Simple harmonic motion problem?

Question

"Simple harmonic" motion is oscillatory motion of the form x=x(o)cos(2pi*t / T), where x(o) is the maximum amplitude of oscillation and T is the period of oscillation.

given:
a= –(2pi / T)^2 * x(o)cos(2pi*t / T).

QUESTION:
If x(o) = 0.1 m and T = 2 s, what are the maximum values of velocity and acceleration?

Answer 1

Amplitude is displacement. Velocity is the derivation of amplitude. Acceleration is the derivation of velocity.

Answer 2

The maximum speed = 0.1 pi m/s = 0.31416... m/s; and the

maximum acceleration = 0.1 (pi)^2 m/s^2 = 0.98696... m/s^2.

Her's how these results were found:

Let (2pi / T) = w (really Greek ' omega ').

Strictly speaking we don't talk about the maximum value of the (vector) velocity, but of the (scalar) speed.

Standard results for SHM are that the maximum value of the SPEED is w (amplitude), and of the modulus of the ACCELERATION w^2 (amplitude).

(The maximum +/- value for the speed is when the particle is passing through x = 0, which occurs when the cosine's argument is pi / 2, 3 pi / 2, 5 pi / 2, etc., i.e. when t = 0.5s, 1.5s, 2.5s, etc. The maximum +/- value for the acceleration occurs at -/+ x(o), when t = 0s, 1s, 2s, 3s, etc.)

w = (2pi / T) = pi /s, and so, with amplitude x(o) = 0.1m and T = 2s:

Maximum speed = 0.1 pi m/s = 0.31416... m/s;

Maximum acceleration = 0.1 (pi)^2 m/s^2 = 0.98696... m/s^2.

Live long and prosper.