The equation of a transverse wave travelling in a rope is given by `y = 5 sin (4t - 0.02 x)`, where y and x are in cm and time t is in second. Then the maximum transverse speed of wave in the rope is

Options

A. 125 cm/s
B. 200 cm/s
C. 250 cm/s
D. 100 cm/s

Explanation

To find the maximum transverse speed of the wave in the rope given by the equation \( y = 5 \sin(4t - 0.02x) \), we will follow these steps:

### Step 1: Identify the parameters from the wave equation
The given wave equation is in the form:
\[ y = A \sin(\omega t - kx) \]
where:
- \( A \) is the amplitude,
- \( \omega \) is the angular frequency,
- \( k \) is the wave number.

From the equation \( y = 5 \sin(4t - 0.02x) \):
- Amplitude \( A = 5 \) cm,
- Angular frequency \( \omega = 4 \) rad/s,
- Wave number \( k = 0.02 \) rad/cm.

### Step 2: Calculate the wave speed
The speed \( v \) of the wave can be calculated using the formula:
\[ v = \frac{\omega}{k} \]

Substituting the values of \( \omega \) and \( k \):
\[ v = \frac{4}{0.02} \]

### Step 3: Perform the calculation
Calculating \( v \):
\[ v = \frac{4}{0.02} = 200 \text{ cm/s} \]

### Step 4: Determine the maximum transverse speed
The maximum transverse speed \( v_{max} \) of a wave is given by:
\[ v_{max} = A \cdot \omega \]

Substituting the values of \( A \) and \( \omega \):
\[ v_{max} = 5 \cdot 4 = 20 \text{ cm/s} \]

### Conclusion
Thus, the maximum transverse speed of the wave in the rope is:
\[ \boxed{20 \text{ cm/s}} \]

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