A metallic rod of length `1m` is rigidly clamped at its end points. Longitudinal stationary waves are setup in the ord in such a way that there six antinodes of displacement wave observed along the rod. The amplitude of the anotinode is `2 xx 10^(-6) m`. Write the equations of the stationary wave and the component waves at the point `0.1 m` from the one end of the rod.
[Young modulus `= 7.5 xx 10^(10)N//m^(2)`, density `= 2500 kg//m^(3)`]

Answer

To solve the problem step-by-step, we will derive the equations for the stationary wave and the component waves at a specific point on the metallic rod.

### Step 1: Determine the Wavelength
Given that there are 6 antinodes along the rod of length \( L = 1 \, \text{m} \), we can determine the wavelength \( \lambda \).

Since the number of antinodes \( n = 6 \), the number of wavelengths \( \lambda \) in the rod is given by:
\[ \text{Number of wavelengths} = \frac{n - 1}{2} = \frac{6 - 1}{2} = 2.5 \]
Thus, the total length of the rod can be expressed as:
\[ L = \text{Number of wavelengths} \times \lambda = 2.5 \lambda \]
From this, we can find the wavelength:
\[ \lambda = \frac{L}{2.5} = \frac{1 \, \text{m}}{2.5} = 0.4 \, \text{m} \]

### Step 2: Calculate the Wave Velocity
The velocity \( v \) of the wave in the rod can be calculated using the formula:
\[ v = \sqrt{\frac{Y}{\rho}} \]
where \( Y = 7.5 \times 10^{10} \, \text{N/m}^2 \) (Young's modulus) and \( \rho = 2500 \, \text{kg/m}^3 \) (density).

Calculating \( v \):
\[ v = \sqrt{\frac{7.5 \times 10^{10}}{2500}} = \sqrt{3 \times 10^{7}} = 5.477 \times 10^{3} \, \text{m/s} \]

### Step 3: Determine the Angular Frequency and Wave Number
The angular frequency \( \omega \) is related to the frequency \( f \) and the wavelength \( \lambda \):
\[ f = \frac{v}{\lambda} = \frac{5.477 \times 10^{3}}{0.4} = 13693 \, \text{Hz} \]
Thus,
\[ \omega = 2 \pi f = 2 \pi \times 13693 \approx 86100 \, \text{rad/s} \]

The wave number \( k \) is given by:
\[ k = \frac{2 \pi}{\lambda} = \frac{2 \pi}{0.4} = 15.707 \, \text{rad/m} \]

### Step 4: Write the Equation of the Stationary Wave
The equation of a stationary wave can be expressed as:
\[ y(x, t) = A \sin(kx) \cos(\omega t) \]
Given the amplitude \( A = 2 \times 10^{-6} \, \text{m} \), the equation becomes:
\[ y(x, t) = 2 \times 10^{-6} \sin(15.707 x) \cos(86100 t) \]

### Step 5: Find the Component Waves at \( x = 0.1 \, \text{m} \)
The component waves can be expressed as:
\[ y_1(x, t) = A \sin(kx + \omega t) \quad \text{and} \quad y_2(x, t) = A \sin(kx - \omega t) \]
Substituting \( x = 0.1 \, \text{m} \):
\[ y_1(0.1, t) = 2 \times 10^{-6} \sin(15.707 \times 0.1 + 86100 t) = 2 \times 10^{-6} \sin(1.5707 + 86100 t) \]
\[ y_2(0.1, t) = 2 \times 10^{-6} \sin(15.707 \times 0.1 - 86100 t) = 2 \times 10^{-6} \sin(1.5707 - 86100 t) \]

### Final Equations
1. **Stationary Wave Equation**:
\[ y(x, t) = 2 \times 10^{-6} \sin(15.707 x) \cos(86100 t) \]
2. **Component Waves at \( x = 0.1 \, \text{m} \)**:
\[ y_1(0.1, t) = 2 \times 10^{-6} \sin(1.5707 + 86100 t) \]
\[ y_2(0.1, t) = 2 \times 10^{-6} \sin(1.5707 - 86100 t) \]