# Simple Harmonic Motion with trigonometry

I need some help with my high school maths question:

A particle is moving in simple harmonic motion has speed 12m/s at the origin. Find the displacement-time equation if it is known that for positive constants a and n:

1. $x=a\cos 8t$
2. $x=16\cos nt$

So far this is what I know: The particle has a speed of 12m/s at the origin so when t=0, v=12. The origin is where the maximum speed occurs (Or is it at the centre of the motion?).

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Do you know the relation of speed and displacement? That would be a good place to start. –  Gerry Myerson Jun 13 '12 at 4:43
@GerryMyerson I'm unsure of that relation. Would you be able to please explain it to me? –  juleszero Jun 13 '12 at 4:48
If you've done calculus, the velocity is the derivative of the displacement. If you haven't done calculus, you've probably been given some law that says, if the displacement is [some formula], then the velocity is [some other formula]; look through your notes/text/whatever to see if you can find that law. –  Gerry Myerson Jun 13 '12 at 6:13

Problem $1.$ "At the origin" presumably refers to $x=0$. At the time $t_1$ when the particle is at the origin, $8t_1=\pi/2$, or $8t_1$ has some other value whose cosine is $0$. The sine of $8t_1$ is therefore $\pm 1$.
The velocity at any time $t$ is $\frac{dx}{dt}$, which is $-8a\sin(8t)$. The speed is the absolute value of this, and is $12$ when we are at the origin. Thus at the origin $-8a\sin(8t)$ has absolute value $12$. Since the sine of $8t_1$ has absolute value $1$, it follows that $8a$ has absolute value $12$. Since $a$ is positive, $a=\frac{12}{8}$. Thus $x=\frac{12}{8}\cos(8t)$.
The reasoning for Problem $2$ is similar.