Alpha-Omega
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#1
Report Thread starter 9 years ago
#1
I know that v2=w2(A2-x2) comes from a=-w2x since a=v(dv/dx), but what is the mathematical proof for a=-w2x ??
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Stonebridge
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It's not proved. It's the definition of SHM. The acceleration of the particle is directly proportional to its displacement from the origin and directed towards it. So a = -kx
The ω² comes from solving a=-kx as a sinusoidal function with angular frequency ω. It turns out the constant k = ω²
You'll see this if you differentiate x = A sin (ωt) twice wrt t
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Alpha-Omega
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(Original post by Stonebridge)
It's not proved. It's the definition of SHM. The acceleration of the particle is directly proportional to its displacement from the origin and directed towards it. So a = -kx
The ω² comes from solving a=-kx as a sinusoidal function with angular frequency ω. It turns out the constant k = ω²
You'll see this if you differentiate x = A sin (ωt) twice wrt t
That make sense

a=-kx=-kAsinωt=-Aω2sinωt

(-kAsinωt)/(-Asinωt)=(-Aω2sinωt)/(-Asinωt)

k=ω2

a=-ω2x
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Stonebridge
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Like this
x = A sin (ωt) ---- (1)
dx/dt = v = A ω cos (ωt)
d2x/dt2 = a = - A ω² sin (wt)
sub from (1) for A sin ωt = x
gives
a = -ω² x
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Alpha-Omega
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#5
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(Original post by Stonebridge)
Like this
x = A sin (ωt) ---- (1)
dx/dt = v = A ω cos (ωt)
d2x/dt2 = a = - A ω² sin (wt)
sub from (1) for A sin ωt = x
gives
a = -ω² x
Thanks
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Varun Ghanta
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#6
Report 1 year ago
#6
(Original post by Stonebridge)
Like this
x = A sin (ωt) ---- (1)
dx/dt = v = A ω cos (ωt)
d2x/dt2 = a = - A ω² sin (wt)
sub from (1) for A sin ωt = x
gives
a = -ω² x
Thank you so much.
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