::1994 Motion Q:: (1 Viewer)

[CrashOveride]

x'' = 3x(x-4)
when t=0, x=0 and v=4sqrt(2)

can be shown v^2 = 2(x^3-6x^2+16)
Velocity is 0 at x=2 and x'' < 0

iv) Briefly describe the motion of the particle after it moves from x=2. How do people do this without resorting to graphing?

 

[jumb]

" Velocity is 0 at x=2 and x'' < 0 "

It is accellerating to the left from rest.

Velocity is 0 when it turns around at the edge, so it has just turned around.

 

[CrashOveride]

maybe i shud be a bit more clear,...ok its easy to see its turned to the left but what i want to know is how do we know if its turning again or whatever its doing, and the graph of its velocity is a bit weird, albeit i had to use graphmatica

 

[Sirius Black]

I think x=2 is actually the centre of the motion. Since x''=0 at x=0 or x=4 , the particle is oscillating b/w x=0 and x=4. when it starts at x=2 from rest and x''<0, it will move to the left and change direction at x=0

 

[CrashOveride]

it doesnt start at x=2, nor from rest. refer back to question.
and its not SHM either

 

[CrashOveride]

actually upon graphing, there does seem to be an oscillatory nature hmmm even tho we cant show it thru the algebra?

 

[Danny11]

i wish graphamatica was allowed in exams, i can't draw

 

[CrashOveride]

Can anyone help out here ^^

 

[Xayma]

Yes well of course there is an osscilating nature. You said it yourself x''<0 at x=2.

Now that means that it accelerates to the left, however, at x<0 it accelerates to the right creating an ossciliating nature. To find between which to points it oscillates it will be x=2 and approx x=-1.5, with the acceleration being greater to the left of the origin.

 

[CrashOveride]

Ok fair enough, how did u get that x=-1.5? Also, i was looking at eg. in the excel book and they reckon good way to find restirction on x is to remember that v^2 >= 0
With that in mind, i get x^3 -6x^2 + 16 >= 0
Dont think thats so easily solvable?

 

[Danny11]

hey i was just doing some revision came across a question exactly like this one in pheonix book. the only difference is that the x'' = 3x(x-2) not -4. so at 2 the particle stops and stays there. maybe its a modify version of the hsc question, where did you get the 1994 paper?

 

[Sirius Black]

Ok fair enough, how did u get that x=-1.5? Also, i was looking at eg. in the excel book and they reckon good way to find restirction on x is to remember that v^2 >= 0
With that in mind, i get x^3 -6x^2 + 16 >= 0
Dont think thats so easily solvable?

remember that you have already proved that at x=2, v=0
solving for x^3 -6x^2 + 16 =0 just simply using long division
f(x)=(x-2)(x²-4x-8)=0

 

[CrashOveride]

yeah it just occured to me then

 

[Xayma]

Well 2(1-&radic;3) if you really want to get technically, so as you can see it isnt exactly SHM, considering that the maximum velocity occurs at x=0 it isnt symmetrical (shown by the stat point of the above)

 

[CrashOveride]

ok so i cud just say:

after it moves from x=2 it moves towards the origin and crosses it where it begins to slow down at stops at 2(1-sqrt3) where it then begins to move to the right again. This motion is oscillatory, but it is not SHM as it is not of the form -n^2(x-b) "

interestingly, the sugested answer in MANSW mentioned nothing about this osciallation