A transverse wave travelling on a taut string is represented by:
`Y=0.01 sin 2 pi(10t-x)`
Y and x are in meters and t in seconds. Then,

Options

A. The speed of the wave is `10m//s`.
B. closet points on the string which differ in phase by `60^(@)` are `(1//6)m` apart
C. maximum particle velocity is `pi//4 m//s`
D. the phase of a certain point on the string changes by `120^(@)` in `(1//20)` seconds.

Explanation

To solve the problem step by step, we will analyze the given wave equation and extract the necessary information to answer the questions.

### Given Wave Equation:
\[ Y = 0.01 \sin(2\pi(10t - x)) \]

### Step 1: Identify Parameters
The wave equation can be rewritten in the standard form:
\[ Y = A \sin(\omega t - kx) \]
where:
- \( A = 0.01 \) m (amplitude)
- \( \omega = 20\pi \) rad/s (angular frequency)
- \( k = 2\pi \) rad/m (wave number)

### Step 2: Calculate Wave Speed
The speed of the wave \( v \) can be calculated using the formula:
\[ v = \frac{\omega}{k} \]
Substituting the values:
\[ v = \frac{20\pi}{2\pi} = 10 \text{ m/s} \]

### Step 3: Find the Distance for a Phase Difference of 60 Degrees
Given that the phase difference \( \Delta \phi = 60^\circ = \frac{\pi}{3} \) radians, we can find the distance \( \Delta x \) between points on the string that differ in phase by this amount using the formula:
\[ \Delta \phi = \frac{2\pi}{\lambda} \Delta x \]
First, we need to find the wavelength \( \lambda \):
\[ \lambda = \frac{2\pi}{k} = \frac{2\pi}{2\pi} = 1 \text{ m} \]
Now substituting \( \lambda \) into the equation:
\[ \frac{\pi}{3} = \frac{2\pi}{1} \Delta x \]
Solving for \( \Delta x \):
\[ \Delta x = \frac{\pi/3}{2\pi} = \frac{1}{6} \text{ m} \]

### Step 4: Calculate Maximum Particle Velocity
The maximum particle velocity \( V_{max} \) is given by:
\[ V_{max} = A \omega \]
Substituting the values:
\[ V_{max} = 0.01 \times 20\pi = 0.2\pi \text{ m/s} \]
Calculating \( V_{max} \):
\[ V_{max} \approx 0.628 \text{ m/s} \]

### Step 5: Determine Time for a Phase Change of 120 Degrees in 1/20 Seconds
Given \( \Delta \phi = 120^\circ = \frac{2\pi}{3} \) radians, we can find the time interval \( \Delta t \) for this phase change:
Using the relationship:
\[ \Delta \phi = \frac{2\pi}{T} \Delta t \]
Where \( T \) is the period of the wave:
\[ T = \frac{2\pi}{\omega} = \frac{2\pi}{20\pi} = \frac{1}{10} \text{ s} \]
Now substituting into the equation:
\[ \frac{2\pi}{3} = \frac{2\pi}{\frac{1}{10}} \Delta t \]
Solving for \( \Delta t \):
\[ \Delta t = \frac{2\pi/3}{20\pi} = \frac{1}{30} \text{ s} \]

### Summary of Results:
1. Wave speed \( v = 10 \text{ m/s} \)
2. Distance for phase difference of 60 degrees \( \Delta x = \frac{1}{6} \text{ m} \)
3. Maximum particle velocity \( V_{max} \approx 0.628 \text{ m/s} \)
4. Time for a phase change of 120 degrees \( \Delta t = \frac{1}{30} \text{ s} \)