The equation of a stationary wave in a metal rod is given by `y=0.92 sin.(pix)/(3)sin1000t`, where x is in cm and t is in second. The maximum tensile stress at a point x = 1 cm is `(npi)/(3)xx10^(8)" dyne cm"^(-2)`. What is the value of n? [Young's modulus of the material of rod is `=8xx10^(11)" dyne cm"^(-2)`]

Answer

To solve the problem, we need to determine the maximum tensile stress at a point in the metal rod given the equation of the stationary wave.

### Step-by-Step Solution:

1. **Identify the Wave Equation**:
The equation of the stationary wave is given as:
\[ y = 0.92 \sin\left(\frac{\pi x}{3}\right) \sin(1000t) \]
where \( x \) is in cm and \( t \) is in seconds.

2. **Differentiate the Wave Equation**:
To find the strain, we need to differentiate \( y \) with respect to \( x \):
\[ \frac{dy}{dx} = 0.92 \sin(1000t) \cdot \frac{d}{dx}\left(\sin\left(\frac{\pi x}{3}\right)\right) \]
Using the chain rule:
\[ \frac{d}{dx}\left(\sin\left(\frac{\pi x}{3}\right)\right) = \cos\left(\frac{\pi x}{3}\right) \cdot \frac{\pi}{3} \]
Therefore,
\[ \frac{dy}{dx} = 0.92 \sin(1000t) \cdot \cos\left(\frac{\pi x}{3}\right) \cdot \frac{\pi}{3} \]

3. **Evaluate at \( x = 1 \) cm**:
Substitute \( x = 1 \) cm into the equation:
\[ \frac{dy}{dx} \bigg|_{x=1} = 0.92 \sin(1000t) \cdot \cos\left(\frac{\pi \cdot 1}{3}\right) \cdot \frac{\pi}{3} \]
The value of \( \cos\left(\frac{\pi}{3}\right) \) is \( \frac{1}{2} \):
\[ \frac{dy}{dx} \bigg|_{x=1} = 0.92 \sin(1000t) \cdot \frac{1}{2} \cdot \frac{\pi}{3} \]
\[ = 0.46 \sin(1000t) \cdot \frac{\pi}{3} \]

4. **Find Maximum Value of \( \frac{dy}{dx} \)**:
The maximum value of \( \sin(1000t) \) is 1, so:
\[ \left(\frac{dy}{dx}\right)_{\text{max}} = 0.46 \cdot \frac{\pi}{3} \]

5. **Calculate the Maximum Tensile Stress**:
Using the relationship between stress, strain, and Young's modulus:
\[ \text{Stress} = Y \cdot \text{Strain} \]
where \( Y = 8 \times 10^{11} \, \text{dyne/cm}^2 \) and strain can be expressed as:
\[ \text{Strain} = \frac{dy}{dx} \]
Thus,
\[ \text{Stress} = 8 \times 10^{11} \cdot \left(0.46 \cdot \frac{\pi}{3}\right) \]
\[ = \frac{8 \times 0.46 \pi}{3} \times 10^{11} \, \text{dyne/cm}^2 \]
\[ = \frac{3.68 \pi}{3} \times 10^{11} \, \text{dyne/cm}^2 \]

6. **Convert Stress to Required Form**:
We need to express the stress in the form \( \frac{n \pi}{3} \times 10^8 \):
\[ \text{Stress} = 3.68 \pi \times 10^{8} \, \text{dyne/cm}^2 \]
To express this in the required form, we can multiply by \( 10^3 \):
\[ = \frac{3680 \pi}{3} \times 10^8 \, \text{dyne/cm}^2 \]

7. **Identify the Value of \( n \)**:
Comparing this with the given form \( \frac{n \pi}{3} \times 10^8 \):
\[ n = 3680 \]

### Final Answer:
The value of \( n \) is \( 3680 \).