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 speed is

Options

A. 10.884 m/s
B. 8.44 m/s
C. 844 m/s
D. none of these

Answer

To find the maximum magnitude of the transverse speed of the wave generated on the string, we can follow these steps:

### Step 1: Identify the given parameters
- Amplitude of the wave, \( A = 1.12 \, \text{cm} = 1.12 \times 10^{-2} \, \text{m} \)
- Frequency of the wave, \( f = 120 \, \text{Hz} \)
- Linear density of the string, \( \mu = 117 \, \text{g/m} = 0.117 \, \text{kg/m} \)
- Mass hanging from the string, \( m = 4.68 \, \text{kg} \)

### Step 2: Calculate the angular frequency \( \omega \)
The angular frequency \( \omega \) is given by the formula:
\[ \omega = 2\pi f \]
Substituting the value of \( f \):
\[ \omega = 2\pi \times 120 = 240\pi \, \text{rad/s} \]

### Step 3: Determine the wave speed \( v \)
The wave speed \( v \) on the string can be calculated using the formula:
\[ v = \sqrt{\frac{T}{\mu}} \]
Where \( T \) is the tension in the string, which can be calculated as:
\[ T = mg = 4.68 \times 9.81 \, \text{N} \approx 45.92 \, \text{N} \]
Now substituting \( T \) and \( \mu \):
\[ v = \sqrt{\frac{45.92}{0.117}} \approx \sqrt{392.24} \approx 19.8 \, \text{m/s} \]

### Step 4: Calculate the maximum transverse speed
The maximum transverse speed \( v_{\text{transverse max}} \) can be calculated using the formula:
\[ v_{\text{transverse max}} = \omega A \]
Substituting the values of \( \omega \) and \( A \):
\[ v_{\text{transverse max}} = (240\pi) \times (1.12 \times 10^{-2}) \]
Calculating this:
\[ v_{\text{transverse max}} = 240 \times 3.14 \times 1.12 \times 10^{-2} \approx 8.44 \, \text{m/s} \]

### Conclusion
The maximum magnitude of the transverse speed is approximately \( 8.44 \, \text{m/s} \).

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