A standing wave is maintained in a homogeneous string of cross-sectional area s and density `rho` . It is formed by the superposition of two waves travelling in opposite directions given by the equation `y_1 = a sin (omega t - kx) " and " y_2 = 2a sin (omeg t + kx)`. The total mechanical energy confined between the sections corresponding to the adjacent antinodes is

Options

A. `(3pi s rho omega^2 a^2)/(2k)`
B. `(pi s rho omega^2 a^2)/(2k)`
C. `(5pi s rho omega^2 a^2)/(2k)`
D. `(2pi s rho omega^2 a^2)/(2k)`

Answer

To solve the problem of finding the total mechanical energy confined between the sections corresponding to adjacent antinodes in a standing wave formed by the superposition of two waves, we can follow these steps:

### Step 1: Identify the Distance Between Adjacent Antinodes
The distance between two adjacent antinodes in a standing wave is given by:
\[ \text{Distance} = \frac{\pi}{k} \]
This is because the distance between two consecutive antinodes corresponds to half the wavelength (\(\lambda/2\)), and since \(\lambda = \frac{2\pi}{k}\), we have:
\[ \frac{\lambda}{2} = \frac{2\pi}{2k} = \frac{\pi}{k} \]

**Hint:** Remember that the distance between antinodes is half the wavelength.

### Step 2: Calculate the Volume of the String Segment
The volume \(V\) of the string segment between two adjacent antinodes can be calculated as:
\[ V = \text{Cross-sectional area} \times \text{Distance} = S \times \frac{\pi}{k} \]

**Hint:** Volume can be found by multiplying the cross-sectional area by the length of the segment.

### Step 3: Determine the Energy Density for Each Wave
The energy density \(U\) for a wave is given by:
\[ U = \frac{1}{2} \rho A^2 \omega^2 \]
For the two waves:
- For \(y_1 = a \sin(\omega t - kx)\), the amplitude \(A_1 = a\):
\[ U_1 = \frac{1}{2} \rho a^2 \omega^2 \]
- For \(y_2 = 2a \sin(\omega t + kx)\), the amplitude \(A_2 = 2a\):
\[ U_2 = \frac{1}{2} \rho (2a)^2 \omega^2 = \frac{1}{2} \rho (4a^2) \omega^2 = 2 \rho a^2 \omega^2 \]

**Hint:** Energy density depends on the square of the amplitude and the density of the medium.

### Step 4: Calculate the Total Energy Density
The total energy density \(U_{\text{total}}\) in the segment is the sum of the energy densities of the two waves:
\[ U_{\text{total}} = U_1 + U_2 = \frac{1}{2} \rho a^2 \omega^2 + 2 \rho a^2 \omega^2 = \frac{1}{2} \rho a^2 \omega^2 + \frac{4}{2} \rho a^2 \omega^2 = \frac{5}{2} \rho a^2 \omega^2 \]

**Hint:** Add the energy densities of both waves to find the total energy density.

### Step 5: Calculate the Total Mechanical Energy
Now, we can find the total mechanical energy \(E\) confined between the sections corresponding to the adjacent antinodes:
\[ E = V \cdot U_{\text{total}} = \left(S \cdot \frac{\pi}{k}\right) \cdot \left(\frac{5}{2} \rho a^2 \omega^2\right) \]
Thus, we have:
\[ E = \frac{5\pi S \rho a^2 \omega^2}{2k} \]

**Hint:** Multiply the volume by the total energy density to get the total energy.

### Final Answer
The total mechanical energy confined between the sections corresponding to the adjacent antinodes is:
\[ E = \frac{5\pi S \rho a^2 \omega^2}{2k} \]