Question

A ball hung from a vertica; spring oscillates in simple harmonic motion with an angular frequency of 2.6 rad/s and an amplitude of 0.075m. What is the maximum acceleration of the ball?

Answer 1

The equation of motion for any SHO is
\[x(t) = Asin(\omega_0 t)\]
where x is the displacement, A is the amplitude, t is time, and omega_0 is the natural frequency of the system. For spring systems, the natural frequency is
\[w_0 = \sqrt{k/m}\]
just for reference...
If you differentiate the equation of motion once, you get the velocity equation. Twice will give you the acceleration equation. There's lots of different ways to get the max accel. You could differentiate 3 times, set equal to 0, solve for t, and throw it into the accel equation to get the max, which is the calculus way. In fact, I suggest you do that just clarify the whole situation. It's the long way, but it'll make so much more sense if you do it...

The quick and dirty way, which you will see if you do it the long way, is that you can reason that the max acceleration occurs when the mass is in the equilibrium position. So set the position equation equal to zero, solve for t, then throw it into the acceleration equation. Just play around with this equation of motion. There are alot of ways to do this problem.

Answer 2

Note, you can use cosine instead of sign if it makes you happy, since there are no initial conditions given in this problem. What is really going on here, is that they both solve the same differential equation
\[x(t) = -\omega_0 ^2 (d^2x/dt^2)\]
ie, x is the function that is equal to the negative of itself. Sine and cosine both meet that requirement.

Answer 3

er, meant to say "x is the function that is equal to the negative second derivative of itself".