# 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. "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."

I take it that you state the equilibrium point is where the mass would be once the whole system is damped to motionlessness (and zero energy).
Yet you state there is a force there.

Is there not zero force at the equilibrium point? Yet you claim a "restoring" force. Should you not be claiming zero force, zero displacement and maximum absolute value of momentum and velocity (in a live oscillating system)?

In SHM, aren't displacement, velocity, acceleration, jerk etc all variants of + or - sine or cosine functions? Thus with positive values for half a cycle and negative for half. Thus there is no "kicking in", no constant anything, everything is changing back and forth.
As for KE and PE, are they not proportional to the aforemention trig functions squared?

Find your mistake, if you have not already done so, and you can move on?  3. Hi DB, welcome,to the forum! As you stated the force is proportional to the displacement. However the momentum is not zero at the equilibrium point, in fact it's at its maximum. The mass accelerates up the equilibrium point and when it goes past it starts slowing down and stops at the turnaround point (or maximum displacement). Only at the turnaround point is the momentum zero.

Where are you up to with calculus - have you studied differential equations?  4. If we are talking about sympathetic harmonic vibrations here - this gets very complicated very quickly. In either sound (molecular movements in airwaves) or in water (waves on the water) or in electromagnetic waves in the air , the prime frequency creates waves in harmonics which are double, triple (and so forth) in frequency. This can be very pleasing in music as if it is in phase with the original signal it adds to the richness to the timbre and tone, but if out of phase it can cause what is called harmonic distortion.  5. Mayflow, we are not discussing the harmonics, just the simple harmonic motion. Best not to complicate at this point.   6. I don't think I am capable of complicating massless strings. I was already struggling with negative speeds of light and capacitors being the same as resitors as so elequently stated on another topic by someone.  7. Speeds are always positive! Velocities are different - there is a direction involved, it is a vector(like an arrow) and has a direction rather than just a value so depending on what angle you take as positive ( your choice and yours alone! ) the velocity can be positive or negative.  8. Oh so much better. Not negative speed. Negative velocity. So not -.87* the speed of light, but -.87* the velocity of light? I negatively see the difference, so which way and at what negative speed am I moving now?

This is like maybe when in high school someone gave me a beat up old motorcycle. I had friends who knew a bit of mechanics and we were able to fix it up to run, but we got something in backwards and it started up fine and we were all happy and when I put it in gear it went backwards. Still, I do not think that is negative speed. Just a speed in the opposite directtion of my intentions.  9. If you are moving away from me we might say that the velocity is positive, but is you are moving towards me then that would be negative. Velocity is aways relative to an origin and has a direction. We can choose that direction, it up to us which direction is the most useful.  10. Originally Posted by Jilan If you are moving away from me we might say that the velocity is positive, but is you are moving towards me then that would be negative. Velocity is aways relative to an origin and has a direction. We can choose that direction, it up to us which direction is the most useful.
This I think is applying a value judgement as to velocity, but to me velocity is just velocity. you are probably far from alone on the forum though for thinking that moving away from me is a positive velocity, but that is not physics, it is mental value judgements.  11. Mayflow, you are thinking of speed. Speed is just speed. It is just a value. Velocity is different. It is a speed in a certain direction.  12. Yeah, tell it to the cop that stops you for going in excessive velocity. You still get ticket unless you are as cute as me.  13. Might be worth a try when you get old and less cute?  14. Originally Posted by Jilan Might be worth a try when you get old and less cute?
This is a good point. So you are an accountant? (PS) It doesn't always work. Some people are bad cops! : So I can see for you that there are probably incomes and expenses and like deficits and stuff like that? I am pretty good with money. We should discuss this more.  15. Originally Posted by DB. Hello all,

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?  16. Thanks for the answers. They have answered the question and I have yet to do differential equations, so I'll do that.  17. Originally Posted by DB. Thanks for the answers. They have answered the question and I have yet to do differential equations, so I'll do that.
Yes, ODEs are an important part of physics.  Posting Permissions
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