A transverse periodic wave ona strin with a linear mass density of 0.200kg/m is described by the following equations
`y=0.05sin(420t-21.0x)`
where x and y in metres and t is in seconds. Tension in the string is

Options

A. 32N
B. 42N
C. 66N
D. 80N

Answer

To find the tension in the string described by the wave equation \( y = 0.05 \sin(420t - 21.0x) \), we can follow these steps:

### Step 1: Identify the angular frequency and wave number
From the wave equation \( y = 0.05 \sin(420t - 21.0x) \), we can identify:
- Angular frequency \( \omega = 420 \, \text{rad/s} \)
- Wave number \( k = 21.0 \, \text{rad/m} \)

### Step 2: Calculate the wave speed
The wave speed \( v \) can be calculated using the formula:
\[ v = \frac{\omega}{k} \]
Substituting the values:
\[ v = \frac{420}{21.0} = 20 \, \text{m/s} \]

### Step 3: Calculate the wavelength
The wavelength \( \lambda \) can be calculated using the wave number:
\[ \lambda = \frac{2\pi}{k} \]
Substituting the value of \( k \):
\[ \lambda = \frac{2\pi}{21.0} \approx 0.299 \, \text{m} \]

### Step 4: Relate wave speed to tension and linear mass density
We know that the wave speed \( v \) is also related to the tension \( T \) and the linear mass density \( \mu \) by the formula:
\[ v = \sqrt{\frac{T}{\mu}} \]
Squaring both sides gives:
\[ v^2 = \frac{T}{\mu} \]

### Step 5: Solve for tension
Rearranging the equation to solve for tension \( T \):
\[ T = \mu v^2 \]
Substituting the known values (\( \mu = 0.200 \, \text{kg/m} \) and \( v = 20 \, \text{m/s} \)):
\[ T = 0.200 \times (20)^2 = 0.200 \times 400 = 80 \, \text{N} \]

### Final Answer
The tension in the string is \( T = 80 \, \text{N} \).

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