The equation of a stationary wave in a metal rod is given by y (in cm ) = 0.002 sin `(pi x )/(3) ` sin 1000 t where x is in cm and tis in second. The maximum tensile stress at a point x =1 cm is `(n pi )/(3) xx 10^(8)` dyne `cm^(-2)` .
What is the value of n? [Young's modulus of the material of rod is = ` 8 xx 10^(11) ` dyne `cm^(-2)` ]

Explanation

To solve the problem, we need to find the value of \( n \) in the expression for the maximum tensile stress at a point \( x = 1 \, \text{cm} \) given the equation of the stationary wave.

### Step-by-Step Solution:

1. **Identify the given wave equation:**
The stationary wave is described by the equation:
\[ y(x, t) = 0.002 \sin\left(\frac{\pi x}{3}\right) \sin(1000t) \]
where \( y \) is in cm, \( x \) is in cm, and \( t \) is in seconds.

2. **Differentiate the wave equation with respect to \( x \):**
We need to find \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = 0.002 \cdot \frac{\pi}{3} \cos\left(\frac{\pi x}{3}\right) \sin(1000t) \]

3. **Evaluate \( \frac{dy}{dx} \) at \( x = 1 \, \text{cm} \):**
Substitute \( x = 1 \) into the derivative:
\[ \frac{dy}{dx} \bigg|_{x=1} = 0.002 \cdot \frac{\pi}{3} \cos\left(\frac{\pi \cdot 1}{3}\right) \sin(1000t) \]
Since \( \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \):
\[ \frac{dy}{dx} \bigg|_{x=1} = 0.002 \cdot \frac{\pi}{3} \cdot \frac{1}{2} \sin(1000t) = \frac{0.001\pi}{3} \sin(1000t) \]

4. **Determine the maximum value of \( \frac{dy}{dx} \):**
The maximum value of \( \sin(1000t) \) is 1, so:
\[ \left(\frac{dy}{dx}\right)_{\text{max}} = \frac{0.001\pi}{3} \]

5. **Calculate the maximum tensile stress using Young's modulus:**
The formula for maximum tensile stress \( \sigma \) is given by:
\[ \sigma = Y \cdot \frac{dy}{dx} \]
where \( Y \) is Young's modulus. Given \( Y = 8 \times 10^{11} \, \text{dyne/cm}^2 \):
\[ \sigma_{\text{max}} = 8 \times 10^{11} \cdot \frac{0.001\pi}{3} \]
Simplifying this:
\[ \sigma_{\text{max}} = \frac{8\pi \times 10^{8}}{3} \, \text{dyne/cm}^2 \]

6. **Compare with the given expression for maximum tensile stress:**
The problem states that the maximum tensile stress is:
\[ \sigma_{\text{max}} = \frac{n\pi}{3} \times 10^{8} \, \text{dyne/cm}^2 \]
Setting the two expressions equal:
\[ \frac{8\pi \times 10^{8}}{3} = \frac{n\pi}{3} \times 10^{8} \]

7. **Solve for \( n \):**
Cancel \( \frac{\pi}{3} \times 10^{8} \) from both sides:
\[ 8 = n \]
Thus, the value of \( n \) is:
\[ n = 8 \]

### Final Answer:
The value of \( n \) is \( 8 \).