What is the maximum acceleration of the particle doing the SHM `gamma=2sin[(pit)/2+phi]` where gamma is in cm?

Options

A. `(pi)/(2) cms^(-2)`
B. `(pi^(2))/(2) cms^(-2)`
C. `(pi)/(4) cms^(-2)`
D. `(pi^(2))/(4) cms^(-2)`

Explanation

To find the maximum acceleration of the particle performing Simple Harmonic Motion (SHM) described by the equation \( \gamma = 2 \sin\left(\frac{\pi t}{2} + \phi\right) \), we can follow these steps:

### Step 1: Identify the parameters from the given equation
The equation of SHM can be compared with the standard form:
\[ y = a \sin(\omega t + \phi) \]
From the given equation \( \gamma = 2 \sin\left(\frac{\pi t}{2} + \phi\right) \), we can identify:
- Amplitude \( a = 2 \) cm
- Angular frequency \( \omega = \frac{\pi}{2} \) rad/s
- Phase constant \( \phi \) (not needed for maximum acceleration)

### Step 2: Write the formula for maximum acceleration
The maximum acceleration \( a_{\text{max}} \) in SHM is given by:
\[ a_{\text{max}} = \omega^2 a \]

### Step 3: Substitute the values of \( \omega \) and \( a \)
Now, substituting the identified values into the formula:
\[ a_{\text{max}} = \left(\frac{\pi}{2}\right)^2 \cdot 2 \]

### Step 4: Calculate \( \omega^2 \)
Calculating \( \omega^2 \):
\[ \omega^2 = \left(\frac{\pi}{2}\right)^2 = \frac{\pi^2}{4} \]

### Step 5: Calculate the maximum acceleration
Now substituting \( \omega^2 \) back into the equation for maximum acceleration:
\[ a_{\text{max}} = \frac{\pi^2}{4} \cdot 2 = \frac{\pi^2}{2} \text{ cm/s}^2 \]

### Conclusion
Thus, the maximum acceleration of the particle is:
\[ \boxed{\frac{\pi^2}{2} \text{ cm/s}^2} \]