The displacement of a particle executing SHM is given by
`y=0.2sin50pi(t+0.01)`metre
Calculate the amplitude,t he period, maximum velocity and the displacement at the start of the motion.

Answer

To solve the problem, we will follow these steps:

### Step 1: Identify the Amplitude
The displacement of the particle is given by the equation:
\[ y = 0.2 \sin(50\pi(t + 0.01)) \]

In the standard form of SHM, \( y = A \sin(\omega t + \phi) \), the amplitude \( A \) is the coefficient of the sine function.

**Solution:**
From the equation, we can see that:
\[ A = 0.2 \, \text{m} \]

### Step 2: Calculate the Angular Frequency \( \omega \)
The angular frequency \( \omega \) is the coefficient of \( t \) in the sine function.

**Solution:**
From the equation:
\[ \omega = 50\pi \, \text{rad/s} \]

### Step 3: Calculate the Time Period \( T \)
The time period \( T \) is related to the angular frequency by the formula:
\[ T = \frac{2\pi}{\omega} \]

**Solution:**
Substituting the value of \( \omega \):
\[ T = \frac{2\pi}{50\pi} = \frac{2}{50} = 0.04 \, \text{s} \]

### Step 4: Calculate the Maximum Velocity \( V_{\text{max}} \)
The maximum velocity in SHM can be calculated using the formula:
\[ V_{\text{max}} = \omega A \]

**Solution:**
Substituting the values of \( \omega \) and \( A \):
\[ V_{\text{max}} = (50\pi)(0.2) = 10\pi \approx 31.4 \, \text{m/s} \]

### Step 5: Calculate the Displacement at the Start of Motion
To find the displacement at the start of motion, we evaluate \( y \) at \( t = 0 \).

**Solution:**
Substituting \( t = 0 \) into the displacement equation:
\[ y(0) = 0.2 \sin(50\pi(0 + 0.01)) = 0.2 \sin(0.5\pi) \]
Since \( \sin(0.5\pi) = 1 \):
\[ y(0) = 0.2 \times 1 = 0.2 \, \text{m} \]

### Summary of Results:
1. Amplitude \( A = 0.2 \, \text{m} \)
2. Time Period \( T = 0.04 \, \text{s} \)
3. Maximum Velocity \( V_{\text{max}} \approx 31.4 \, \text{m/s} \)
4. Displacement at the start of motion \( y(0) = 0.2 \, \text{m} \)

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