# Homework Help: Mass Spring System

1. Sep 24, 2005

### 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($$\omega$$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?

2. Sep 24, 2005

### 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.

3. Sep 25, 2005

### frozen7

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

How should I find the phi?

4. Sep 25, 2005

### 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.

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

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

5. Sep 25, 2005

### 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)

6. Sep 25, 2005

### 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 ?

7. Sep 25, 2005

### frozen7

when t = 0
x = A

Any more clues?

8. Sep 26, 2005

### mukundpa

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: Sep 26, 2005
9. Sep 26, 2005

### 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.

10. Sep 26, 2005

### frozen7

OK...Thanks...:)