Simple Harmonic Motion Formula

The SHM general formula is this:

$y(t) = A\sin(\omega t + \alpha) +B$

I have two questions about it

1. As far as I know, there is the cosine formula for when the particle starts at P and sine for when the particle starts at O. But, does it mean that with this general formula above, it doesn't matter where it starts due to α?

2. What does the constant B mean? Like where is it represented in a motion? (Please be as explicit and simple as possible for this as I've tried reading many different websites to no avail)

Thanks

$\alpha$ represents the phase. You are correct that if you include $\alpha$ you can use either $\cos$ or $\sin$ to represent the motion-it will just shift $\alpha$ by $\frac \pi 2$. You can expand your sine wave into $y(t)=A\sin(\omega t) \cos (\alpha) + A\cos(\omega t)\sin \alpha) +B$ Substituting in $t=0$ shows that $y(0)=A\sin (\alpha)+B, y'(0)=A \cos(\alpha)$ so you can use your initial conditions to evaluate $A$ and $\alpha$
$B$ is an offset from zero. If the oscillation is centered at a point other than zero, $B$ is non-zero. For example, $y(t)=\sin(\omega t)+1$ ranges from $0$ to $2$, not from $-1$ to $1$
• It depends on your choice of coordinates. In your example, assuming it starts at rest, it is oscillating around $y=4$, so $B=4$. One way to compute $B$ is to average the maximum excursion in each direction. In your example it will go from $y=0$ to $y=8$ – Ross Millikan Apr 17 '15 at 15:06