A metallic rod of length `1m` has one end free and other end rigidly clamped. Longitudinal stationary waves are set up in the rod in such a way that there are total six antinodes present along the rod. The amplitude of an antinode is `4 xx 10^(-6) m`. young's modulus and density of the rod are `6.4 xx 10^(10) N//m^(2)` and `4 xx 10^(3) Kg//m^(3)` respectively. Consider the free end to be at origin and at `t=0` particles at free end are at positive extreme.
The equation describing stress developed in the rod is

Options

A. `140.8pi xx 10^(4) cos((11)/(2)pi x+pi)cos(22 pi xx 10^(3)t)`
B. `140.8pi xx 10^(4) sin((11)/(2)pi x+pi)cos(22 pi xx 10^(3)t)`
C. `128 pi xx 10^(4) cos(5 pix+pi)cos(20 pi xx 10^(3)t)`
D. `128 pi xx 10^(4) sin(5 pix+pi)cos(20 pi xx 10^(3)t)`

Explanation

To solve the problem, we need to derive the equation describing the stress developed in a metallic rod that is clamped at one end and has longitudinal stationary waves set up in it. Here’s a step-by-step solution:

### Step 1: Understanding the Problem
We have a metallic rod of length \( L = 1 \, \text{m} \) with one end free and the other end clamped. There are 6 antinodes in the rod, and the amplitude of the antinode is given as \( A = 4 \times 10^{-6} \, \text{m} \). The Young's modulus \( Y \) is \( 6.4 \times 10^{10} \, \text{N/m}^2 \) and the density \( \rho \) is \( 4 \times 10^3 \, \text{kg/m}^3 \).

### Step 2: Equation of the Stationary Wave
The equation for the stationary wave can be expressed as:
\[ y(x, t) = A \cos(kx) \cos(\omega t) \]
where \( k \) is the wave number and \( \omega \) is the angular frequency.

Given that there are 6 antinodes, the wave number \( k \) can be calculated as:
\[ k = \frac{n\pi}{L} = \frac{6\pi}{1} = 6\pi \, \text{rad/m} \]
Thus, the equation becomes:
\[ y(x, t) = 4 \times 10^{-6} \cos(6\pi x) \cos(\omega t) \]

### Step 3: Finding the Angular Frequency
To find \( \omega \), we can use the relationship between the speed of sound \( v \), Young's modulus \( Y \), and density \( \rho \):
\[ v = \sqrt{\frac{Y}{\rho}} \]
Calculating \( v \):
\[ v = \sqrt{\frac{6.4 \times 10^{10}}{4 \times 10^3}} = \sqrt{1.6 \times 10^7} \approx 4000 \, \text{m/s} \]
Now, using the relationship \( \omega = vk \):
\[ \omega = 4000 \times 6\pi = 24000\pi \, \text{rad/s} \]
Thus, the equation of the stationary wave is:
\[ y(x, t) = 4 \times 10^{-6} \cos(6\pi x) \cos(24000\pi t) \]

### Step 4: Finding the Strain
Strain \( \epsilon \) is defined as the change in length per unit length, which can be obtained by differentiating the displacement \( y(x, t) \) with respect to \( x \):
\[ \epsilon = \frac{\partial y}{\partial x} \]
Calculating the derivative:
\[ \epsilon = -4 \times 10^{-6} \cdot 6\pi \sin(6\pi x) \cos(24000\pi t) \]
Thus, the strain is:
\[ \epsilon = -24\pi \times 10^{-6} \sin(6\pi x) \cos(24000\pi t) \]

### Step 5: Finding the Stress
Stress \( \sigma \) is given by:
\[ \sigma = Y \cdot \epsilon \]
Substituting the values:
\[ \sigma = 6.4 \times 10^{10} \cdot (-24\pi \times 10^{-6} \sin(6\pi x) \cos(24000\pi t)) \]
Calculating the stress:
\[ \sigma = -1.536 \times 10^5 \pi \sin(6\pi x) \cos(24000\pi t) \]

### Final Equation
Thus, the equation describing the stress developed in the rod is:
\[ \sigma = -1.536 \times 10^5 \pi \sin(6\pi x) \cos(24000\pi t) \]