Harmonic Motion Maximum Acceleration?

[wmrunner24]

Homework Statement

A mass-spring system oscillates with an amplitude of 0.024m. The spring constant is 250 N/m and the mass is 0.494kg.

Calculate the maximum acceleration of the mass-spring system.

Homework Equations

F=-kx
Ue=1/2kx^2
F=ma

The Attempt at a Solution

So, I really don't have much of an idea of how to do this, but I based it off of calculus, which is not included in this course, so there must be another way to do it. If there is an easier way, please point it out, but also verify or deny this way for my own curiosity please.

So, I said that F_g-F_s=ma

From there:
mg-kx=ma

Solve for a.

$$\frac{mg-kx}{m}$$=a

Then, I differentiated the function in order to determine a maximum.
a'=$$\frac{-k*dx/dt}{m}$$

And from here, I'm stuck. I don't know how to determine dx/dt, or even if I've correctly obtained the derivative. Could someone set me straight? Any help is appreciated.

 

[denverdoc]

As to the calculus, right in theory--you want to find where v=0. But in the meantime you have the eqn f=-kx which suggests the force will be maximal at the extremes of displacement.

 

[denverdoc]

As to the math, this would be appropriate strategy if you have x(t) = ...
In this case, you actually have dx^2/dt^2 + k/m= 0. The solution can be guessed at or found using j, the sqrt (-1), and Euler's identities. It can be expressed as
x(t) = A Sin (wt + phi), the phase angle can be found from initial conditions. Now if you were to differentiiate this with respect to t, and solve for zero, your approach would be spot on. But the question can be answered w/o any of this,.

 

[wmrunner24]

Unfortunately, I haven't had enough calc yet to understand your above post. Is there a way to approach this without it?

 

[denverdoc]

yes: f=-kx when will this be maximal? Hint: think of definition of SHM--what happens to x?

 

[wmrunner24]

Right, when x is at is minimum is when acceleration is at its maximum. Because
mg-kx=ma. So, how do I determine a minimal displacement? If its just 0 (which makes no sense), then g=a. Or are you saying that since x is negative, the negatives cancel so you have mg+kx=ma. So then, when the magnitude of x is a maximum, acceleration is a max?

 

[denverdoc]

Very good.

So this is a hanging mass? If so you have an additional complication where the amplitude varies about the initial stretch you'd have in static equilibrium. Otherwise (horizontal) the displacement will be symmetric and equal to the initial stretch.

 

[wmrunner24]

Yes, hanging, and as this as only an introductory course, there's no variation to amplitude after the initial. And it makes a lot of sense that x has to be maximized for acceleration to hit a max as at x's max it switches directions (v=0) and that was what we determined from the calculus earlier. It's amazing how it all ties together that way.

Anyway, back to the solution, it gives us an amplitude of 0.024m. So, it's:

(0.494*9.81+250*0.024)/0.494=a
And a=21.9557m/s^2
Correct?

 

[denverdoc]

Yes, the assumption was it was in a static equilibrium and then displaced--so we really don't need to worry about that aspect.

The answer is close. The force mg is actually offset by the spring's initial displacement, x_o: this gives rise to
delta x*k=ma. Using delta x allows us to ignore mg.

mg=kXo delta x = x-x_o where x_o is the initial displacement.

Another tip, if the displacement is vertical, use the variable y or h instead of x as in x-y axes.

 

[wmrunner24]

Wait...if the downward direction is positive as indicated, wouldn't that make the displacement positive? so you'd still have mg-kx=ma, and so x's max would achieve a minimum acceleration. I'm confused again.

 

[denverdoc]

The question is asking for maximal acceleration--in this situation the accelerations will be equal and opposite at the extremes of displacement. The magnitude will be the same and the direction will be towards the equilibrium position. Think of this as a horizontal spring.

 

[wmrunner24]

Okay. I think I follow. The system accepted my answer as correct, now I'm just making sure I understand.

So you're saying that the acceleration at the extremes of displacement (max and min) yields a maximum acceleration. So we found it at the lowest point. Why is the force of gravity neglible, exactly?

 

[diazona]

"Maximum acceleration" can be in either direction, up or down. So even if the acceleration reaches a peak in the negative direction, what you're calling a minimum, that's still a "maximum acceleration" for the purposes of the question.

To be completely proper, it really should say "maximum magnitude of acceleration" but we're sort of stuck with the slightly confusing usage.

EDIT: ah, looks like you posted while I was typing...

 

[ideasrule]

wmrunner: I think you're overthinking this.

Imagine the mass at its equilibrium position. Not surprisingly, net force = 0. Move the mass up by "d" meters. Net force will be kd, since the only force that changed is the spring force. Move the mass up by its amplitude, 0.024 m. Net force will be k*0.024 m. The same applies if you move the mass downwards by 0.024 m, only the net force will be in the opposite direction.

 

[denverdoc]

Its not that it is neglible--we cheat be changing the coordinate system ever so slightly: we pretend that displacement is occurring about a zero point when in fact it is oscillating about some positive vertical displacement downward.

The magnitude of this initial displacement offsets the mass.

It is present throughout the oscillation cycle and therefore offsetting the weight throughout the cycle. You can treat this problem w/o doing this, but the results will quantitatively be the same. Look http://www.phy.ntnu.edu.tw/ntnujava/index.php?topic=383.0" The springs oscillation is displaced downward relative to y=0, but velocity is symmetric as is acceleration. That help?

 

[wmrunner24]

Ok. That makes sense. I understand a lot better now. My teacher didn't explain anything beyond Hooke's law is F=-kx, so...this is much needed and appreciated.

Thank you guys a bunch. Happy Holidays.

 

[denverdoc]

You bet, and hey don't stop thinking about calculus, in many problems the approach you suggested would be right on! Personally I think non-calculus physics is harder and more contrived in many ways. Happy holidays to you too.