A standing wave `xi= a sin kx. Cos omegat ` is maintained in a homogeneous rod with cross `-` sectional area `S` and density `rho`. Find the total mechanical energy confined between the sections corresponding to the adjacent displacement nodes.

Explanation

To find the total mechanical energy confined between the sections corresponding to adjacent displacement nodes of a standing wave in a homogeneous rod, we can follow these steps:

### Step 1: Identify the Length Between Nodes
The standing wave is given by the equation:
\[ \xi = A \sin(kx) \cos(\omega t) \]
The distance between adjacent displacement nodes is half the wavelength, which can be expressed as:
\[ L = \frac{\lambda}{2} \]

### Step 2: Determine the Kinetic Energy
The kinetic energy \(dT\) of an element of length \(dx\) in the rod can be expressed as:
\[ dT = \frac{1}{2} m v^2 = \frac{1}{2} \rho S dx \left(\frac{\partial \xi}{\partial t}\right)^2 \]
Calculating the time derivative of \(\xi\):
\[ \frac{\partial \xi}{\partial t} = -A \omega \sin(kx) \sin(\omega t) \]
Thus,
\[ \left(\frac{\partial \xi}{\partial t}\right)^2 = A^2 \omega^2 \sin^2(kx) \sin^2(\omega t) \]
Substituting this back into the expression for kinetic energy:
\[ dT = \frac{1}{2} \rho S dx \cdot A^2 \omega^2 \sin^2(kx) \sin^2(\omega t) \]

### Step 3: Integrate Kinetic Energy Over One Node Length
To find the total kinetic energy \(T\) between two nodes, integrate \(dT\) from \(0\) to \(\frac{\lambda}{2}\):
\[ T = \int_0^{\frac{\lambda}{2}} \frac{1}{2} \rho S A^2 \omega^2 \sin^2(kx) \sin^2(\omega t) \, dx \]
Using \(k = \frac{2\pi}{\lambda}\), we can rewrite the integral:
\[ T = \frac{1}{2} \rho S A^2 \omega^2 \sin^2(\omega t) \int_0^{\frac{\lambda}{2}} \sin^2(kx) \, dx \]
The integral \(\int_0^{\frac{\lambda}{2}} \sin^2(kx) \, dx\) evaluates to \(\frac{\lambda}{4}\):
\[ T = \frac{1}{2} \rho S A^2 \omega^2 \sin^2(\omega t) \cdot \frac{\lambda}{4} \]

### Step 4: Determine the Potential Energy
The potential energy \(dU\) can be expressed as:
\[ dU = \frac{1}{2} \rho S \left(\frac{\partial \xi}{\partial x}\right)^2 dx \]
Calculating the spatial derivative:
\[ \frac{\partial \xi}{\partial x} = A k \cos(kx) \cos(\omega t) \]
Thus,
\[ \left(\frac{\partial \xi}{\partial x}\right)^2 = A^2 k^2 \cos^2(kx) \cos^2(\omega t) \]
Substituting this into the expression for potential energy:
\[ dU = \frac{1}{2} \rho S dx \cdot A^2 k^2 \cos^2(kx) \cos^2(\omega t) \]

### Step 5: Integrate Potential Energy Over One Node Length
To find the total potential energy \(U\) between two nodes, integrate \(dU\):
\[ U = \int_0^{\frac{\lambda}{2}} \frac{1}{2} \rho S A^2 k^2 \cos^2(kx) \cos^2(\omega t) \, dx \]
The integral \(\int_0^{\frac{\lambda}{2}} \cos^2(kx) \, dx\) also evaluates to \(\frac{\lambda}{4}\):
\[ U = \frac{1}{2} \rho S A^2 k^2 \cos^2(\omega t) \cdot \frac{\lambda}{4} \]

### Step 6: Total Mechanical Energy
The total mechanical energy \(E\) is the sum of kinetic and potential energy:
\[ E = T + U \]
Substituting the expressions for \(T\) and \(U\):
\[ E = \left(\frac{1}{2} \rho S A^2 \omega^2 \sin^2(\omega t) \cdot \frac{\lambda}{4}\right) + \left(\frac{1}{2} \rho S A^2 k^2 \cos^2(\omega t) \cdot \frac{\lambda}{4}\right) \]

### Final Expression
Using \(k = \frac{2\pi}{\lambda}\) and simplifying, we can express the total mechanical energy confined between the sections corresponding to the adjacent displacement nodes.