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CrashOveride

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Show that the time required to travel from the centre of motion to a point at a distance from the centre equal to half the amplitude is 1/12 of the period.
 

pc_wizz

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i end up with a funky answer .. :p

if x = a cos (nt + b)

then amplitude = a
and period = (2pi)/n

its given that x = a/2

:. nt + b = arccos (1/2)
= pi/3
= (2pi - 6b)/6

:. t = (2pi - 6b)/6n
= (2pi/6n) - b/n
= 2(2pi/12n) - b/n

thats the closest i got .. =/


pcwizz
 

CrashOveride

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ok i duno waht u did...but here is me:

Period: 2pi / n
To show time 1/12 of period, t = pi / 6n **

x = a cos (nt + &)
When t=0, x=0 (say)

0 = acos(&)
cos(&) = 0
& = Pi/2

When x = a/2:

a/2 = a cos (nt + pi/2)
1/2 = cos (nt + pi/2)
nt + pi/2 = Pi/3
nt = -Pi/6
t = -Pi / 6n **

If i take absolute value then i get t = Pi / 6n, as required.

Just the thing is...shouldn't it fall out as a +ve time anyway, i shudnt need to take abs value ?
 
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Let x = a sin (nt)
When x=0, t=0
When x= a/2
a/2 = a sin (nt)
nt = pi/6 (justify blah blah)
t= pi/6n

You have a negative value because you're using cos(nt + pi/2), at t=0 the particle is moving in a negative direction, so to get to a positive x value we are talking about -pi/6n -> 0. Draw a graph of the displacement if you don't understand what I mean.
 

CrashOveride

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Oh ok, so you can say stuff like x = a sin (nt) ?

I thought x = acos(nt + &) was the universal thing, my textbook doesnt say anything about doing it that way.
 
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We can define the centre of motion to be at any time t. As long as the displacement function represents SHM (i.e. x'' = -n^2 x) it can be whatever you like.
 

CrashOveride

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another thing, if i got x'' = -18 (2x - 1) ...and im wanting to show its in SHM, my teacher said u can just let 2x - 1 = X and that solves it. But arn't we just defining displacement in terms of displacement then ? :S So if X was 0 (the x'' would be 0 ) but 2x - 1 = 0 so x = 1/2. So im saying letting displacement be zero means dispalcement is 1/2 ?! Im confused here.
 

CrashOveride

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-36(x - 1/2)

When x is zero (i.e. zero displacement, at the origin) x'' = 18

Property of SHM is x'' = 0 when displacement is zero (x=0)
 

BFEPNS

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omfg

shm is soo hard

how the hell do u guys understand this crap???
 

CrashOveride

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Exactly....and when we're at the centre of motion, we have x = 0 as there is no displacement (x is defined as displacement from centre of motion, nay? )
 
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Originally posted by CrashOveride
Exactly....and when we're at the centre of motion, we have x = 0 as there is no displacement (x is defined as displacement from centre of motion, nay? )
no.
 

CrashOveride

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x is defined as the distance of point P from the origin O. So i'm tending to stick with my arguement unless u can shed some light onto this....
 

CrashOveride

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but we know x'' = 0 at centre of motion.
And my book says when x=0, x''= 0. That is, acceleration iz zero at the origin. And from the diagram in the book, it seems to be definign x as displacement from the centre of motion. So i would assume its being synonomous with origin and centre of motion :S
 

Heinz

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Originally posted by CrashOveride

And my book says when x=0, x''= 0. That is, acceleration iz zero at the origin.
This is only true when the centre of motion is at the origin. i.e. acceleration is -n<sup>2</sup>x and not -n<sup>2</sup>(x-a)
 

CrashOveride

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Ok so is 'x' the displacement from the centre of motion or origin?

So for eg. if something was going from 3 to 5 your saying at x=4 we have x'' = 0.

What book are you guys using ? I dont like the way mine explains things, its the crappy Couchman book
 

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