Question

A spring to which is attached a 2.5kg object is stretched 4cm from its equilibrium position and released to perform simple harmonic motion. The velocity of the object is 2m/s as it passes through the equilibrium position. what is the value of spring constant.

Answer 1

F = -k x = m a
let x = A sin (2 pi f t)
then v = A (2pi f) cos (2 pi f t)
a = -A (2 pi f)^2 sin (2 pi f t)

max extension when sin = 1 or (2 pi f t) = pi/2 , pi/2 + pi, etc
then at max extension A = x = 4 cm = 0.04 meter
then v = 4 (2pi f) cos (2 pi f t)
that v is max when the cos = 1 or -1 etc = 2 m/s
2 = 4 (2 pi f)
so
2 pi f = 0.5
so then
x = .04 sin 0.5 t
v = .04*0.5 cos 0.5 t
a = -.04 (0.25)sin 0.5 t
now back to F = m a
kx = -m a
k (.04 sin .05 t) = - 2.5 * -.04(0.25) sin 0.5 t
k = 2.5 * 0.25