The acceleration of a particle is `a = - 100x + 50`. It is released from `x = 2`. Here, `a` and `x` are in SI units

Options

A. (a)the particle will perform SHM of amplitude `2m`
B. (b)the particle will perform SHM of amplitude `1.5 m`
C. (c)the particle will perform SHM of time period `0.63 s`
D. (d)the particle will have a maximum velocity of `15 m//s`

Explanation

To solve the problem step by step, we will follow the instructions provided in the video transcript and derive the necessary quantities for the simple harmonic motion (SHM) of the particle.

### Step 1: Understanding the Acceleration Equation
The acceleration of the particle is given by the equation:
\[ a = -100x + 50 \]
This equation indicates that the acceleration depends on the position \( x \) of the particle.

### Step 2: Finding the Mean Position
In SHM, the mean position is where the acceleration is zero. We can set the acceleration equation to zero to find this position:
\[ 0 = -100x + 50 \]
Solving for \( x \):
\[ 100x = 50 \]
\[ x = 0.5 \, \text{m} \]

### Step 3: Determining the Amplitude
The particle is released from \( x = 2 \, \text{m} \). The amplitude \( A \) of the SHM is the maximum displacement from the mean position. The amplitude can be calculated as:
\[ A = \text{Initial position} - \text{Mean position} \]
\[ A = 2 - 0.5 = 1.5 \, \text{m} \]

### Step 4: Finding the Angular Frequency \( \omega \)
We can compare the given acceleration equation with the standard form of acceleration in SHM:
\[ a = -\omega^2 x + K \]
From our equation:
\[ -100x + 50 \]
We can identify:
\[ \omega^2 = 100 \]
Thus,
\[ \omega = \sqrt{100} = 10 \, \text{rad/s} \]

### Step 5: Calculating the Time Period \( T \)
The relationship between angular frequency \( \omega \) and the time period \( T \) is given by:
\[ \omega = \frac{2\pi}{T} \]
Rearranging gives:
\[ T = \frac{2\pi}{\omega} \]
Substituting \( \omega = 10 \):
\[ T = \frac{2\pi}{10} = \frac{\pi}{5} \approx 0.628 \, \text{s} \]

### Step 6: Finding the Maximum Velocity \( V_{max} \)
The maximum velocity in SHM is given by:
\[ V_{max} = \omega A \]
Substituting the values we found:
\[ V_{max} = 10 \times 1.5 = 15 \, \text{m/s} \]

### Summary of Results
1. **Amplitude**: \( 1.5 \, \text{m} \)
2. **Time Period**: \( 0.628 \, \text{s} \) (approximately \( 0.63 \, \text{s} \))
3. **Maximum Velocity**: \( 15 \, \text{m/s} \)