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

Maximum velocity ,` v=A omega = 40 ms^(-1)`
`A= 20 m`
`omega = ( 40)/( 20 ) = 2 rad s^(-1)`
Timer period, `T= ( 2pi)/(omega ) = ( 2pi)/( 2) =3.14 s`
Force of attraction towards centre `=` Restoring force
`= m omega^(2) A`
Time period `T = ( 2pi)/( omega) = (2pi)/(2 ) = 3.14 s`
Force of attraction towards centre `=` Restoring force
`m omega^(2)A`
`= 2 xx 2 xx 2xx 20`
`= 160 N`