How Do Oscillator Configurations Affect Energy and Damping?

In summary, the conversation discusses the relationship between a mass hung from a spring and the restoring force, as well as the periods and energy of two harmonic oscillators with the same amplitude. The following questions address topics such as effective spring constant, damping, and critical damping. The conversation also delves into the concept of Q and the displacement and velocity of a damped oscillator at a specific time.
  • #1
razor108
3
0

Homework Statement


problemuy7.jpg


For a mass m hung vertically from a spring with spring constant s, the restoring Force is F=-sx where x is the displacement from equilibrium.

a)What are the periods of the two harmonic oscillators i) and ii).
b)Assuming eaqch oscillator is moving with the same Amplitude A which has the greater total energy of oscillation and by what factor is it greater?

Homework Equations



F=-sx
omega=(s/m)^1/2


The Attempt at a Solution



I just don't really know if there is any difference between i) and ii) but there has to be since all the following questions build up on it. But how can I prove that mathematically?
 
Last edited:
Physics news on Phys.org
  • #2
The first thing you need to do is figure out the effective spring constant of those two spring configurations. Analyze how the force depends on the displacement: F = k' x.
 
  • #3
Yes, I got it now.
The series one adds as in series therefore 1/stotal=1/s1+1/s2 and the other one like parallel therefore stotal=s1+s2.
Wouldn't have thought this because experience actually told me that it shouldn't be so much of a difference... I mean ii) has 4 times more total energy than i).

But now I'm stuck at a new problem.

A damped oscillator of mass m=1,6 kg and spring constant s=20N/m
has a damped frequency omega' htat is 99% of the undamped frequency omega.

a) What is the damping constant r?
Attempt:
As far as I know \omega ' = \sqrt{ \omega ^2 - \frac{b}{2m}}
But were to go from here?
edith: Ok, got that now.

therefore b = (0.99^2-1)*(s/m)*(2m)*(-1) which in this case is 0.796.

b)What is the Q of the system?
Attempt:
Q=\sqrt{mass * Spring constant} / r
But again I would have to find r from a) which I can't really figure out.
Ok, with a) answered I got Q to beeing 7.1066

c) Confirm that the system is lightly damped.
I think a system is lightly damped if omega' is about euqal to omega, but this can't really be the answer here. Because that's what's stated anyways since 99% is "about equal" isn't it?

d) What new damping constant r_{new} is required to make the system critically damped?
Again, I couldn't find any definition on the web or in my notes what critically damped means.

f)Using r_{new} calculate the displacement at t=1s given that the displacement is zero and the velocity is 5 ms^-1 at t=0.
No clue here.

Note:How can I insert Tex in this forum?
 
Last edited:

FAQ: How Do Oscillator Configurations Affect Energy and Damping?

What is a spring?

A spring is a mechanical device that is designed to store and release energy through its elastic properties. It typically consists of a coiled or compressed metal or plastic material that can stretch or compress under force.

How does a spring vibrate?

A spring vibrates when it is disturbed from its equilibrium position, causing it to oscillate back and forth. This vibration is due to the restoring force of the spring, which tries to return the spring to its original position.

What is the relationship between a spring's stiffness and its vibration frequency?

The stiffness of a spring is directly proportional to its vibration frequency. This means that as the stiffness of a spring increases, its vibration frequency also increases.

How does the mass attached to a spring affect its vibration?

The mass attached to a spring affects its vibration by changing the natural frequency of the spring. A heavier mass will result in a lower natural frequency, while a lighter mass will result in a higher natural frequency.

Can the length of a spring affect its vibration?

Yes, the length of a spring can affect its vibration. Longer springs tend to have a lower natural frequency, while shorter springs have a higher natural frequency. This is because the length of the spring determines the distance it needs to stretch or compress to complete one full cycle of vibration.

Back
Top