Oscillatory Motion

dekoi

1.) A cube 1.50 cm on edge mounted on the end of strip that lies in vertical plane. Mass of strip is neglible, but the length of the strip is much larger than cube. The other end of strip is clamped on to a stationary frame. A horizontal force of 1.43 N applied to the cube is required to hold it 2.75 cm away from equilibrium. If the cube is released, what is the frequency of vibration?

Absolutely any hints would be appreciated.


2.) A 2kg object is attached to a spring and placed on a horizontal, smooth surface. A horizontal force of 20N is required to hold the object at rest when it is pulled 0.2 m from equil. Object is released from rest with initial position of .2 m . Find:
Force Konstant
Frequency
Max speed.
Max acceleration.
Total Energy
Velocity when position is equal to one third of maximum value is 1.33 m/s.
Acceleration of object when its position is equal to one third of the maximum value.

The questions also had other parts. For these, i found that:
k= 100 N/m
f = 10 Hz
v = 1.4142 m/s
a = 9.99 m/s^2
The velocity when position is equal to one third of maximum value is 1.33 m/s.

However, how do i find the acceleration of object when its position is equal to one third of the maximum value?

Also, it wouldbe nice for someone to check my answers.



Thank you very much.

HallsofIvy

I don't see any way to give useful hints if you don't tell us what you do understand about this problem or what you have tried. Clearly the "strip" is being considered like a spring. Do you know how to find the "spring constant"? If this were part of a differential equations course, you might then be expected to set up and solve the differential equation of motion. If not then you may have been given a formula for period based on the spring constant.

It would have been simpler if you had told us what "k", "f", etc. mean!
I guess that "k" is what you called the "frequncy Konstant" and I would call the "spring constant".
Yes, 20 N/.2 m= 100 N/m.
No, I do NOT get 10Hz for the frequency but I do get "1.4142" (√(2)) for the maximum speed. And I get 10 m/s², not "0.99" for the maximum acceleration.

Since you found the velocity when the position is 1/3 of its maximum, I presum that you found the time, t, when that happens (and I would argue that the question is ambiguous- there are two different answers to that question). You should be able to just plug that into the equation for acceleration.

If you could do all that for problem 2, why aren't you able to do problem 1? It's essentially the same.