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Simple Harmonic motion (1 Viewer)

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Bit confused with shm...how do you do this question?
A particle moving with SHM starts from rest at a distance of 8m from the centre of oscillation. If the period is 4 seconds, find the initial acceleration.
 

Timothy.Siu

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Bit confused with shm...how do you do this question?
A particle moving with SHM starts from rest at a distance of 8m from the centre of oscillation. If the period is 4 seconds, find the initial acceleration.
period = 2pi/n=4

n=pi/2

i dont know if u can do this but
a=-n^2(x-8)
x=0
a=2pi^2
 

Aerath

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Hmmmm, Tim, (x-8) implies centre of oscillation is at x = 8, which is not true, I don't think. I would just sub in a = -n^2 (x), n = pi/2 and x = 8. And since it doesn't specify which side the particle is on, the acceleration could be either positive or negative.
 

tommykins

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Bit confused with shm...how do you do this question?
A particle moving with SHM starts from rest at a distance of 8m from the centre of oscillation. If the period is 4 seconds, find the initial acceleration.
I'll cheat a little bit.

T = 2pi/w = 4
w = pi/2

Now, a(max) = Aw^2 = 8*(pi/2)^2 = 2pi^2

i'll let you figure out how to do the direction.

full solution below

x = Acos(nt) t = 0, x = 8

so, 8 = A

x = 8cos(nt)

T = 2pi/n = 4
n = pi/2 (notice how w is like n)

so,

v = -8(pi/2)sin(nt)
a = -8(pi/2)^2cos(nt)

t= 0

a = -2pi^2
 
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clintmyster

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Period, T = 2pi/n = 4
n = pi/2

when t=0 x = 8
therefore a = -n^2 x (as SMH)
a = - pi^2/4 x (8)
= - 2pi^2
 

Aerath

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Just wondering, do we need to do that solution that tommykins spoilerised in an exam? Or can we just do the shorter way that clintmyster suggested?
 
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i assume the book's answer is wrong( -2pi m/s^2)? since no one got the same thing.
 

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