Hello superdude Originally Posted by
superdude what does the $\displaystyle \delta$ do in the formula $\displaystyle x=Acos(wt+\delta)$
can someone give me an example of a question when it's not 0? Does it have to do with if at the start of the probelm the wave is not at equillibrium?
Study the diagram I've attached.
You'll see that there are 4 cosine waves, all with different phase angles, based on the following formula $\displaystyle x = 2\cos\big(\tfrac{\pi}{4}t+d\big)$
with values of $\displaystyle d$:$\displaystyle d=0,\; \frac{\pi}{4},\; \frac{\pi}{2},\;-\frac{\pi}{2}$
These all represent fundamentally the same SHM; they all have amplitude $\displaystyle 2$ and period $\displaystyle 8$. But you'll see that they all start in a different position.
When $\displaystyle d= 0$ the motion starts at $\displaystyle (0,2)$ - this corresponds to the moving point starting at the positive end-point of the motion moving back towards the equilibrium position.
$\displaystyle d=\frac{\pi}{4}$ is the same wave as $\displaystyle d = 0$, but pushed $\displaystyle 1$ unit to the left: one-eighth of a complete cycle. It starts nearer to the centre of the motion (the equilibrium position), moving back towards the centre.
$\displaystyle d = \frac{\pi}{2}$ pushes the wave another unit to the left, so it's now one-quarter of a cycle out of phase with the original wave. In this case, the moving point starts at the equilibrium position, and is moving towards the negative end of the motion.
$\displaystyle d=-\frac{\pi}{2}$ pushes the wave $\displaystyle 2$ units - a quarter of a cycle - to the right. This is where the moving point starts at the equilibrium position, and is moving towards the positive end.
Does that help to make it clear?
Grandad