A transverse wave described by
`y=(0.02m)sin[(1.0m^-1)x+(30s^-1)t]`
propagates on a stretched string having a linear mass density of `1.2xx10^-4 kgm^-1`. Find the tension in the string.

Explanation

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

### Step 1: Identify the wave parameters
The wave equation is given as:
\[ y = (0.02 \, \text{m}) \sin[(1.0 \, \text{m}^{-1})x + (30 \, \text{s}^{-1})t] \]

From this equation, we can identify:
- The wave number \( k = 1.0 \, \text{m}^{-1} \)
- The angular frequency \( \omega = 30 \, \text{s}^{-1} \)

### Step 2: Calculate the wavelength
The wavelength \( \lambda \) can be calculated using the relationship:
\[ k = \frac{2\pi}{\lambda} \]
Rearranging gives:
\[ \lambda = \frac{2\pi}{k} = \frac{2\pi}{1.0} = 2\pi \, \text{m} \]

### Step 3: Calculate the frequency
The frequency \( f \) can be calculated from the angular frequency using:
\[ \omega = 2\pi f \]
Rearranging gives:
\[ f = \frac{\omega}{2\pi} = \frac{30}{2\pi} \]

### Step 4: Calculate the wave speed
The wave speed \( v \) can be calculated using the relationship:
\[ v = f \lambda \]
Substituting the values we have:
\[ v = \left(\frac{30}{2\pi}\right)(2\pi) = 30 \, \text{m/s} \]

### Step 5: Use the wave speed to find tension
The relationship between wave speed \( v \), tension \( T \), and linear mass density \( \mu \) is given by:
\[ v = \sqrt{\frac{T}{\mu}} \]
Rearranging gives:
\[ T = \mu v^2 \]

### Step 6: Substitute the values
We are given the linear mass density:
\[ \mu = 1.2 \times 10^{-4} \, \text{kg/m} \]
Now substituting the values:
\[ T = (1.2 \times 10^{-4}) (30)^2 \]
\[ T = (1.2 \times 10^{-4}) (900) \]
\[ T = 1.08 \times 10^{-1} \, \text{N} \]
\[ T = 10.8 \times 10^{-2} \, \text{N} \]

### Final Answer
The tension in the string is:
\[ T = 1.08 \times 10^{-1} \, \text{N} \]

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