What is the effect of the buoyant force on the harmonic oscillator problem?

PhilT
Messages
1
Reaction score
0
I have the following homework problem that I am having trouble with. Any guidance would be appreciated. Thank you in advance.

Consider an object hanging on a spring, immersed in a cup of water. The water
exerts a linear viscous force -bv on the object, where v is the speed of the object
relative to the liquid.
a. The object is displaced vertically from its equilibrium, and is then
released. The cup remains at rest. Find the equation of motion for the
position of the object.
b. Now, the cup is moved up and down. Assuming the motion of the cup is
simple harmonic motion, find the equation of motion for the position of
the object hanging from the spring. Give your answer in terms of the mass
of the object, the spring constant, the damping constant b, the amplitude
through which the cup is moved, and the frequency with which the cup is
moved. Remember that the viscous force depends on the speed of the
object relative to the liquid.
c. What is the steady-state amplitude of oscillation for the object?
d. Now, in addition to moving the cup up and down, we also drive the object
by moving the suspension end of the spring (the opposite end from the one
that the object is attached to). This drive has the same frequency as the
drive that is moving the cup, but it has a different phase. Find the
equation of motion for the object in this case.
e. If the suspension end of the spring is driven with a particular amplitude
and phase, the object will be motionless in steady state. Find the
amplitude and phase for which this occurs.1. Homework Statement

At equilibrium, when the object is at rest in the water the forces acting on it are
mg down
ky and pVg up, with pVg being a bouyant force that is always acting up.
mg-ky-pVg=0 -> mg=pVg+ky

When it starts to move
mg-k(y+x)-bx'+pVg=mx'' -> x''+γx'+ω^2x=2pVg

I am not confident in the effect that the bouyant force has on the mass. Half of the time it is acting as a damping force, half the time as a restoring force, which makes it seem like it would cancel itself out.

So if the effect of the water is just incorporated into the values of the damping force, γ, then the equation of motion would be the same as it is in air, x''+γx'+ω^2x=0.

please let me know if I am missing something. Thanks again!

Phil
 
Physics news on Phys.org
Hormonic Oscillator Problem
Hormones going up and down aye? I sympathize... have you tried cold showers or playing a sport? ;)

I am not confident in the effect that the bouyant force has on the mass. Half of the time it is acting as a damping force, half the time as a restoring force, which makes it seem like it would cancel itself out.
... how does the buoyant force act as a damping force? Doesn't it always act upwards just like gravity always acts downwards?
 
PhilT said:
mg-k(y+x)-bx'+pVg=mx''

To expand on what Simon said. Here your buoyant force is all of a sudden acting in the same direction as gravity wheras in your previous work, it was acting against... So, yes, it will cancel out if you do things correctly.

The big difference to the case of just air is the value of the damping.
 
Thread 'Need help understanding this figure on energy levels'
This figure is from "Introduction to Quantum Mechanics" by Griffiths (3rd edition). It is available to download. It is from page 142. I am hoping the usual people on this site will give me a hand understanding what is going on in the figure. After the equation (4.50) it says "It is customary to introduce the principal quantum number, ##n##, which simply orders the allowed energies, starting with 1 for the ground state. (see the figure)" I still don't understand the figure :( Here is...
Thread 'Understanding how to "tack on" the time wiggle factor'
The last problem I posted on QM made it into advanced homework help, that is why I am putting it here. I am sorry for any hassle imposed on the moderators by myself. Part (a) is quite easy. We get $$\sigma_1 = 2\lambda, \mathbf{v}_1 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \sigma_2 = \lambda, \mathbf{v}_2 = \begin{pmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \\ 0 \end{pmatrix} \sigma_3 = -\lambda, \mathbf{v}_3 = \begin{pmatrix} 1/\sqrt{2} \\ -1/\sqrt{2} \\ 0 \end{pmatrix} $$ There are two ways...
Back
Top