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April 18, 2014

Posted by **sir** on Sunday, January 30, 2011 at 5:57pm.

N

- physics -
**Damon**, Sunday, January 30, 2011 at 6:22pmForce = 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.

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