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shm agn (1 Viewer)

fullonoob

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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

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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

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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

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got it now thx
kcqn a>0, since a is amp :)
 
Last edited:

fullonoob

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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

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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
 

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