A transverse sinusoidal wave is generated at one end of a long horizontal string by a bar that moves with an amplitude of 1.12 cm . The motion of the bar is continuous and is repeated regularly 120 times per second. The string has linear density of 117g/m. The other end of the string is attached to a mass 4.68 kg. The string passes over a smooth pulley and the mass attached to the other end of the string hangs freely under gravity.
The maximum magnitude of the transverse component of tension in the string is

Options

A. zero
B. 19.56 N
C. 37.7N
D. 377 N

Answer

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

### Step 1: Determine the Tension in the String
The tension \( T \) in the string is provided by the weight of the mass attached to the other end of the string. The weight can be calculated using the formula:
\[ T = m \cdot g \]
where:
- \( m = 4.68 \, \text{kg} \) (mass)
- \( g = 9.8 \, \text{m/s}^2 \) (acceleration due to gravity)

Calculating this gives:
\[ T = 4.68 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 45.864 \, \text{N} \]

### Step 2: Convert Amplitude to SI Units
The amplitude \( A \) is given as \( 1.12 \, \text{cm} \). We need to convert this into meters:
\[ A = 1.12 \, \text{cm} = 1.12 \times 10^{-2} \, \text{m} \]

### Step 3: Calculate the Linear Density in SI Units
The linear density \( \mu \) is given as \( 117 \, \text{g/m} \). We need to convert this into kilograms per meter:
\[ \mu = 117 \, \text{g/m} = 117 \times 10^{-3} \, \text{kg/m} \]

### Step 4: Calculate the Angular Frequency
The frequency \( f \) is given as \( 120 \, \text{Hz} \). The angular frequency \( \omega \) can be calculated using the formula:
\[ \omega = 2 \pi f \]
Calculating this gives:
\[ \omega = 2 \pi \times 120 \, \text{Hz} = 240 \pi \, \text{rad/s} \]

### Step 5: Calculate the Maximum Transverse Component of Tension
The maximum magnitude of the transverse component of tension can be calculated using the formula:
\[ T_{\text{max}} = T \cdot A \cdot \omega \]
Substituting the values we have:
\[ T_{\text{max}} = 45.864 \, \text{N} \cdot 1.12 \times 10^{-2} \, \text{m} \cdot 240 \pi \, \text{rad/s} \]

Calculating this step-by-step:
1. Calculate \( 240 \pi \):
\[ 240 \pi \approx 753.98 \, \text{rad/s} \]
2. Now calculate \( T_{\text{max}} \):
\[ T_{\text{max}} = 45.864 \cdot 1.12 \times 10^{-2} \cdot 753.98 \]
\[ T_{\text{max}} \approx 45.864 \cdot 0.0112 \cdot 753.98 \approx 19.56 \, \text{N} \]

### Final Answer
The maximum magnitude of the transverse component of tension in the string is approximately:
\[ \boxed{19.56 \, \text{N}} \]