pls help (SHM Q, 2002 past hsc) (1 Viewer)

[smelnizzle]

part iii. confuses me.
A particle, whose is x, moves in shm such that a = - 16x. At t =0, x =1, and v=4.

i. Show that, for all positions of the particle, |v| = 4 sqroot (2-x^2)
ii. what is the particle's greatest displacement?
ii. Find x as a function of t. You may assume the general form for x.

in the back of success one the ans is x = sqroot (2) cos (4t - pi/4)
so they took the angle cos theta to be - pi/4. (which i did not get)

meanwhile, the back of another past paper book is x = sqroot (2) cos (4t + pi/4)
and this is what i got.


so which one is correct?! :s and whichever one it is.. why is it that instead of the neg/posi value?

 

[Michaelmoo]

Ok. First of all, the success answer is the right one and I'll tell you why.

If you differentiate YOUR x as a function of t, the data with the question does not satisfy the equation. (i.e. when t=0, x(dot) is not equal to 4). I think I know why.

When you looked to determine the angle (alpha) in your equation by plugging in your data, you got cos(alpha) = 1/(root)2. However, you assumed alpha was in the first quadrant, and went ahead and solved (alpha) = pi/4. You cannot assume this, (alpha) can be either in the first or fourth quadrant. You have to find out which quadrant satisfies the given data:.

Basically, you differentiate your x as a function of t to get x(dot). That is, you differentiate x = (root)2cos[4t + (alpha)]. When you get your velocity function, sub in the given data. Simplify and you will end up with sin(alpha) = -1/(root)2.Now we have enough data to find precisely what alpha. We have two equations:

cos(alpha) = 1/(root)2

sin(alpha) = -1/(root)2

Obviously, the angle is pi/4. However, we choose the quadrant where cos is positive and sin is negative, that is the fourth quadrant. So the angle is -pi/4.

Sorry, I use explorer and BoS won't let me format my replies in any way.

Good luck.

 

[smelnizzle]

thanks ya! i get it now.
and let me help you with formatting you just chuck in < br > (and minus the gaps) at the end of each line you want to create a new one.

and good luck to you too.