Amplitude of oscillations

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1. Jun 20, 2016

cdummie

1. The problem statement, all variables and given/known data
A little amount of sand is spilt over horizontal membrane that oscillates with frequency f=500Hz in vertical plane. If sand grains are jumping to the height h=3mm with respect to the equilibrium position, find amplitude of oscillation of membrane.

2. Relevant equations
ω=2πf

3. The attempt at a solution
I know that there are two forces acting on the sand when it's on the membrane, first one is gravity and the second one is force between sand and membrane. I know that at some point, sand will jump off the membrane, but i don't actually know when. Is it going to happen in the moment when membrane reaches maximum elongation (amplitude) or before and why?

2. Jun 20, 2016

RUber

have you transfered the frequency information into a vertical displacement function yet?
Once you do that, you should be able to see when the membrane and sand particle positions diverge.

Last edited: Jun 20, 2016
3. Jun 20, 2016

cdummie

I am sorry but i don't know how to transfer frequency into vertical displacement function. Can you help me with that?

4. Jun 20, 2016

RUber

Normally it will look like
$A\sin(\omega t)$
Where A is the amplitude you are looking for.

5. Jun 20, 2016

haruspex

While it is on the membrane, they have the same velocity. At what point will the velocities start to diverge? Think about accelerations.

6. Jun 21, 2016

cdummie

Oh, that's what you meant, i didn't knew it's called vertical displacement function, i always refer to it as elongation in function of time, anyway, what's with the starting phase, is it zero or what?

7. Jun 21, 2016

cdummie

Maybe when they reach equilibrium point, but i am not sure, could you explain it to me?

8. Jun 21, 2016

RUber

I don't think it really matters. You are given the maximum height of the bounce, so I would look for maximum upward velocity.
Your position function for the sand would be:
**edited** $p_0 +v_0t-4.9t^2$
where p_0 and v_0 refer to the position and velocity when the sand leaves the membrane.

To find the point of divergence, you want to look at the acceleration of the membrane compared to the acceleration due to gravity.
$a_m = \frac{d^2}{dt^2} A\sin(\omega t)$
And find when $a_m < g$, that will be the point where the two diverge.

For simplicity, I recommend assuming that since omega is large, it will be almost immediately after reaching maximum velocity.
(this can be backed up semi-rigorously by looking at approximations of $\sin x$ for small x).

Thus, use maximum upward velocity as your starting point.
Find out what velocity is required to move something 3mm. Work backward from there to find your amplitude of oscillation.

Last edited: Jun 22, 2016
9. Jun 21, 2016

cdummie

So when $a_m-g<0$ they diverge, since $a_m=-Aω^2sinωt$ it means that $-Aω^2sinωt<g$ but what i am supposed to do next, we will have maximum velocity at equilibrium point, since after reaching equilibrium point it begins slowing down, right? Now, i should, as you said, use maximum velocity as starting point, but i don't know how to use it when i don't know what velocity it is, how can i determine it? I mean i could use the fact that, in order to move something to some height, i need to have force that is equal to sum of gravitational force and inertia force. but i don't know the value of inertia force so i am not sure what can i do here.

10. Jun 21, 2016

RUber

Velocity of the membrane is the first derivative w.r.t. time and will be $A\omega\cos (\omega t)$ which has a maximum of $A\omega$ when $A\sin(\omega t) = 0$.
Impart your particle with initial velocity of $A\omega$ starting from an initial height of 0m. Solve for A.
Time to peak: $v_0 - 9.8t = 0 m/s$
Height at time t: $s(t) = v_0t -4.9t^2 =.003 m$

11. Jun 21, 2016

ehild

Still the membrane exerts an upward force (normal force, N) on the sand grains, they stay on the membrane and move together with it. Find the condition when N becomes zero: at that moment the grains and the membrane separate, as the membrane can not pull downward the sand. Determine the velocity of the sand grains at that moment.

12. Jun 22, 2016

cdummie

Ok, let's see, N will be zero when gravity loses it's effect on the sand, which means, when velocity of the sand (along with the membrane) is such, that it's acceleration is equal to gravity acceleration, just with opposite direction. Is that right?

13. Jun 22, 2016

RUber

Right, except that I would not say "when the velocity ... is such that." It is simply when the acceleration upward is equal to gravity downward.