Queries on Damped Harmonic Motion


by LameGeek
Tags: damped, damped oscillation, harmonic, motion, queries
LameGeek
LameGeek is offline
#1
Dec5-13, 08:56 PM
P: 6
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
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AlephZero
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#2
Dec5-13, 10:18 PM
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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##.
LameGeek
LameGeek is offline
#3
Dec5-13, 10:55 PM
P: 6
I'm not really familiar with complex numbers (other than i^2 = -1) but your explanation does makes some sense to me. Thank You!! =)


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