Question

Given a damped mass system in which the mass is subject to harmonic forcing of Fo*sin(wt), show that the peak amplitude of the vibration occurs at a frequency wp given by the expression.

wp/wn = [1 - 2(Z^2)]^(1/2) where Z is the damping ratio

Answer 1

@IrishBoy123

Answer 2

wp = peak frequency

Answer 3

wn = natural frequency

Answer 4

yeah, i got it!!

so we need to solve a 2nd order non-hom DE

it's all been transcribed on Wiki , thankfully

Answer 5

I am going to upload a picture of my notes. I want to know if we apply these steps.

Answer 6

sure

but you know what the story is here?

you're solving a 2nd order non-hom DE ?

Answer 7

and then doing a load of really shιtty algebra to match stuff up

Answer 8

So the homogenous part goes to zero and we just take the derivative of the particular solution.

Answer 9

you will always have a complementary solution and a particular solution.

Answer 10

For a long period of time, the complementary goes to zero though.

Answer 11

no!

the complimentary is the perpetual oscillation for a spring-mass system. the eternal exchange of energy.

it is then conditioned by the damping term. which is the loss of energy

Answer 12

which can also be included in the complementary....so you are right

Answer 13

Okay.

Answer 14

The natural frequency goes to zero.

Answer 15

I am working on it now. I will show you my solution when I am done.

Answer 16

I will post it up. I will brb and then tag you.

Answer 17

i will try work a solution too

Answer 18

What I have so far.

Answer 19

Your stuff looks good and ends I think here:

\(x(t) = - \dfrac{F_o c \omega}{ ( c \omega )^2 + (k - m \omega^2)^2} \cos \omega t + \dfrac{F_o (k - m \omega^2))}{(c \omega )^2 + (k - m \omega^2)^2} \sin \omega t \)



if we replace with \(z = \dfrac{c}{2m}\) and \(\omega_n = \sqrt{\dfrac{k}{m}}\)

it becomes

\(x(t) = - \dfrac{F_o 2mz \omega}{ ( 2mz \omega )^2 + (m \omega_n^2 - m \omega^2)^2} \cos \omega t + \dfrac{F_o (m \omega_n^2 - m \omega^2))}{(2mz \omega )^2 + (m \omega_n^2 - m \omega^2)^2} \sin \omega t \)

and


\(x(t) = - \dfrac{2 z \omega F_o /m}{ ( 2z \omega )^2 + ( \omega_n^2 - \omega^2)^2} \cos \omega t + \dfrac{ ( \omega_n^2 - \omega^2) F_o/m }{(2z \omega )^2 + ( \omega_n^2 - \omega^2)^2} \sin \omega t \)

In this form it is easy to turn into a single sinusoid using an addition formula \(\sin (A - B) = \sin A \cos B - \sin B \cos A\)

Looks more obvious this way....

\(x(t) =\dfrac{F_o}{m} \dfrac{1}{ \sqrt {( 2z \omega )^2 + ( \omega_n^2 - \omega^2)^2}} \left( - \dfrac{2 z \omega }{\sqrt{ ( 2z \omega )^2 + ( \omega_n^2 - \omega^2)^2}} \cos \omega t \\ + \dfrac{ ( \omega_n^2 - \omega^2) }{\sqrt { (2z \omega )^2 + ( \omega_n^2 - \omega^2)^2}} \sin \omega t \right)\)

Making it:

\(x_p=\dfrac{F_o}{m} \dfrac{1}{ \sqrt {( 2z \omega )^2 + ( \omega_n^2 - \omega^2)^2}} \sin (\omega t - \psi) \qquad \star\)

That way you can read the amplitude straight off.

If you differentiate the non-sinusoid bit of \(\star\) wrt \(\omega\) you should then get \(\omega\) for the max amplitude of the steady state oscillation.

Without some IV's I don't see how you can say much about the transient decaying oscillation that is your complimentary solution. Which again I would also be tempted to write in this combined form as

\(x_c = A e^{- zt } \sin (\omega_n t + \phi)\)

Furthermore I get a slightly different answer to the one you are looking for so maybe I am goofing along the way.

Answer 20

Guess what... All she wanted was to show a proof... Lol

Answer 21

yes but there's no proof

or her equation is wrong

or i'm a fucκwit. most likely outcome.