A particle of mass 2 kg is moving with SHM. Its greatest velocity is `40 ms^(-1)` and its amplitude is 20m. Find the time period and the force of attraction towards the centre when the particles is at its greates distance.
Hint `:` Maximum velocity ,`A omega = 40ms^(-1) ,A = 20 cm`, find `omega ` .
Then find `T= ( 2pi)/( omega)` Force of attractino towards centre `=` Restoring force `= m omega^(2) A`

Explanation

To solve the problem step by step, we will find the time period and the force of attraction towards the center when the particle is at its greatest distance (which is equal to the amplitude).

### Step 1: Find Angular Frequency (ω)

Given:
- Maximum velocity (V_max) = 40 m/s
- Amplitude (A) = 20 m

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