# Queries on Damped Harmonic Motion

1. Dec 5, 2013

### LameGeek

So we know that SHM can be described as:
x(t) = Acos(ωt + ϕ)
v(t) = -Aω sin(ωt + ϕ)
a(t) = -Aω^2 cos(ωt + ϕ)

it can be easily said that the max acceleration (in terms of magnitude) a SHM system can achieve is Aω^2

In Damped Harmonic Motion we know that:
x(t) = (A)(e^(-bt/2m))cos(ωt + ϕ)

given that:
A' = (A)(e^(-bt/2m))
ω' = sqrt( (ω^2) - (b/2m)^2 )

Is it true that the max acceleration at any given time is (A')(ω')^2?

My intuition tells me that the above statement is not true =/
because differentiating the function x(t) = (A)(e^(-bt/2m))cos(ωt + ϕ) gives me a complex function (which has sine & cosine in it) & it doesn't really give me anything close to the (A')(ω')^2 term

Last edited: Dec 5, 2013
2. Dec 5, 2013

### AlephZero

This is easier using complex numbers.

For damped motion, $x(t)$ = the real part of $Ae^{(-s + i\omega')t}$ where $s$ is your $b/2m$. Note, $A$ is a complex constant (to account for your phase angle $\varphi$) and of course $e^{i\theta} = \cos\theta + i\sin\theta$.

So $a(t)$ = the real part of $(-s + i\omega')^2 x(t)$

Your intuition is right, but if the damping is small, $(-s + i\omega')^2$ is close to $-\omega^2$.

3. Dec 5, 2013

### LameGeek

I'm not really familiar with complex numbers (other than i^2 = -1) but your explanation does makes some sense to me. Thank You!! =)