# Harmonic Oscillation problem

1. Sep 12, 2004

### Dracovich

Ok so this is propably borderline college :) But it is the first college course i'm in so don't bash me if this is too basic for this forum. (btw hi i'm new)

Well the question in the book is as follows:

"A mass m is attatched to the end of a helical spring (spring constant k) which hangs vertically from a fixed support. Show that the mass executes a simple harmonic motion with the T=2*pi*sqrt(m/k) about a point whose displacement below the unextende position of the spring is (m*g)/k".

So i'm suppose to show that the harmonic oscillation is independant of g, and i've been trying a couple of different things but not been very successful. Mostly been rewriting the formulas to see if anything comes to me. Thought perhaps i had something when i had x=(T^2*g)/(4*pi^2)=(mg)/k in which g goes out, but i don't really think that shows anything.

Plus i tried writing up the forces when the spring is fully extended (maximum of x) in each direction which gives F=mg+kx=0 and F=mg-kx=0 but that just basicly gives me the same mg=-kx which was used to begin with to derive x=(mg)/k so i'm not seeing a whole lot of help in that either :/

So perhaps someone here could give me a hint as to what direction i should go in, it would be greatly appreciated :)

2. Sep 12, 2004

### Tide

Does this look familiar?
$$\frac {d^2x}{dt^2} + \frac {k}{m} x = -g$$

3. Sep 12, 2004

### Dracovich

Yeah i at least know the first part, although now i am a bit confused (i should've caught and asked about this in class). Since the textbook says:
$$\frac {d^2x}{dt^2} = a = -kx$$
But isn't -kx suppose to give F not a ? And i'm assuming you took
$$\frac {d^2x}{dt^2} + kx= mg$$
And got
$$\frac {d^2x}{dt^2} + \frac {k}{m} x = -g$$
But i'm afraid i don't see how those two amount to the same, since
$$\frac {d^2x}{dt^2} + kx= 0$$
but mg=F

I just hope i got all the latex correct not very used to that

4. Sep 12, 2004

### Dracovich

Oh wait i think i just got it. So you got

$$\frac {d^2x}{dt^2} + kx = 0$$

As the formula for simple harmonic oscillation, and then the resulting forces are mg+kx=0 which leads to what you just said. Although i'm afraid i don't see where that leads me :/

5. Sep 12, 2004

### Tide

Let's back up. Newton's law of motion says force = mass times acceleration. In your situation the two forces are the spring force (Hooke's Law) and gravity. So
$$m \frac {d^2x}{dt^2} = -kx -mg$$
The k here is the spring constant. Now divide both sides by m to get my previous equation.
$$\frac {d^2x}{dt^2} + \frac {k}{m} x = -g$$
If the g weren't there, could you solve the equation? (Hint: It's a simple harmonic oscilator!) The presence of the constant g on the right side does alter the nature of the differential equation. For example, using the simple change of variables
$$y = x + \frac {mg}{k}$$
gives
$$\frac {d^2y}{dt^2} + \frac {k}{m} y = 0$$

6. Sep 12, 2004

### Dracovich

Yup the solving without the g would be the same way that x(t) was derived through the first differential equation (Asin(wt+theta), and hence obviously the second equation. I'm with you up until you change the variables, i think i get it but i'm still a bit perplexed by it.

Buuut i wrote it all down and will look it over with my studygroup tomorrow, i know they were having problems as well but perhaps this will spark some good conversation which will help me understand a bit further. Thanks a bunch :)

7. Sep 12, 2004

### Tide

Just replace x with
$$y - \frac {mg}{k}$$
in the DE and remember that the second part is a constant!

8. Sep 12, 2004

### Dracovich

Right'o :) Thx a bunch i really appreciate it