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1. Hi, I'm revising Simple Harmonic Motion in edexcel M3. Whilst I can derive and know how to use acceleration=-w^2x and v^2=w^2(a^2-x^2), I just don't know what difference it makes if I use x=asin(wt) or x=acos(wt). I realise that in deriving them, it's whether x is 0 or 1 at t=0, but when you are not given the initial position but have to work out time (eg the time it takes for a particle to move between two points), I get too confused .

Basically the question is, "when do I use x=asin(wt) and when do I use x=acos(wt)?"

2. For example in this question's part c, I'm supposed to use asinwt; but WHY?
3. (Original post by TommyBrecon)
Hi, I'm revising Simple Harmonic Motion in edexcel M3. Whilst I can derive and know how to use acceleration=-w^2x and v^2=w^2(a^2-x^2), I just don't know what difference it makes if I use x=asin(wt) or x=acos(wt). I realise that in deriving them, it's whether x is 0 or 1 at t=0, but when you are not given the initial position but have to work out time (eg the time it takes for a particle to move between two points), I get too confused .

Basically the question is, "when do I use x=asin(wt) and when do I use x=acos(wt)?"

First off, it's not if x = 0 or 1, it's if your displacement is zero at then you use or if your displacement is (amplitude) at .

Second off, even if the intial positin is not give, you need to decide where you want to start measuring time from. If you think the problem is made easier by starting the time or the imaginary stopwatch as I like to think of it when the particle is at the origin, then use sine. If you think the problem is made easier by starting the time/stopwatch when the particle is at the amplitude, then use cosine.

I'll let some of the better people at mech give an example of this!
4. (Original post by TommyBrecon)
For example in this question's part c, I'm supposed to use asinwt; but WHY?
See, in this example, you want the time taken to travel from the origin to 0.4 - your life would be made infinitely easier by starting your counter when the particle is at the origin, hence you'd use .
5. Effectively, the solutions to the differential equation governing SHM are of the form y=Acos(wt+P) (amongst other forms e.g complex exponentials Aexp(iwt) for complex A). There are two arbitrary constants (because the SHM equation is a 2nd order linear ODE), the amplitude A and the phase P. These are determined on initial conditions, generally given as x(0) and v(0) i.e the conditions of the oscillator as how you set it going. More generally you can be given two pieces of information at any time t, but it is common to be given velocity and displacement at t=0, these allow you to solve for A and P. If you set it oscillating at equilibrium (i.e at y=0, t=0) then you can see that P=+/-pi/2 which gives y=+/-Asin(wt) (v(0) needed to determine if the solution is a negative or positive sin). If you set it oscillating at (positive) amplitude A at t=0 (y=A,t=0) then P=0 and A=cos(wt).

In short, it depends on how you set the oscillator going.
6. (Original post by TommyBrecon)
Hi, I'm revising Simple Harmonic Motion in edexcel M3. Whilst I can derive and know how to use acceleration=-w^2x and v^2=w^2(a^2-x^2), I just don't know what difference it makes if I use x=asin(wt) or x=acos(wt). I realise that in deriving them, it's whether x is 0 or 1 at t=0, but when you are not given the initial position but have to work out time (eg the time it takes for a particle to move between two points), I get too confused .

Basically the question is, "when do I use x=asin(wt) and when do I use x=acos(wt)?"

This is mostly to do with the nature of sinusoidal curves - we know that the range of and go from -1 to 1 and for , sin is at 0 while cos is at maximum.

Because of this, we use for when our oscillation begins from maximum amplitude. This is because when .

Similarly for when it starts from minimum amplitude - when .

Please read through pages 49-51 if you own an Edexcel M3 textbook.

(Original post by Zacken)
...
Help me on the M3 thread. Just did my first IAL paper and it wasn't that bad - definitely more of an algebraic slog than UK equivalents.
7. Ooh I see it now thank you very much everyone
8. (Original post by TommyBrecon)
Ooh I see it now thank you very much everyone
You're welcome.

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