The equation of a transverse wave propagating in a string is `y=0.02sin(x-30t)`. Hwere x and y are in metre and t is in second. If linear density of the string is `1.3xx10^(-4)kg//m`, then the tension in the string is :

Options

A. 0.12 N
B. 1.2 N
C. 12 N
D. 120 N

Explanation

To find the tension in the string given the wave equation and linear density, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the wave equation**: The given wave equation is
\[ y = 0.02 \sin(x - 30t) \]
Here, \(x\) and \(y\) are in meters, and \(t\) is in seconds.

2. **Extract parameters from the wave equation**:
- The wave number \(k\) can be identified from the equation as the coefficient of \(x\). Thus,
\[ k = 1 \, \text{m}^{-1} \]
- The angular frequency \(\omega\) is the coefficient of \(t\), so
\[ \omega = 30 \, \text{s}^{-1} \]

3. **Calculate the wavelength \(\lambda\)**: The wavelength can be calculated using the formula
\[ \lambda = \frac{2\pi}{k} \]
Substituting the value of \(k\):
\[ \lambda = \frac{2\pi}{1} = 2\pi \, \text{m} \]

4. **Calculate the frequency \(f\)**: The frequency can be calculated using the relation
\[ f = \frac{\omega}{2\pi} \]
Substituting the value of \(\omega\):
\[ f = \frac{30}{2\pi} \approx \frac{30}{6.2832} \approx 4.77 \, \text{Hz} \]

5. **Calculate the wave speed \(v\)**: The wave speed can be calculated using the relation
\[ v = f \cdot \lambda \]
Substituting the values of \(f\) and \(\lambda\):
\[ v = 4.77 \cdot 2\pi \approx 30 \, \text{m/s} \]

6. **Relate wave speed to tension and linear density**: The wave speed is also related to tension \(T\) and linear density \(\mu\) by the formula
\[ v = \sqrt{\frac{T}{\mu}} \]
Rearranging this gives us
\[ T = \mu v^2 \]

7. **Substitute the values to find tension**: Given the linear density \(\mu = 1.3 \times 10^{-4} \, \text{kg/m}\) and the calculated wave speed \(v = 30 \, \text{m/s}\):
\[ T = (1.3 \times 10^{-4}) \cdot (30)^2 \]
\[ T = (1.3 \times 10^{-4}) \cdot 900 \]
\[ T = 1.17 \times 10^{2} \, \text{N} = 117 \, \text{N} \]

### Final Answer:
The tension in the string is \(117 \, \text{N}\).