Simple Harmonic Motion Question

[theneedtoknow]

Homework Statement

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.

Homework Equations

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

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

 

[neutrino]

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

 

[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

 

[neutrino]

How did the mg come in?

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

 

[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

 

[neutrino]

Well isn't mg the force of gravity on the mass?
So the enet force would be force of gravity + resporing force Fnet = mg + xk

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.

And how come Fnet = d^2/dt^2? I am missing the link between the two , can you help me out to relate them?

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

cause if Fnet is indeed equal to that, then the relationship should be

mg + xk = -w^2x and then i guess the final relationship between those would be

k = (-w^2x - mg )/x
m = (-w^2x - xk ) / g
w^2 = (-mg - xk )/ x

or...no?

No.

 

[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 I am curious...
Would it be mg + xk = -mw^2x?

 

[neutrino]

OK so then for a horizontal string :
kx = -mw^2x

There is a rather important detail missing.

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

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.

 

[theneedtoknow]

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

 

[neutrino]

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

Exactly.

 

[theneedtoknow]

ahhh excellent thanks a lot for all the help