Resonance of a cube floating on water under external force

My attempt is approaching this problem like the mass spring model. Considering the buoyancy force as spring force. By doing so, we can have the typical mass-spring equation

𝑚 (𝑑^2 𝑥)/(𝑑𝑡^2 )+Fbuoyancy = 𝐹_𝑒𝑥𝑡

Then I can assuming the displacement a will be the sinusoidal function

a=a_𝑀cos⁡(𝜔𝑡+𝜑)

The only problem that I am confusing how to represent Fbuoyancy as a function of displacement a. If I can do so, I think I can finish the problem.

Hope to hear from you guys soon.

What did Archimedes say?

Hill said:

Archimedes

Gordianus said:

What did Archimedes say?

Thank you for quick reply. This is my attempt to represent Fb as a function of displacement a

But I am not sure is it right or not. Cause If I substitute Fb onto mass-spring equation. Due to additional mg, the result is not as I expected.
Hope to hear from you guys opinion.
Thank you.

k makes things worse. Choose a reference system such that "a" carries the right sign

Gordianus said:

k makes things worse. Choose a reference system such that "a" carries the right sign

Thank you, this is my attempt. Not sure it make sense or not ...

Assuming the force ##F_{ext}## is upward, the force balance is $$F_{ext}-500g+(1)(1000) dg=500\frac{d^2x}{dt^2}$$where d is the submerged depth and x=-d. At equilibrium, without the external force present, we have: $$-500g+(1)(1000) d_eg=0$$or $$d_e=0.5$$Letting ##y=d-d_e##, the equation reduces to $$F_{ext}+1000gy=-500\frac{d^2y}{dt^2}$$where y is the downward displacement relative to the equilibrium depth.

Pedantry note: I doubt the cube would float upright in its equilibrium position without some horizontal constraints.

Chestermiller said:

Assuming the force ##F_{ext}## is upward, the force balance is $$F_{ext}-500g+(1)(1000) dg=500\frac{d^2x}{dt^2}$$where d is the submerged depth and x=-d. At equilibrium, without the external force present, we have: $$-500g+(1)(1000) d_eg=0$$or $$d_e=0.5$$Letting ##y=d-d_e##, the equation reduces to $$F_{ext}+1000gy=-500\frac{d^2y}{dt^2}$$where y is the downward displacement relative to the equilibrium depth.

Following up on this derivation, if we substitute the given values for F_{ext} and g into the final equation above, we obtain $$\frac{d^2y}{dt^2}+20 y=-0.2\cos{(0.1t)}$$According to this, the natural frequency of oscillation would be ##\sqrt{20}=4.48/ radians/sec## while the forced frequency is only 0.1 radians/second. This tells us that the forcing in changing very slowly, and the cube motion is going to be nearly in phase with the external force and very close to the quasi steady state displacement ##y=0.01 \cos{(0.1t)}##. Let's see how this plays out. After a very short transient, the oscillatory steady state can be represented by $$y=A\cos{(0.1t)}+B\sin{(0.1t)}$$Substituting this into the differential equation, we obtain: $$-0.01 A\cos{(0.1t)}-0.01 B\sin{(0.1t)}+20 A\cos{(0.1t)}+20B\sin{(0.1t)}=-0.2\cos{(0.1t)}$$This gives us B+0 and $$A=-\frac{0.2}{20-0.01}=-0.01$$