Question

A body oscillates with simple harmonic motion according to the equation a=-(6.00m)(3pi rad/s)^2 cos [(3pi rad/s)t + pi/3 rad]

At t = 2s, what are (a) the displacement, (b) the velocity, and (c) the acceleration of the motion?
Also, what are the (d) amplitude, (e) phase angle, (f) frequency (g) and period of the motion?

Answer 1

\[a = - 18π \cos (3πt + \frac{π}{3})\]

is this what you mean?

Answer 2

Is it possible to multiply the -(6.00)(3π rad/s)^2 ..that is different in unit?

Answer 3

Ah, i see a mistake in the formula. The formula is :

\[a = - (6) (3π)^{2} \cos (3πt + \frac{π}{3})\]
now to make sense of it all. the standard solution to simple harmonic motion that you will find in every elementary physics text, where x is the displacement, is:
\[x = A cost (wt + ∆)\]
this gives:
\[v = \frac{dx}{dt} = -wAsin(wt + ∆)\]
and
\[a = \frac{dv}{dt} = -w^{2}Acos(wt + ∆)\]

if you compare that last equation to what you have been given, you will see that they are trying to steer you to the answer.

w is the angular velocity of the motion. T (the period) = 2π/w, and frequency f = 1/T.

A is the amplitude, ∆ is the phase angle.

Everything that you now need is in this post, i think. is that OK?

Answer 4

I still got confuse . so you mean this picture that is A and its displacement is = a=−18πcos(3πt+π3)

Answer 5

no A is the amplitude. 3π is the angular velocity. just match the numbers up with the letters and it should become clear.

Answer 6

so in the given equation the A amplitude is −(6)(3π)2 ?

Answer 7

no!!!

A = 6.

(3π)^2 = w^2

∆ = π/3