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 \(U_{\text {ave }}(x)=T(\kappa A)^{2}\left(\cos ^{2} \kappa x\right)\)

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: \(v(x, t) = -2 A \omega \sin(\omega t) \sin(\kappa x)\)

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: \(a(x, t) = -2 A \omega^2 \cos(\omega t) \sin(\kappa x)\)

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: \(K(x, t) = 2 \mu A^2 \omega^2 \sin^2(\omega t) \sin^2(\kappa x)\)

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: \(K_{\text{ave}}(x) = \mu \omega^2 A^2 \sin^2(\kappa x)\)

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: \(U(x, t) = 2 T \kappa^2 A^2 \cos^2(\omega t) \sin^2(\kappa x)\)

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 \(U_{\text {ave }}(x)=T(\kappa A)^{2}\left(\cos ^{2} \kappa x\right)\).