The equation of a transverse wave propagating in a string is given by
`y = 0.02 sin (x + 30t)`
where, `x and y` are in second.
If linear density of the string is `1.3 xx 10^(-4)kg//m`, then the tension in the string is

Options

A. `0.12N`
B. `1.2N`
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 parameters**:
The wave equation is given as:
\[ y = 0.02 \sin(x + 30t) \]
From this equation, we can identify:
- The angular frequency \(\omega = 30 \, \text{rad/s}\)
- The wave number \(k = 1 \, \text{rad/m}\) (since the coefficient of \(x\) is 1).

2. **Calculate the speed of the wave**:
The speed \(v\) of the wave can be calculated using the formula:
\[ v = \frac{\omega}{k} \]
Substituting the values:
\[ v = \frac{30}{1} = 30 \, \text{m/s} \]

3. **Use the relationship between speed, tension, and linear density**:
The speed of a wave on a string is also given by the formula:
\[ v = \sqrt{\frac{T}{\mu}} \]
where \(T\) is the tension in the string and \(\mu\) is the linear density. We can rearrange this formula to solve for tension \(T\):
\[ T = v^2 \cdot \mu \]

4. **Substitute the known values**:
We know:
- \(v = 30 \, \text{m/s}\)
- \(\mu = 1.3 \times 10^{-4} \, \text{kg/m}\)

Now substituting these values into the tension formula:
\[ T = (30)^2 \cdot (1.3 \times 10^{-4}) \]

5. **Calculate the tension**:
\[ T = 900 \cdot (1.3 \times 10^{-4}) = 0.117 \, \text{N} \]

6. **Final result**:
Rounding to two decimal places, we find:
\[ T \approx 0.12 \, \text{N} \]

### Final Answer:
The tension in the string is approximately **0.12 N**.