0
$\begingroup$

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

$\endgroup$
1
$\begingroup$

$\alpha$ 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 $1$

$\endgroup$
  • $\begingroup$ (1) was very, very clear and I understand now. However for (2) how do you know the point of oscillation is not = 0? For example a spring hanging with mass at equilibrium of extension 4 and oscillates about the extension, starting a when extension = 0, what will B be? Is it possible to tell immediately or must I work it out with the other factors? Thank you $\endgroup$ – LucasCK Apr 17 '15 at 14:54
  • $\begingroup$ 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 $y=8$ $\endgroup$ – Ross Millikan Apr 17 '15 at 15:06
  • $\begingroup$ I get it now! Finally! Cannot thank you enough!! :) $\endgroup$ – LucasCK Apr 17 '15 at 15:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.