A 5 kg collar is attached to a spring of spring constant 500 N `m^(-1)`. It slides without friction over a horizontal rod. The collar is displaced from its equilibrium position by 10.0 cm and released. Calculate
(a) the period of oscillation,
(b) the maximum speed and
(c) maximum acceleration of the collar.

Explanation

(a) The period of oscillation as given by Eq. (14.21) is,
`T=2pisqrt(m/k)=2pisqrt((5.0kg)/(500Nm^(-1)))`
`=(2pi//10)s`
`=0.63s`
(b) The velocity of the collar executing SHM is given by,
`v(t)=Aomega`
`=0.1xxsqrt(k/m)`
`=0.1xxsqrt((500Nm^(-1))/(5kg))`
`=1ms^(-1)`
and it occurs at x = 0
(c) The acceleration of the collar at the displacement x (t ) from the equilibrium is given by,
`a(t)=-omega^(2)xt`
`=-k/mx(t)`
Therefore, the maximum acceleration is,
`a_("max")=omega^(2)A`
`(500Nm^(-1))/(5kg)xx0.1m`
`=10ms^(-2)`
and it occurs at the extremities.