fdsadsfa said:
What do you mean by y''(t)? Did you mean x(t)" ??
Yes, that was a typo. I apologise.
fdsadsfa said:
I also don't know what the chain rule has to do with this problem but the instruction was to utilize this rule when solving the problem =<
Could you please think a little bit more about the relevance of the chain rule to this problem?
Hmm, maybe you have to use the chain rule with something like
[tex]\frac{da}{dx} \frac{dx}{dt}[/tex]
fdsadsfa said:
Also,
Is there any reason that you included both cosine and sine fuction in the fuction of x(t)?
Originally I intended to use x(t) = Asin([tex]\omega[/tex]t + [tex]\phi[/tex])...
Well, my reason was the following (rather technical point): x'' = a x is a second order equation (because there are two derivatives), so there must be two independent solutions. In this case, they can be easily found to be sin(x) and cos(x). The general solution to the equation is then A sin(x) + B cos(x), with A and B two constants which have to be determined from initial conditions (for example, at t = 0, x must equal 23 and x' must equal 0). You need two conditions to fix A and B.
In the expression you gave, there are also two undetermined constants, [itex]\omega[/itex] and [itex]\phi[/itex]. If you take [itex]\sin(\omega t + \phi)[/itex] you can use the two conditions to fix these constants.
It doesn't really matter which of the two you take, in the end they are equivalent. In fact, a cosine is nothing more than a sine which is shifted a bit horizontally so you can always write cos(t) in the form [itex]\sin(t + \phi)[/itex] by choosing [itex]\phi[/itex] properly.
I don't know how much you know about differential equations and related topics, but I hope that makes sense. I'll stick to [itex]\sin(\omega t + \phi)[/itex]
