Need help with Oscillation Vertical Spring Problem

Homework Statement Homework Equations

ΣF , ω=√(k/m), f=w/(2pi)

The Attempt at a Solution

I figured since the only forces acting on the block is the Spring force and a normal force in the downward position. Not a hundred percent sure of this though.

For Disc:
ΣFy= Fn - Fg - fs (static) = ma = 0

For Spring:
ΣFy= fs-Fs (Spring) - Fn = ma = 0
-> fs-Fs=Fn

Equating both Spring and Disc:
fs-Fs-Fg-fs = 0
--> ky = mg
k= mg / y= (.002kg)(9.8m/s^2) / (.08m) = .245 N/m

w= sqrt (k/m) = sqrt ((.245N/m)/(.010 kg+.002kg)) = 4.52 rad/s

f= .719 Hz

Is this correct for part a? Can somebody please assist me, thank you!

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"If the oscillation speed is slowly increased..." What on earth does this mean? There is no excitation indicated, so how does the system oscillate at any frequency other than its natural frequency? Not a clear problem statement.

Also, I fail to see any place where the friction between the two masses is relevant at all.

"If the oscillation speed is slowly increased..." What on earth does this mean? There is no excitation indicated, so how does the system oscillate at any frequency other than its natural frequency? Not a clear problem statement.

Also, I fail to see any place where the friction between the two masses is relevant at all.

Hm, well if we assume that the system oscillates at its natural frequency, would my steps be correct in tackling this problem?

Hm, well if we assume that the system oscillates at its natural frequency, would my steps be correct in tackling this problem? I mean, I agree with you since my static frictions that I've calculated in the problem cancelled out...

haruspex
Homework Helper
Gold Member
Vertical plane could be interpreted as side-to-side. Friction would matter then, but the rest would be nonsense.
The only way I can make sense of this question is as follows:
A spring+block+disc oscillates in a vertical line with an amplitude of 8cm.
If the oscillation were any faster than it is, the disc would lose contact with the block.
How fast is the oscillation?​
(and the answer is > 1 Hz).
For part b, the obvious answer is zero. Maybe it means that you should now take the frequency to be some arbitrary value ##\omega## greater than the critical value calculated in part a and obtain an expression for the recontact displacement in terms of that.

For part b, the obvious answer is zero. Maybe it means that you should now take the frequency to be some arbitrary value ω\omega greater than the critical value calculated in part a and obtain an expression for the recontact displacement in terms of that.
I'm sorry, I don't follow what you're saying.

haruspex
Homework Helper
Gold Member
I'm sorry, I don't follow what you're saying.
In part a), we find an oscillation frequency (no idea why it says maximum; minimum would make more sense) that is just sufficient to make the disc lose contact with the block. This will happen at a particular point in the oscillation (where?). If it is only just sufficient then recontact will be almost immediate, so the displacement is effectively zero.
To make sense of part b), I think you need to assume an unknown frequency, ##\omega##, greater than that found in part a), and obtain an expression for the recontact displacement as a function of ##\omega##.

Where does this shockingly worded question come from?

In part a), we find an oscillation frequency (no idea why it says maximum; minimum would make more sense) that is just sufficient to make the disc lose contact with the block. This will happen at a particular point in the oscillation (where?). If it is only just sufficient then recontact will be almost immediate, so the displacement is effectively zero.
To make sense of part b), I think you need to assume an unknown frequency, ##\omega##, greater than that found in part a), and obtain an expression for the recontact displacement as a function of ##\omega##.

Where does this shockingly worded question come from?
Well think of it this way. If you were holding a folder with your hands, and for a brief moment you quickly let go of your hands...where does the folder go? Acceleration is then gravity (9.81m/s^2). Similar goes to this problem.
Okay so you have this spring and its oscillating up and down right? and theres a block attached to it with the disc on top so when the spring oscilates at its highest point it'll come back down where the disc will lose contact with the block for a brief moment so at the brief moment when the disc is away from the block as it oscillates down we have to look for how far the disc will displace to meet again with the block Turns out a=g. Got about 1.7 Hz for part a.

haruspex
Homework Helper
Gold Member
Well think of it this way. If you were holding a folder with your hands, and for a brief moment you quickly let go of your hands...where does the folder go? Acceleration is then gravity (9.81m/s^2). Similar goes to this problem.
Is this in response to my claim that at the critical frequency the recontact displacement will be zero? If so, I don't get the analogy.
Turns out a=g
Right. But a moment later, what will the acceleration of the block be?
Got about 1.7 Hz for part a.
I get 1.76.

Is this in response to my claim that at the critical frequency the recontact displacement will be zero? If so, I don't get the analogy.

Right. But a moment later, what will the acceleration of the block be?

I get 1.76.

Right. But a moment later, what will the acceleration of the block be?
At what point later?
Right. But a moment later, what will the acceleration of the block be?
Are you asking of the acceleration of the block as it reaches its max from the spring or...?