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.

A 5kg collar is attached to a spring of spring constant 500 Nm^(-1). I

1.	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.
Answer» Solution :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`