave per unit length of the string.