A 5kg collar is attached to a spring . It slides without friction over a horizontal surface. It is displaced from its equilibrium position by 10 cm and released , its maximum speed is `1 ms^(-1)` Calculate
(a) Spring constant
(b) The period of oscillation
(c ) Maximum acceleration of the collar
Hint `: m = 5 kg , A= 0.1 m ,k =?, T =? `or `a=? , v_(m) =1 ms^(-1)`
`v_(m) = Aomega =A sqrt((k)/(m))`
Find k,
` T = 2pi sqrt((m)/(k))`
` a_(max) = omega^(2) A = (k)/(m) A`

Explanation

`m = 5 kg , A = 0.1 m , V_(m) = 1ms^(-1) , k= ? , T =? , a =?`
(i) `V_(m) = A omega = A sqrt((k)/( m))`
`1= 0.1 xx sqrt((k)/(5))`
`100 = (k)/(5)`
` k = 500 Nm^(-1)`
(ii) `T =2pi sqrt((m)/(k)) = 2 xx 3.14 xx sqrt((5)/(500)) =0.63s`
(iii) Acceleration of the collar at displacement `x (t)` from the equilibrium is given by
`a(t) = - omega^(2) x(t) = - (k)/(m) x(t)`
Maximum acceleration is
`a_(max) = omega^(2) A = ( 500 Nm^(-1))/( 5 kg) xx 0.1 m= 10 ms^(-1)`
and it occurs at the extermities.