# Mass Spring System

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($$\omega$$t + $$\phi$$)
0.015 = 0.12cos(25.65t + $$\phi$$)
$$\phi$$ = -1.38

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

Is it correct?

mukundpa
Homework Helper
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?

mukundpa
Homework Helper
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.

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

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

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

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

when t = 0
x = A

Any more clues?

mukundpa
Homework Helper
The same solution can be written as 3pi/2 - wt = 4.712 - 2.821 = 1.891 this is given in your text book.

Last edited:
mukundpa
Homework Helper
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.

OK...Thanks...:)