Phase and Initial Conditions

The general solution of SHM is:

$x(t) = A\cos(\omega t + \phi)$

$\phi$ is the phase constant — it encodes where in the cycle the oscillator starts at $t = 0$.

Starting at maximum: $\phi = 0$

Released from rest at $x = +A$. This is the simplest case — pure cosine:

SpringMass amplitude 80 k 100 m 1 phase 0
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At $t=0$: $x = A$, $v = 0$. Graph starts at the peak.

Starting at negative extreme: $\phi = \pi$

Released from rest at $x = -A$:

SpringMass amplitude 80 k 100 m 1 phase 3.14159
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At $t=0$: $x = -A$, $v = 0$. Graph starts at the trough. This is antiphase with $\phi=0$.

Starting at centre moving right: $\phi = -\pi/2$

Pushed through equilibrium with maximum velocity:

SpringMass amplitude 80 k 100 m 1 phase -1.5708
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At $t=0$: $x = 0$, $v = +A\omega$ (maximum positive). Graph starts as a sine, not a cosine.

Starting at centre moving left: $\phi = +\pi/2$

SpringMass amplitude 80 k 100 m 1 phase 1.5708
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At $t=0$: $x = 0$, $v = -A\omega$ (maximum negative). Graph is a negative sine.

Finding $\phi$ from initial conditions

Given $x_0 = x(0)$ and $v_0 = v(0)$:

$A = \sqrt{x_0^2 + \left(\frac{v_0}{\omega}\right)^2}$

$\cos\phi = \frac{x_0}{A}, \quad \sin\phi = \frac{-v_0}{A\omega}$

Use both equations together to determine the correct quadrant of $\phi$.

Reading phase from a graph

Graph at $t=0$	Slope at $t=0$	$\phi$
Starts at $+A$	zero	$0$
Starts at $0$	positive	$-\pi/2$
Starts at $0$	negative	$+\pi/2$
Starts at $-A$	zero	$\pi$