Standing wave produced in a metal rod of length `1m` is represented by the equation `y=10^(-6)sin.(pix)/(2)sin200pit` where `x` is in metre and t is in seconds. The maximum tensile stress at the mid point of the rod is (Young's modulus of material of rod `=10^(12)N//m^(2)`)

Options

A. `(pi)/(2)xx10^(6)N//m^(2)`
B. `2pixx10^(6)N//m^(2)`
C. `(pi)/(2sqrt(2))xx10^(6)N//m^(2)`
D. `(2pi)/(sqrt(3))xx10^(6)N//m^(2)`

Explanation

To solve the problem of finding the maximum tensile stress at the midpoint of a metal rod described by the standing wave equation \( y = 10^{-6} \sin\left(\frac{\pi x}{2}\right) \sin(200 \pi t) \), we will follow these steps:

### Step 1: Understand the wave equation
The wave equation given is:
\[ y = 10^{-6} \sin\left(\frac{\pi x}{2}\right) \sin(200 \pi t) \]
Here, \( y \) represents the displacement of the wave at position \( x \) and time \( t \).

### Step 2: Identify the parameters
- The amplitude \( A \) of the wave is \( 10^{-6} \) m.
- The wave number \( k \) is \( \frac{\pi}{2} \) (from \( \sin\left(\frac{\pi x}{2}\right) \)).
- The angular frequency \( \omega \) is \( 200 \pi \) rad/s (from \( \sin(200 \pi t) \)).

### Step 3: Calculate the strain
The strain \( \epsilon \) in the rod can be derived from the displacement \( y \):
\[ \epsilon = \frac{dy}{dx} \]
To find \( \frac{dy}{dx} \), we differentiate \( y \) with respect to \( x \):
\[ \frac{dy}{dx} = 10^{-6} \cdot \frac{\pi}{2} \cos\left(\frac{\pi x}{2}\right) \sin(200 \pi t) \]

### Step 4: Evaluate at the midpoint
The midpoint of the rod is at \( x = \frac{L}{2} = \frac{1}{2} \) m. We substitute \( x = \frac{1}{2} \) into the derivative:
\[ \frac{dy}{dx}\bigg|_{x=\frac{1}{2}} = 10^{-6} \cdot \frac{\pi}{2} \cos\left(\frac{\pi}{4}\right) \sin(200 \pi t) \]
Since \( \cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \):
\[ \frac{dy}{dx}\bigg|_{x=\frac{1}{2}} = 10^{-6} \cdot \frac{\pi}{2} \cdot \frac{1}{\sqrt{2}} \sin(200 \pi t) \]

### Step 5: Find maximum tensile stress
The tensile stress \( \sigma \) is given by:
\[ \sigma = Y \cdot \epsilon \]
where \( Y \) is Young's modulus. Given \( Y = 10^{12} \, \text{N/m}^2 \):
\[ \sigma = 10^{12} \cdot 10^{-6} \cdot \frac{\pi}{2\sqrt{2}} \sin(200 \pi t) \]
The maximum tensile stress occurs when \( \sin(200 \pi t) = 1 \):
\[ \sigma_{\text{max}} = 10^{12} \cdot 10^{-6} \cdot \frac{\pi}{2\sqrt{2}} = \frac{\pi \cdot 10^6}{2\sqrt{2}} \, \text{N/m}^2 \]

### Final Answer
Thus, the maximum tensile stress at the midpoint of the rod is:
\[ \sigma_{\text{max}} = \frac{\pi}{2\sqrt{2}} \times 10^6 \, \text{N/m}^2 \]

---