The linear density of a vibrating string is `1.3 xx 10^(-4) kg//m` A transverse wave is propagating on the string and is described by the equation `y= 0.021 sin (x + 30 t)` where x and y are measured in meter and t`t` in second the tension in the string is :-

Options

A. 0.12 N
B. 0.48 N
C. 1.20 N
D. 4.80 N

Explanation

To find the tension in the vibrating string, we can follow these steps:

### Step 1: Identify the given parameters
- Linear density (ρ) of the string: \( \rho = 1.3 \times 10^{-4} \, \text{kg/m} \)
- Wave equation: \( y = 0.021 \sin(x + 30t) \)

### Step 2: Extract wave properties from the wave equation
From the wave equation \( y = 0.021 \sin(x + 30t) \), we can identify:
- Wave number \( k \) and angular frequency \( \omega \):
- \( k = 1 \, \text{m}^{-1} \) (coefficient of \( x \))
- \( \omega = 30 \, \text{s}^{-1} \) (coefficient of \( t \))

### Step 3: Calculate the wave velocity (v)
The velocity \( v \) of the wave can be calculated using the formula:
\[ v = \frac{\omega}{k} \]
Substituting the values:
\[ v = \frac{30}{1} = 30 \, \text{m/s} \]

### Step 4: Relate wave velocity to tension and linear density
The relationship between wave velocity \( v \), tension \( T \), and linear density \( \rho \) is given by:
\[ v = \sqrt{\frac{T}{\rho}} \]
Squaring both sides gives:
\[ v^2 = \frac{T}{\rho} \]
Rearranging this equation to find tension \( T \):
\[ T = \rho v^2 \]

### Step 5: Substitute the values to find tension
Now substituting the values of \( \rho \) and \( v \):
\[ T = (1.3 \times 10^{-4}) \times (30)^2 \]
Calculating \( 30^2 \):
\[ 30^2 = 900 \]
Now substituting this back:
\[ T = (1.3 \times 10^{-4}) \times 900 \]

### Step 6: Perform the multiplication
Calculating the tension:
\[ T = 1.3 \times 900 \times 10^{-4} = 1170 \times 10^{-4} = 0.117 \, \text{N} \]

### Final Answer
The tension in the string is approximately:
\[ T \approx 0.12 \, \text{N} \]