The equation of a wave on a string of linear mass density `0.04 kgm^(-1)` is given by
`y = 0.02(m) sin[2pi((t)/(0.04(s)) -(x)/(0.50(m)))]`.
Then tension in the string is

Options

A. 4.0 N
B. 12.5 N
C. 0.5 N
D. 6.25 N

Answer

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

### Step 1: Identify the wave equation parameters
The wave equation is given as:
\[ y = 0.02 \, \text{m} \sin\left[2\pi\left(\frac{t}{0.04 \, \text{s}} - \frac{x}{0.50 \, \text{m}}\right)\right] \]

From this equation, we can identify:
- Amplitude \( A = 0.02 \, \text{m} \)
- Angular frequency \( \omega = 2\pi \left(\frac{1}{0.04}\right) \)
- Wave number \( k = 2\pi \left(\frac{1}{0.50}\right) \)

### Step 2: Calculate angular frequency \( \omega \) and wave number \( k \)
Calculating \( \omega \):
\[ \omega = 2\pi \left(\frac{1}{0.04}\right) = \frac{2\pi}{0.04} = 50\pi \, \text{rad/s} \]

Calculating \( k \):
\[ k = 2\pi \left(\frac{1}{0.50}\right) = \frac{2\pi}{0.50} = 4\pi \, \text{rad/m} \]

### Step 3: Calculate the wave velocity \( v \)
The velocity of the wave \( v \) can be calculated using the relationship:
\[ v = \frac{\omega}{k} \]
Substituting the values of \( \omega \) and \( k \):
\[ v = \frac{50\pi}{4\pi} = \frac{50}{4} = 12.5 \, \text{m/s} \]

### Step 4: Use the wave velocity to find tension \( T \)
The tension \( T \) in the string is related to the wave velocity \( v \) and the linear mass density \( \mu \) by the formula:
\[ T = \mu v^2 \]
Given that the linear mass density \( \mu = 0.04 \, \text{kg/m} \), we can substitute the values:
\[ T = 0.04 \, \text{kg/m} \times (12.5 \, \text{m/s})^2 \]
Calculating \( v^2 \):
\[ v^2 = (12.5)^2 = 156.25 \, \text{m}^2/\text{s}^2 \]
Now substituting back into the tension formula:
\[ T = 0.04 \times 156.25 = 6.25 \, \text{N} \]

### Final Answer
The tension in the string is:
\[ \boxed{6.25 \, \text{N}} \]

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