A string wave equation is given y = 0.002 `sin(300t-15x)` and mass density is `(mu=(0.1kg)/(m))`. Then find the tension in the string

Options

A. 30 N
B. 20 N
C. 40 N
D. 45 N

Explanation

To find the tension in the string given the wave equation \( y = 0.002 \sin(300t - 15x) \) and mass density \( \mu = 0.1 \, \text{kg/m} \), we can follow these steps:

### Step 1: Identify the parameters from the wave equation
The wave equation is given in the form:
\[ y = A \sin(\omega t - kx) \]
From the given equation \( y = 0.002 \sin(300t - 15x) \), we can identify:
- Amplitude \( A = 0.002 \, \text{m} \)
- Angular frequency \( \omega = 300 \, \text{rad/s} \)
- Wave number \( k = 15 \, \text{rad/m} \)

### Step 2: Calculate the wave velocity
The wave velocity \( v \) can be calculated using the formula:
\[ v = \frac{\omega}{k} \]
Substituting the values:
\[ v = \frac{300 \, \text{rad/s}}{15 \, \text{rad/m}} = 20 \, \text{m/s} \]

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

### Step 4: Substitute the values to find tension
Now substituting the values of \( v \) and \( \mu \):
\[ T = (20 \, \text{m/s})^2 \cdot 0.1 \, \text{kg/m} \]
\[ T = 400 \cdot 0.1 = 40 \, \text{N} \]

### Final Answer
The tension in the string is \( T = 40 \, \text{N} \).

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