Resonance of a cube floating on water under external force

[tanhanhbi]

Homework Statement

The problem discusses a scenario involving a rigid plastic cube floating on water, with specific assumptions made:

-The cube's dimensions are 1m x 1m x 1m, and its mass is 500kg.
-Only vertical motion is considered.
-The buoyancy force is determined by the cube's submerged volume and gravity (g = 10 m/s^2).
-The water pool is assumed to be infinite, so the water surface remains level as the cube moves.
-Friction is neglected.
-Displacement (a) is measured relative to the neutral position, where a = 0 indicates the cube is at -equilibrium (zero net force).

Find the displacement as a function of time when subjected to an external force of F(t)=100cos(0.1t) N in the vertical direction.

Relevant Equations

𝑚 (𝑑^2 𝑥)/(𝑑𝑡^2 )+mg=𝐹_𝑒𝑥𝑡
Fbuoyancy = -pgV

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.

 

[Hill]

The only problem that I am confusing how to represent Fbuoyancy as a function of displacement a.

Archimedes

 

[Gordianus]

What did Archimedes say?

 

[tanhanhbi]

Archimedes

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.

 

[Gordianus]

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

 

[tanhanhbi]

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 ...

 

[Chestermiller]

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.

 

[haruspex]

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

 

[Chestermiller]

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 $$A=-\frac{0.2}{20-0.01}=-0.01$$