A transverse wave propagating on a stretched string of linear density `3xx10^-4 kg-m^-1` is represented by the equation
`y=0.2sin(1.5x+60t)`
Where x is in metre and t is in second. The tension in the string (in Newton) is

Options

A. 0.24
B. 0.48
C. 1.2
D. 1.8

Answer

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

### Step 1: Identify the given parameters
From the wave equation \( y = 0.2 \sin(1.5x + 60t) \):
- The angular frequency \( \omega = 60 \, \text{rad/s} \)
- The wave number \( k = 1.5 \, \text{rad/m} \)
- The linear density \( \mu = 3 \times 10^{-4} \, \text{kg/m} \)

### Step 2: Calculate the wave velocity
The velocity \( v \) of the wave can be calculated using the relationship between angular frequency and wave number:
\[ v = \frac{\omega}{k} \]
Substituting the values:
\[ v = \frac{60}{1.5} = 40 \, \text{m/s} \]

### Step 3: Use the formula for tension
The tension \( T \) in the string can be calculated using the formula:
\[ T = \mu v^2 \]
Substituting the values of \( \mu \) and \( v \):
\[ T = (3 \times 10^{-4}) \times (40)^2 \]
Calculating \( (40)^2 \):
\[ (40)^2 = 1600 \]
Now substituting this back into the equation for tension:
\[ T = 3 \times 10^{-4} \times 1600 \]
Calculating the product:
\[ T = 4.8 \times 10^{-1} = 0.48 \, \text{N} \]

### Final Answer
The tension in the string is \( T = 0.48 \, \text{N} \).

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