shm agn (1 Viewer)

fullonoob

fail engrish? unpossible!
Joined
Jul 19, 2008
Messages
465
Gender
Male
HSC
2010
A particle is moving with SHM of period pi seconds and max velo 8m/s. If the particle started from x = a, find a , then find the velo when the particle is distant 3m from the mean position.
x.. = -n^2x or x.. = -n^2(x-x0)

damn i fail at this topic :haha:
 

kcqn93

Member
Joined
Jul 27, 2008
Messages
473
Gender
Male
HSC
2010
what's the answer??

i got like a=-4 and v = 5.3 m/s at x=3

i'm not too sure on this one
 

mirakon

nigga
Joined
Sep 18, 2009
Messages
4,222
Gender
Male
HSC
2011
well if period is pi

2pi/n=pi

Therefore n=2

For SHM

v^2= n^2(a^2-x^2) where a=amplitude in this case

Max velocity = 8

therefore obviously from above this is when x=0

64=n^2a^2

we know n=2

Therefore

a=4 where a is amplitude

I don't have more time atm, will post tomorrow, but perhaps this data helps in some way...........
 

fullonoob

fail engrish? unpossible!
Joined
Jul 19, 2008
Messages
465
Gender
Male
HSC
2010
got it now thx
kcqn a>0, since a is amp :)
 
Last edited:

fullonoob

fail engrish? unpossible!
Joined
Jul 19, 2008
Messages
465
Gender
Male
HSC
2010
i need to confirm something
prove that v^2 = n^2 (a^2 - x^2)
x.. = -n^2x
1/2 v^2 = -(n^2 x^2)/ 2 + C/2
v^2 = -(n^2. x^2) + C
when v = 0, x=a
therefore C = n^2 . a^2
v^2 = -(n^2x . x^2) + n^2 . a^2
hence v^2 = n^2 (a^2 - x^2)

is this proof correct?
and do you have derive each time or can you just state it in exams
 

shaon0

...
Joined
Mar 26, 2008
Messages
2,029
Location
Guess
Gender
Male
HSC
2009
i need to confirm something
prove that v^2 = n^2 (a^2 - x^2)
x.. = -n^2x
1/2 v^2 = -(n^2 x^2)/ 2 + C/2
v^2 = -(n^2. x^2) + C
when v = 0, x=a
therefore C = n^2 . a^2
v^2 = -(n^2x . x^2) + n^2 . a^2
hence v^2 = n^2 (a^2 - x^2)

is this proof correct?
and do you have derive each time or can you just state it in exams
Yeah looks okay to me.
Alternatively, You could just say:
Assume, v^2=n^2(a^2-x^2)
2v dv/dx=-2n^2.x
v dv/dx=-n^2.x
x..=-n^2.x
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top