Mass Spring System: Solving for Motion Over Time

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Homework Help Overview

The discussion revolves around a vertical mass-spring system where a mass is attached to a spring with a known spring constant. Participants are exploring the equation of motion for the system, particularly focusing on the phase angle and the conditions at specific times.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the application of Hooke's Law and the derivation of the motion equation. Questions arise regarding the balance of forces at equilibrium and the interpretation of the phase angle in relation to the motion of the mass.

Discussion Status

There is an ongoing exploration of the phase angle, with some participants providing different values and questioning the assumptions made in the calculations. Guidance is offered regarding the relationship between displacement and time, but no consensus on the correct phase angle has been reached.

Contextual Notes

Participants note the importance of understanding the equilibrium position and the implications of the mass's initial conditions on the phase angle. There is mention of textbook solutions that differ from individual calculations, highlighting potential discrepancies in interpretation.

frozen7
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In a vertical mass spring system, spring with spring constant 250 N/m vibrates with an amplitude of 12cm when 0.38kg hangs from it.
What is the equation describing this motion as a function of time? ( assume the mass passes through the equilibrium point, towards the positive x(upward), at t = 0.110s)

I do it in this way:
Hooke`s Law: F = kx
(0.38)(9.8)= 250 N/m (x)
x = 0.015(m)

[tex]\omega[/tex] = [tex]\sqrt{k/m}[/tex]
[tex]\omega[/tex] = 25.65

x = Acos([tex]\omega[/tex]t + [tex]\phi[/tex])
0.015 = 0.12cos(25.65t + [tex]\phi[/tex])
[tex]\phi[/tex] = -1.38

So, the equation : x = 0.12cos(25.65t -1.38)

Is it correct?
 
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1. Here the force is not mg, Actually mg is balanced by initial streach of the spring when it come in equilibrium position after the mass is attached.

2. F represents the restoring force trying to bring the mass back in the equilibrium position.
 
mukundpa said:
2. F represents the restoring force trying to bring the mass back in the equilibrium position.

The spring extent because of mg. Then why not F = mg?

How should I find the phi?
 
What is the resultant force when the syatem is in equilibrium?

Actually the force F in S.H.M. is the restoring force on the mass when it has displacement x from equilibrium position.

your equation
[tex]x = A cos (\omega t + \phi )[/tex]
is correct.

When x = 0 ; t = 0.110s
put these values to get [tex]\phi[/tex]
 
Ya, I did it in this way before also but the value I get is [tex]\phi[/tex] equal to -1.25

But it is actually should be about +1.89 ( from given answer)
 
[tex]cos \theta = cos \alpha[/tex] has general solution
[tex]2n \pi \pm \alpha[/tex]

which sign is to be taken?
where was the mass at t = 0 ?
 
when t = 0
x = A

Any more clues?
 
The same solution can be written as 3pi/2 - wt = 4.712 - 2.821 = 1.891 this is given in your textbook.
 
Last edited:
The solutions of the equation are
[tex]\pi /2 - 2.821[/tex] and [tex]3 \pi /2 - 2.821[/tex]
= -1.25 and 1.891 radians respectively.
Now think, when t=0.11s; x=0 means at t = 0 the particle is yet to reach the equilibrium position [Where the phase angle (wt + phi) is pi/2] and x is - ve because the time period is 0.225s and 0.11s this is a bit less then half time period. For that the phase angle should be more then pi which is given by the value of phi = 1.891.
 
  • #10
OK...Thanks...:)
 

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