The Student Room Group
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
(edited 12 years ago)
Reply 2
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
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
Reply 4
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
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.
Reply 6
Isn't the formula F(restoring force)=-KxAs then we can write,ma=-Kxa=(-Kx)/m and we know that K/m=w²a=-w²x

Latest