A particle executes SHM on a straight line path. The amplitude of oscialltion is 3 cm. Magnitude of its acceleration is eqal to that of its velocity when its displacement from the mean position is 1 cm. Find the time period of S.H.M.
Hint `: A= 3cm`
When`x=1 cm` , magnitude of velocity `=`magnitude of acceleration
i.e., `omega(A^(2) - x^(2))((1)/(2)) = omega^(2) x`
Find `omega`
`T = ( 2pi)(omega)`

Explanation

To solve the problem step by step, we will use the given information about the particle executing simple harmonic motion (SHM).

### Step 1: Understand the given data
- Amplitude \( A = 3 \, \text{cm} \)
- Displacement from the mean position \( x = 1 \, \text{cm} \)

### Step 2: Write the equations for velocity and acceleration in SHM
The magnitude of velocity \( v \) at a displacement \( x \) is given by:
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