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

Explanation

Spring constant `k= 500 Nm^(-1)`
Mass of collar `m = 5kg`
Amplitude `A = 10 cm = 0.1 m`
The period of oscillation of collar attached at the end of spring.
`T= 2pi sqrt((m)/(k))`
`= 2 xx3.14 xx sqrt((5)/(500)) = 6.28 xx (1)/(sqrt(1000))`
`=(6.28)/(10) = 0.628 approx 0.63s`