A transverse sinusoidal wave is generted at one end of long, horizontal string by a bar that moves up and down through a distance of `1.00 cm`. The motion is continuous and is repreated regularly `120` times per second. The string has linear density `90 gm//m` and is kept under a tension of `900 N`. Find :
The maximum value of the transverse component of the tension (in newton)

Options

A. `1.8 pi`
B. `10.8 pi`
C. `9`
D. `18 pi`

Answer

To find the maximum value of the transverse component of the tension in the string, we can follow these steps:

### Step 1: Identify the Given Data
- Amplitude of the wave, \( A = 1.00 \, \text{cm} = 0.01 \, \text{m} \) (since \( 1 \, \text{cm} = 0.01 \, \text{m} \))
- Frequency of the wave, \( f = 120 \, \text{Hz} \)
- Linear density of the string, \( \mu = 90 \, \text{g/m} = 0.09 \, \text{kg/m} \) (since \( 90 \, \text{g} = 0.09 \, \text{kg} \))
- Tension in the string, \( T = 900 \, \text{N} \)

### Step 2: Calculate Angular Frequency
The angular frequency \( \omega \) is given by:
\[ \omega = 2 \pi f \]
Substituting the value of \( f \):
\[ \omega = 2 \pi \times 120 \approx 753.98 \, \text{rad/s} \]

### Step 3: Calculate Wave Speed
The wave speed \( v \) can be calculated using the formula:
\[ v = \sqrt{\frac{T}{\mu}} \]
Substituting the values of \( T \) and \( \mu \):
\[ v = \sqrt{\frac{900}{0.09}} = \sqrt{10000} = 100 \, \text{m/s} \]

### Step 4: Calculate Wave Number \( k \)
The wave number \( k \) is given by:
\[ k = \frac{\omega}{v} \]
Substituting the values of \( \omega \) and \( v \):
\[ k = \frac{753.98}{100} \approx 7.5398 \, \text{rad/m} \]

### Step 5: Calculate Maximum Transverse Component of Tension
The maximum transverse component of the tension \( T_y \) can be calculated using:
\[ T_y = T \cdot \frac{dy}{dx} \]
For a sinusoidal wave, the maximum value of \( \frac{dy}{dx} \) is given by:
\[ \frac{dy}{dx} = A \cdot k \]
Thus,
\[ T_y = T \cdot A \cdot k \]
Substituting the values:
\[ T_y = 900 \cdot 0.01 \cdot 7.5398 \]
Calculating this gives:
\[ T_y = 900 \cdot 0.01 \cdot 7.5398 \approx 67.8582 \, \text{N} \]

### Step 6: Final Answer
The maximum value of the transverse component of the tension is approximately:
\[ T_y \approx 67.86 \, \text{N} \]