Harmonic Oscillator - Acceleration

a_{x} = - ω^{2} A cos (ω \cdot t + ϕ)

$a_x = - omega ^2 "A" cos( omega * "t" + phi )$

(ω) Angular Velocity

$(omega)"Angular Velocity"$

(A) Amplitude

$(A)"Amplitude"$

(t) Time

$(t)"Time"$

(ϕ) Phase Angle

$(phi)"Phase Angle"$

Inputs

$ω$ $omega$ - the angular velocity
A - amplitude, the maximum displacement of the oscillator
$ϕ$ $phi$ - the phase angle defining the starting displacement at time t = 0
t - the time at which we compute the acceleration

Definition

If the phase angle $ϕ$ $phi$ is zero, then $x_{0} = A \cdot cos (0) = A$ $x_0 = A*cos(0) = A$ . So, $ϕ = 0$ $phi = 0$ , means the oscillator is at maximum displacement.

The acceleration as a function of time can be found by taking the simple derivative of velocity:

$a_{x} = d \frac{v_{x}}{d t} = \frac{d^{2} x}{{d t}^{2}} = - ω^{2} \cdot A cos (ω \cdot t + ϕ)$ $a_x = dv_x/dt = (d^2x) /dt^2 = - omega^2 * A cos(omega*t + phi)$ , since

$x = A \cdot cos (ω \cdot t + ϕ)$ $x = A*cos(omega*t +phi)$ ^[¹^]

$\frac{d x}{d t} = - ω \cdot A \cdot sin (ω \cdot t + ϕ)$ $dx/dt = -omega*A*sin(omega*t + phi)$ and $\frac{d^{2} x}{{d t}^{2}} = - ω^{2} \cdot A cos (ω \cdot t + ϕ)$ $(d^2x) /dt^2 = - omega^2 * A cos(omega*t + phi)$

Reference

Young, Hugh and Freeman, Roger. University Physics With Modern Physics. Addison-Wesley, 2008. 12th Edition, (ISBN-13: 978-0321500625 ISBN-10: 0321500628 ) Pg 426, Equation #13.16

^ University Physics, page 425, eq. 13.13