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I'm trying to derive an equation for the velocity and acceleration of an object undergoing simple harmonic motion.

I have the equation for displacement: $x = A\sin (2 \pi ft)$

If I differentiate the equation with respect to $t$ then I should get an equation for the velocity and if I differentiate again, I should get an equation for the acceleration, right?

So if I differentiate once, I get: $v = 2 \pi f\cos (2 \pi ft)$

If I differentiate again, I get: $a = -2 \pi fsin (2 \pi ft)$

However, I don't think these equations are correct. Am I making a silly mistake?

N.b. This isn't homework, I'm making extra revision notes :P

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You should put the minus sign before the 2, or it will be quite confusing – gerrit Dec 18 '12 at 20:58
Whoops, I missed this when editing! – Todd Davies Dec 18 '12 at 20:59
And you lost the amplitude $A$. Acceleration has $(2\pi f)^2$. – Vladimir Kalitvianski Dec 18 '12 at 20:59
and maybe a $2\pi f$ missing in the expression for a – twistor59 Dec 18 '12 at 21:00
It is noteworthy that acceleration actually depends upon position. ie, $a(t) = -(2 \pi f)^2x(t)$ This is indicative of simple harmonic motion. – Cactus BAMF Dec 19 '12 at 0:00
up vote 1 down vote accepted

Your method is correct, but your calculations of the derivative are not.

$v = \frac{dx}{dt}$ where $x=A sin(2 \pi f t)$ would be $A 2 \pi f cos(2 \pi f t)$. Then $a = \frac{dv}{dt} = \frac{d^2x}{dt^2} = - A (2 \pi f)^2 sin(2 \pi f t)$.

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$x=A sin(\omega t)$

$v=-A\omega cos(\omega t)$

$a=-A\omega^2 sin(\omega t)$

where $\omega=2\pi f$

$d/dx (sin u) = cos u (du/dx)$ $d/dx cos u = -sin u (du/dx)$ See for derivative of sin u and cos u.

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Your 2nd and 3rd equations are incorrect. – gerrit Dec 18 '12 at 21:03
The 2nd equation has a rogue minus sign, but the 3rd equation looks correct. Am I missing an obvious mistake? – John Rennie Dec 19 '12 at 7:58

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