A simple harmonic oscillator of mass ...?

Question

A simple harmonic oscillator of mass 60 kg is subjected to a net restoring force F = - 5000 N/m * x at displacement x from equilibrium. What is the period of its motion?

Answer 1

F = - 5000 N/m * x
x = displacement from equilibrium. period of its motion?
the general equation of motion of simple harmonic oscillator is given by
x = a sin (wt) ---(1)
where x is displacement, a is its amplitude (maximum displ), and w is its angular speed, t is time. Now

w = 2*pi*f (f-frequency) f = 1/T (T-time period of motion)
w= 2*pi / T ----(2)

its velocity v = dx/dt = a w cos (wt)
its acceleration a = dv/dt = a w^2 [ - sin (wt)] = - w^2 [x] from (1)
the restoring force on this=
F = m*a = - m w^2 [x] -(3)
we are given F = - 5000 x (4) restoring force

Also, this F=ma from equation of motion (sine wave) is the restoring force = - k * x (where equivalent force constant is
k = 5000 (N/m) comparing (3) and (4)

k = m w^2 or
w = sqrt (k/m) = 2*pi / T
T = 2*pi sqrt (m / k) = 2*pi sqrt (60 / 5000)

T = 2*3.14* sqrt (60 / 5000)
=0.6789 secs

Answer 2

f= - 5000x
ma=-5000x
a=-5000x/60 (m=60)
a=-(250/3)* x
compare it with a= -w^2*x
w^2=250/3
time period T=2(pi)/w ,calculate w from above and put it in it