4—Differential Equations 86Solve forAandBto getA=−B=v 0 /(2iω 0 ). Then
x(t) =v 0
2 iω 0[
eiω^0 t−e−iω^0 t]
=
v 0
ω 0sinω 0 tAs a check on the algebra, use the first term in the power series expansion of the sine function to see howx
behaves for smallt. The sine factor issinω 0 t≈ω 0 t, and thenx(t)is approximatelyv 0 t, just as it should be.
Also notice that despite all the complex numbers, the final answer is real. This is another check on the algebra.
Damped Oscillator
If there is damping, but not too much, then theα’s have an imaginary partand a negative real part. (Is it
important whether it’s negative or not?)
α=−b±i√
4 km−b^2
2 m=−
b
2 m±iω′, where ω′=√
k
m−
b^2
4 m^2(8)
This represents a damped oscillation and has frequency a bit lower than the one in the undamped case. Use the
same initial conditions as above and you will get similar results (letγ=b/ 2 m)
x(t) =Ae(−γ+iω′)t
+Be(−γ−iω′)tx(0) =A+B=0, vx(0) = (−γ+iω′)A+ (−γ−iω′)B=v 0 (9)The two equations for the unknownsAandBimplyB=−Aand
2 iω′A=v 0 , so x(t) =v 0
2 iω′e−γt[
eiω′t
−e−iω′t]
=v 0
ω′e−γtsinω′t (10)