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 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

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

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

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

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

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

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

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

Key concepts

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