A transverse wave described by equation `y=0.02sin(x+30t)` (where x and t are in meters and seconds, respectively) is travelling along a wire of area of cross-section `1mm^2` and density `8000kg//m^(2)`. What is the tension in the string?

Options

A. (a)20 N
B. (b)7.2 N
C. (c)30 N
D. (d)14.4 N

Explanation

To find the tension in the string described by the wave equation \( y = 0.02 \sin(x + 30t) \), we can follow these steps:

### Step 1: Identify the wave parameters
The wave equation is given as:
\[ y = 0.02 \sin(x + 30t) \]
From this equation, we can identify:
- Amplitude \( A = 0.02 \, \text{m} \)
- Angular frequency \( \omega = 30 \, \text{rad/s} \)

### Step 2: Calculate the wave velocity
The wave velocity \( v \) can be determined from the angular frequency using the formula:
\[ v = \frac{\omega}{k} \]
Where \( k \) is the wave number. In this case, since the wave equation is in the form \( y = A \sin(kx + \omega t) \), we can see that \( k = 1 \, \text{m}^{-1} \) (since the coefficient of \( x \) is 1).

Thus, the wave velocity is:
\[ v = \frac{\omega}{k} = \frac{30}{1} = 30 \, \text{m/s} \]

### Step 3: Find the linear density \( \mu \)
The linear density \( \mu \) is defined as:
\[ \mu = \frac{\text{mass}}{\text{length}} = \text{density} \times \text{cross-sectional area} \]
Given:
- Density \( \rho = 8000 \, \text{kg/m}^3 \)
- Cross-sectional area \( A = 1 \, \text{mm}^2 = 1 \times 10^{-6} \, \text{m}^2 \)

Now, calculate \( \mu \):
\[ \mu = \rho \times A = 8000 \, \text{kg/m}^3 \times 1 \times 10^{-6} \, \text{m}^2 = 0.008 \, \text{kg/m} \]

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

Now substituting the values:
\[ T = 0.008 \, \text{kg/m} \times (30 \, \text{m/s})^2 \]
\[ T = 0.008 \times 900 = 7.2 \, \text{N} \]

### Final Answer
The tension in the string is \( T = 7.2 \, \text{N} \).

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