A weight of 50.0 N is suspended from a spring that has a force constant of 190 N/m. The system is undamped and is subjected to a harmonic driving force of frequency 10.0 Hz, resulting in a forced-motion amplitude of 4.00 cm. Determine the maximum value of the driving force.

Question

N

Answer 1

Force = m a
w = 2 pi f = 62.8 radians/second
Forcing function -kx = m d^2x/dt^2
let forcing function = F sin wt
so F sin w t = k x + m d^2x/dt^2
let resulting motion be x = a sin wt + b cos w t
then
d^2x/dt^2 = - a w^2 sin wt - bw^2 cos wt
so (s is sin and c is cos)
Fsin wt=k(a s wt+b c wt)-m(a w^2 s wt-b w^2 c wt)

sin terms
F = k a - m a w^2
F = a (k-w^2 m)
but
a = .04 meters
k = 190 N/m
m = 50/9.8 = 5.1 kg
w^2 = 3944 rad^2/s^2
so
F = .04(190 - 3944*5.1)
= -797N

the cos terms give you the natural frequency motion at w = sqrt(k/m)
note we are driving at about 60 rad/sec whereas the natural frequency is about 6 rad/sec so it is not going to move much for a large force.