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 displacements of particles about their mean positions is.

Options

A. `s-4 xx 10^(6) cos((11 pi)/(2)x)cos(22 pi xx 10^(3)t)`
B. `s=4 xx 10^(-6)cos((11 pi)/(2)x)sin(22 pi xx 10^(3)t)`
C. `s=4 xx 10^(-6)cos(5pi x)cos (20 pi xx 10^(3)t)`
D. `s=4 xx 10^(-6)cos(5 pi x)sin (20 pi xx 10^(3)t)`

Explanation

To find the equation describing the displacements of particles about their mean positions in a metallic rod with longitudinal stationary waves, we can follow these steps:

### Step 1: Understand the Setup
We have a metallic rod of length \( L = 1 \, \text{m} \) with one end free and the other end clamped. There are 6 antinodes present along the rod. The amplitude of the antinode is given as \( A = 4 \times 10^{-6} \, \text{m} \).

### Step 2: Determine the Wavelength
In a rod with \( n \) antinodes, the relationship between the number of antinodes and the wavelength \( \lambda \) is given by:
\[ L = \frac{n \lambda}{2} \]
For \( n = 6 \):
\[ L = \frac{6 \lambda}{2} \implies \lambda = \frac{2L}{6} = \frac{L}{3} = \frac{1}{3} \, \text{m} \]

### Step 3: Calculate the Speed of the Wave
The speed of longitudinal waves in the rod can be calculated using the formula:
\[ v = \sqrt{\frac{Y}{\rho}} \]
where \( Y \) is the Young's modulus and \( \rho \) is the density of the rod. Given:
- \( Y = 6.4 \times 10^{10} \, \text{N/m}^2 \)
- \( \rho = 4 \times 10^3 \, \text{kg/m}^3 \)

Calculating the speed:
\[ v = \sqrt{\frac{6.4 \times 10^{10}}{4 \times 10^3}} = \sqrt{1.6 \times 10^7} \approx 4000 \, \text{m/s} \]

### Step 4: Calculate the Frequency
The frequency \( f \) can be calculated using the relationship:
\[ f = \frac{v}{\lambda} \]
Substituting the values:
\[ f = \frac{4000}{\frac{1}{3}} = 12000 \, \text{Hz} \]

### Step 5: Determine the Wave Number \( k \)
The wave number \( k \) is given by:
\[ k = \frac{2\pi}{\lambda} \]
Substituting the value of \( \lambda \):
\[ k = \frac{2\pi}{\frac{1}{3}} = 6\pi \, \text{m}^{-1} \]

### Step 6: Determine Angular Frequency \( \omega \)
The angular frequency \( \omega \) is related to the frequency by:
\[ \omega = 2\pi f \]
Calculating \( \omega \):
\[ \omega = 2\pi \times 12000 \approx 24000\pi \, \text{rad/s} \]

### Step 7: Write the Equation of the Stationary Wave
The general form of the equation of a stationary wave is:
\[ y(x, t) = A \cos(kx) \sin(\omega t + \phi) \]
Given that at \( t = 0 \), the particles at the free end (origin) are at their positive extreme, we have \( \phi = \frac{\pi}{2} \).

Thus, the equation becomes:
\[ y(x, t) = 4 \times 10^{-6} \cos(6\pi x) \sin(24000\pi t + \frac{\pi}{2}) \]

Using the identity \( \sin(\theta + \frac{\pi}{2}) = \cos(\theta) \):
\[ y(x, t) = 4 \times 10^{-6} \cos(6\pi x) \cos(24000\pi t) \]

### Final Equation
The final equation describing the displacements of particles about their mean positions is:
\[ y(x, t) = 4 \times 10^{-6} \cos(6\pi x) \cos(24000\pi t) \]

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