# Thread: Explaining simple harmonic motion

1. Hello all,
I've been trying to understand SHM, especially with a spring. Suppose we exert a force and pull the spring by a displacement x. The spring then takes t seconds (let's say 4 secs) to return to equilibrium. However at equilibrium, although the restoring force -kx is 0, the object moves past the equilibrium point making x negative, and oscillates. My source of information says that the momentum of the spring is non-zero at equilibrium point and thus continues past.
This makes physical sense to me but when I try to apply mathematics I'm not sure of what to do. I've tried to analyse it like this:

t=0 when the spring is displaced by x. The spring takes four seconds to return to equilibrium. This is when the restoring force kicks in.
At t = 0, the velocity of the mass m is 0, and so is the momentum p.
dp/dt = F the force, so p = the integral from 0 to t (Sorry about this. Still trying to figure out how to insert math symbols) Fdt, where 0 is when t = 0, so at and t is the time we are interested in. The whole is the interval over which we apply the force. therefore ∆p = Ft, the impulse momentum theorem.
However, we have F = -kx, so ∆p = -kxt. This gives p = -kxt + u, where u is the initial momentum and p is the actual momentum.
Using this, we can say that at t = 3, p = -3kx[t], where x[t] is the position at t = 3. The initial momentum would be zero since at t= 0, the momentum was 0.
However at x = 0, which is the equilibrium point at t = 4, applying the above formula gives p = 0. This seems impossible, so there must be something wrong with the approach, or the calculations.
Could anyone please tell me what went wrong and what would be a correct approach to analyse SHM in terms of momentum?

Thanks

2.

3. Originally Posted by DB.
I've tried to analyse it like this:

t=0 when the spring is displaced by x. The spring takes four seconds to return to equilibrium. This is when the restoring force kicks in.
At t = 0, the velocity of the mass m is 0, and so is the momentum p.
dp/dt = F the force, so p = the integral from 0 to t (Sorry about this. Still trying to figure out how to insert math symbols) Fdt, where 0 is when t = 0, so at and t is the time we are interested in. The whole is the interval over which we apply the force. therefore ∆p = Ft, the impulse momentum theorem.
However, we have F = -kx, so ∆p = -kxt. This gives p = -kxt + u, where u is the initial momentum and p is the actual momentum.
Using this, we can say that at t = 3, p = -3kx[t], where x[t] is the position at t = 3. The initial momentum would be zero since at t= 0, the momentum was 0.
However at x = 0, which is the equilibrium point at t = 4, applying the above formula gives p = 0. This seems impossible, so there must be something wrong with the approach, or the calculations.
Could anyone please tell me what went wrong and what would be a correct approach to analyse SHM in terms of momentum?

Thanks
Yes, your math went haywire early on. Let's fix it:

is the corrected equation of motion.

It is an ordinary differential equation degree 2 with the solution:

where

Now, let's use the initial conditions in order to find A and B:

A is a little more complicated:

, so

This says that:

Armed with that we can get the correct function for the momentum:

As you can see, the correct function is very different from what you got. Can you figure out what you did wrong?[/QUOTE]

4. Ah right. Thanks for the correction. I didn't integrate my original function properly. I actually just started learning about solving ODEs, so I suppose I haven't solved such an equation yet. Anyway, I'll continue with ODEs then. Thanks

5. Originally Posted by DB.
Ah right. Thanks for the correction. I didn't integrate my original function properly. I actually just started learning about solving ODEs, so I suppose I haven't solved such an equation yet. Anyway, I'll continue with ODEs then. Thanks
You are welcomed.

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