Damped Oscillations

Consider a simple harmonic oscillator that has friction, then the equations of motion must be changed with the addition of a friction term. So we write m ⅆ 2 x ⅆ t 2 = − k x − b ⅆ x ⅆ t m ⅆ 2 x ⅆ t 2 = − k x − b ⅆ x ⅆ t where b ⅆ x ⅆ t b ⅆ x ⅆ t is the friction term. Rearranging we obtain: m ⅆ 2 x ⅆ t 2 + b ⅆ x ⅆ t + k x = 0 m ⅆ 2 x ⅆ t 2 + b ⅆ x ⅆ t + k x = 0 or ⅆ 2 x ⅆ t 2 + γ ⅆ x ⅆ t + ω 0 2 x = 0 ⅆ 2 x ⅆ t 2 + γ ⅆ x ⅆ t + ω 0 2 x = 0 Where γ = b m γ = b m and ω 0 2 = k m ω 0 2 = k m Assume a solution of form x = A e i ( p t + α ) x = A e i ( p t + α ) substitute into equation and get ( − p 2 + i p γ + ω 0 2 ) A e i ( p t + α ) = 0 ( − p 2 + i p γ + ω 0 2 ) A e i ( p t + α ) = 0 so − p 2 + i p γ + ω 0 2 = 0 − p 2 + i p γ + ω 0 2 = 0 p p must have real and imaginary parts, so rewrite: p = ω + i s p = ω + i s p 2 = ω 2 + 2 i ω s − s 2 p 2 = ω 2 + 2 i ω s − s 2 So the equation − p 2 + i p γ + ω 0 2 = 0 − p 2 + i p γ + ω 0 2 = 0 becomes upon substitution: − ω 2 − 2 i ω s + s 2 + i ω γ − s γ + ω 0 2 = 0 − ω 2 − 2 i ω s + s 2 + i ω γ − s γ + ω 0 2 = 0 This equation implies that the real and imaginary parts are each zero.Separate the real and imaginary partsImaginary parts give: − 2 ω s + ω γ = 0 s = γ 2 − 2 ω s + ω γ = 0 s = γ 2 From Real parts get − ω 2 + s 2 − s γ + ω 0 2 = 0 − ω 2 + s 2 − s γ + ω 0 2 = 0 or − ω 2 + γ 2 4 − γ 2 γ + ω 0 2 = 0 − ω 2 + γ 2 4 − γ 2 γ + ω 0 2 = 0 − ω 2 − γ 2 4 + ω 0 2 = 0 − ω 2 − γ 2 4 + ω 0 2 = 0 Which rearranges to ω 2 = ω 0 2 − γ 2 4 ω 2 = ω 0 2 − γ 2 4 Thus the solution becomes x = A e − γ t / 2 e i ( ω t + α ) x = A e − γ t / 2 e i ( ω t + α ) where ω = ω 0 2 − γ 2 4 ω = ω 0 2 − γ 2 4 Note that this has assumed a frictional damping force. For a more complicated damping force, the result would be different.