Question

A body of mass 500g suspended from the end of a spiral spring which obeys Hooks law, produced an extension of 10cm. If the mass is pulled down a distance of 5cm and released, calculate
1 the force constant of the spring
2 the frequency of oscillation
3 the period of oscillation
4 the angular speed of the body

Answer 1

m = 500 g = 0.500 kg
x = 10 cm = 0.10 m
g = 9.81 m/s^2

F = k x
0.5 * 9.81 = k * 0.10
k = 49 Newtons / meter

F = m a
-k x = m a
if x = A sin (2 pi f t)
v = A (2 pi f) cos (2 pi f t)
a = -A (2 pi f)^2 sin (2 pi f t) = - (2 pi f)^2 x
so
-k = -(2 pi f)^2
(2 pi f)^2 =49
2 pi f = 7 = omega by the way
f = 7 / 2pi = 1.11 Hz
T =1/f
omega = 2 pi f = 7 radians/ second