Given a harmonic oscillator with mass m, and spring constant

[Futurestar33]

Homework Statement

Given a harmonic oscillator with mass m, and spring constant k, is subject to damping force F= cdx/dt and driven by an external force of the form F[ext]= FoSin(wt).

A) Find the steady state solution.
B) Find the amplitude and the phase.

Homework Equations

F=-kx

the steady state is usually in the form of X(t)= Acos(wt+Φ)

The Attempt at a Solution

\
So i came up with this equation for the Fnet force.

F[net]= -kx+c(dx/dt)+FoSin(wt)[/B]

 

[rude man]

Homework Statement

Given a harmonic oscillator with mass m, and spring constant k, is subject to damping force F= cdx/dt and driven by an external force of the form F[ext]= FoSin(wt).

A) Find the steady state solution.
B) Find the amplitude and the phase.

Homework Equations

F=-kx

the steady state is usually in the form of X(t)= Acos(wt+Φ)

The Attempt at a Solution

\
So i came up with this equation for the Fnet force.

F[net]= -kx+c(dx/dt)+FoSin(wt)[/B]

What should be the sign of your c dx/dt term? Think about it.

 

[Futurestar33]

The sign of my dx/dt term should be X(dot)
or are you saying it should be negative instead of positive?

Once I have that should I move all the signs to the other side

 

[BvU]

Hello star,

Rudy wants you to think about the sign: a + or a -
In other words: if dx/dt > 0, which way does a damping force point ?

Usually we take the damping coefficient (in your case c) as a positive value and therefore we need a + or - sign in the equation of motion to let the force point the correct way.

 

[Futurestar33]

When dx/dt >0 then the force has displacement to the right

Well I rearranged the equation of motion into
X(doubledot)=-(k/m)-cx(dot)+Fosin(wt)

which then turns into X(doubledot)= -Wo^2X-2γx(dot)+(Fo/m)Sin(wt)

Then I balanced the equation by bringing everything to the otherside
and using known equation for the angular frequence Wo=√(k/m) and C = 2γm
we receive

X(doubledot)= -Wo^2X-2γx(dot)+(Fo/m)Sin(wt)
X(doubledot)+Wo^2X+2γx(dot)=(Fo/m)Sin(wt)

Then in order to find the steady state solution we must assume the gamma is equal to 0 but why do we consider gamma as equal to zero?

 

[BvU]

$$\ddot x + 2\gamma\;\dot x + \omega_0^2\; x = {F_0\over m} \sin(\omega t)$$ looks good to me. But the

Then in order to find the steady state solution we must assume the gamma is equal to 0

does not. As you say: "why consider $\gamma = 0$ ?". What is your perception of this steady state solution you are looking for ?

the steady state is usually in the form of X(t)= A cos(wt+Φ)

is indeed correct. What it tells you (should tell you) is that after an initial reponse that dampens out (-- thanks to the non-zero gamma! --), the oscillator will oscillate with the period of the driving force and with a certain amplitude; there will also be (or may be) a phase difference between the driving force and the oscillator.

That the given form is actually a solution can be shown by substituting it in the equation. That will also help you on your way to find A and $\phi$.