Acceleration of a particle in SHM at displacement `x=10 cm` (from the mean position is `a =-2.5 cm//s^(2)`). Find time period of oscillations.

Explanation

To find the time period of oscillations for a particle in simple harmonic motion (SHM) given the acceleration and displacement, we can follow these steps:

### Step 1: Write down the formula for acceleration in SHM
The acceleration \( a \) of a particle in SHM is given by the formula:
\[ a = -\omega^2 x \]
where \( \omega \) is the angular frequency and \( x \) is the displacement from the mean position.

### Step 2: Substitute the known values
From the problem, we know:
- Acceleration \( a = -2.5 \, \text{cm/s}^2 \)
- Displacement \( x = 10 \, \text{cm} \)

Substituting these values into the formula gives:
\[ -2.5 = -\omega^2 \cdot 10 \]

### Step 3: Simplify the equation
Removing the negative signs from both sides, we have:
\[ 2.5 = \omega^2 \cdot 10 \]

### Step 4: Solve for \( \omega^2 \)
Rearranging the equation to isolate \( \omega^2 \):
\[ \omega^2 = \frac{2.5}{10} \]
\[ \omega^2 = 0.25 \]

### Step 5: Calculate \( \omega \)
Taking the square root of both sides to find \( \omega \):
\[ \omega = \sqrt{0.25} = 0.5 \, \text{rad/s} \]

### Step 6: Use \( \omega \) to find the time period \( T \)
The time period \( T \) of oscillation is given by:
\[ T = \frac{2\pi}{\omega} \]
Substituting the value of \( \omega \):
\[ T = \frac{2\pi}{0.5} = 4\pi \, \text{seconds} \]

### Final Answer
The time period of oscillations is:
\[ T = 4\pi \, \text{seconds} \]

---