The equation of a transverse wave travelling on a rope given by `y=10sinpi(0.01x-2.00t)` whrer y and x are in cm and t in second .This maximum traverse speed of a particle in the rope is about

Options

A. 63 cm/s
B. 75cm/s
C. 100cm/s
D. 121cm/s

Answer

To find the maximum transverse speed of a particle in the rope given the wave equation \( y = 10 \sin(\pi(0.01x - 2.00t)) \), we can follow these steps:

### Step 1: Identify the parameters from the wave equation
The standard form of a transverse wave equation is:
\[ y = A \sin(kx - \omega t) \]
From the given equation \( y = 10 \sin(\pi(0.01x - 2.00t)) \), we can identify:
- Amplitude \( A = 10 \) cm
- Wave number \( k = \pi \times 0.01 \) (which is \( 0.01\pi \) cm\(^{-1}\))
- Angular frequency \( \omega = 2\pi \) rad/s

### Step 2: Determine the maximum transverse speed
The maximum transverse speed \( V_{p \text{ max}} \) of a particle in the wave can be calculated using the formula:
\[ V_{p \text{ max}} = A \cdot \omega \]
Substituting the values we found:
- \( A = 10 \) cm
- \( \omega = 2\pi \) rad/s

### Step 3: Calculate \( V_{p \text{ max}} \)
Now, substituting the values into the formula:
\[ V_{p \text{ max}} = 10 \cdot (2\pi) = 20\pi \text{ cm/s} \]
Using \( \pi \approx 3.14 \):
\[ V_{p \text{ max}} \approx 20 \cdot 3.14 = 62.8 \text{ cm/s} \]

### Final Answer
Thus, the maximum transverse speed of a particle in the rope is approximately:
\[ \boxed{62.8 \text{ cm/s}} \]

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