Mass Spring System: Solving for Motion Over Time

[frozen7]

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)

\omega = \sqrt{k/m}
\omega = 25.65

x = Acos(\omegat + \phi)
0.015 = 0.12cos(25.65t + \phi)
\phi = -1.38

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

Is it correct?

 

[mukundpa]

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.

 

[frozen7]

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?

 

[mukundpa]

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
x = A cos (\omega t + \phi )
is correct.

When x = 0 ; t = 0.110s
put these values to get \phi

 

[frozen7]

Ya, I did it in this way before also but the value I get is \phi equal to -1.25

But it is actually should be about +1.89 ( from given answer)

 

[mukundpa]

cos \theta = cos \alpha has general solution
2n \pi \pm \alpha

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

 

[frozen7]

when t = 0
x = A

Any more clues?

 

[mukundpa]

The same solution can be written as 3pi/2 - wt = 4.712 - 2.821 = 1.891 this is given in your textbook.

 

[mukundpa]

The solutions of the equation are
\pi /2 - 2.821 and 3 \pi /2 - 2.821
= -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.

 

[frozen7]

OK...Thanks...:)