The linear density of a vibrating string is `10^(-4) kg//m`. A transverse wave is propagating on the string, which is described by the equation `y=0.02 sin (x+30t)`, where x and y are in metres and time t in seconds. Then tension in the string is

Options

A. 0.09 N
B. 0.36 N
C. 0.9 N
D. 3.6 N

Explanation

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

### Step 1: Identify the wave equation parameters
The wave is described by the equation:
\[ y = 0.02 \sin (x + 30t) \]
From this equation, we can identify:
- Angular frequency \( \omega = 30 \, \text{rad/s} \)
- Wave number \( k \) can be determined from the wave equation format \( y = A \sin(kx - \omega t) \). Here, since we have \( (x + 30t) \), we can rewrite it as \( (x - (-30)t) \), which gives \( k = 1 \, \text{rad/m} \).

### Step 2: Calculate the wave velocity
The wave velocity \( v \) can be calculated using the relationship:
\[ v = \frac{\omega}{k} \]
Substituting the values we found:
\[ v = \frac{30}{1} = 30 \, \text{m/s} \]

### Step 3: Use the relationship between wave velocity, tension, and linear density
The wave velocity is also related to the tension \( T \) in the string and the linear density \( \mu \) by the formula:
\[ v = \sqrt{\frac{T}{\mu}} \]
Rearranging this gives:
\[ T = v^2 \cdot \mu \]

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

Now substituting these values into the equation for tension:
\[ T = (30)^2 \cdot (10^{-4}) \]
\[ T = 900 \cdot 10^{-4} \]
\[ T = 0.09 \, \text{N} \]

### Final Answer
The tension in the string is \( 0.09 \, \text{N} \).

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