Question

A 1 kg object is attached to a spring and stretches it 0.2 m on its own. There is no damping in the system, but an external force is present, described by the function F(t) = 6 cos omega t. The object is initially displaced 30 cm downward from equilibrium with no initial velocity and system experiences resonance. Find the displacement of the object at any time t.

Answer 1

Given that the spring is stretched by x = 0.2 m the only force acting on it is mg = 1*9.8 = 9.8 N

So, we have -

kx = 9.8 ,

=> k = 9.8/0.2 = 49 N/m

Now, when the spring oscillates about its equilibrium position the period of oscillation T = $2\pi \sqrt{m/k}$

or $\omega =\sqrt{k/m}$ = sqrt[49] = 7 rads/s

Now when the external force is acting on the spring it is resonance, hence w = 7

Again, the object is initially dispalced by 30 cm , the amplitude A = 30cm

hence the motion of the object is given by

y = 30Cos(7t)   , at t=0   y = 30cm, amplitude

There is mistake in the above, The resonance force which adds up the amplitude is not considered

F = 6Cos(wt)

m =1 kg

So, acceleration, a = F/m = 6Cos(wt)

at t=0   a =6 as the force is in resonance with the object SHM w² = 49

The motion due to the force can be described as

y₁ = ACos(wt)

a = d²y₁/dt² = -Aw²Cos(wt) = -6

A = 6/49 = 0.12 Cos(wt)

Therefore, the resultant motion under the influence of the force

                y    = 30Cos(7t) + 0.12Cos(7t)

                       = 30.12Cos(7t)

So, the requisite displacement of the object at any time t is -

y = 30.12*cos(7t)