# Simple Harmonic Motion Question

1. Oct 22, 2008

### theneedtoknow

1. The problem statement, all variables and given/known data

A weight of mass m is placed on a string. as a result it exerts a force equal to xk that tends to reduce x, where k is the spring constand and x is the distance from the equilibrium position of the spring.

1. Confirm that x = Asin(wt+ f) is indeed a solution of theequation of SHM. (given below)
2. Write down the relationship between k, m, and w^2.

2. Relevant equations

d^2 / dt ^ 2 = -w^2 * x

3. The attempt at a solution

1. Well i got this part by differentiating x = Asin(wt+f) and indeed it is a solution
2.

I have no idea how to do this part...
All i can think of is that at the peaks and troughs of oscillation a is zero so the forces mg and xk must be equal but that doesn't really help me relate them in general

2. Oct 22, 2008

### neutrino

Think in terms of Newton's second law of motion.

3. Oct 22, 2008

### theneedtoknow

Well i'm tryin to...but not sure how....
Fnet = mg + xk = ma

(ma - mg) / k = x

well this is tthe relationship between m and k and... I doubt its even right... and i have no idea how to go to relate to w^2 of all things

4. Oct 22, 2008

### neutrino

How did the mg come in?

Anyway, Fnet is also equal to m(d2x/dt2).

5. Oct 22, 2008

### theneedtoknow

d^x / dt^2 = a(t) right?
so would it be f = ma
so Fnet = m * d^2/dt^2?

crap, i misread the first time lol

Last edited: Oct 22, 2008
6. Oct 22, 2008

### neutrino

It is not mentioned that the spring is hanging vertically. You must therefore assume that is placed on a smooth horizontal surface. In that case, the force due to gravity and the normal force due to the surface cancel each other.

Please look at the equation I typed once more. If it is still not clear, read your textbook (on Newton's second law).

No.

7. Oct 22, 2008

### theneedtoknow

OK so then for a horizontal string :
kx = -mw^2x
so they are related by

k = -mw^2 (i think)

But can you help me figure out what the answer would be with a vertical spring as well cause now im curious...
Would it be mg + xk = -mw^2x?

8. Oct 22, 2008

### neutrino

There is a rather important detail missing.

Again, the same detail is missing here. But as it turns out, the relation between these three quantities (omega, k and m) is the same no matter how the spring is oriented. This of course assumes ideal conditions, such as a negligible mass for the spring, etc.

9. Oct 22, 2008

### theneedtoknow

is it that its a restorive force so it shud be k = mw^2?

10. Oct 22, 2008

### neutrino

Exactly.

11. Oct 22, 2008

### theneedtoknow

ahhh excellent thanks a lot for all the help