The Steady-state motion of a forced oscillator

[RJLiberator]

Homework Statement

Solve for the steady-statem otion of a forced oscillator if the forcing is F_0*sin(wt) instead of F_0*cos(wt). Use complex representations.

Homework Equations

e^(i*x) = cos(x)+isin(x)

The Attempt at a Solution

I assume, First, steady-state means without damping.
Next, we have the equation
m*x'' = -kx + F_0*sin(wt)
where x'' is the second derivative of x position with respect to time. Also known as acceleration.

So, we have:

z'' + w_0^2*z = F_0/m * (-i*e^(iwt)) where Re(z) = x

So far so good? I think I changed it correctly, normally when you do F_0*cos(wt) as forcing, you just switch it to e^(iwt)

Now, if everything is looking good up to here, we take z = A*(-ie^[i(wt-φ)])

After taking the second derivative of this and inputting it into our equation, our equation looks like:

i*w^2*A*e^[i(wt-φ)]-w_0^2*A*i*e^[i(wt-φ)] = F_0/m * (-i*e^(iwt))

dividing through by -i*e^(iwt) nets us

F_0/m = A(w_0^2 - w^2)*e^(-iφ)
Simplifying: F_0/m *e^(iφ) = A(w_0^2-w^2)

But now I seem to have run into a problem. The real part of this is F_0/m*cos(φ)
But this is the exact same solution to the forcing with F_0*cos(wt).
I thought the problem wants me to reach Re(z) = F_0*sin(wt)

Any help? Did I go wrong somewhere?

 

[RJLiberator]

Wait,

Did I reach the answer here:

i*w^2*A*e^[i(wt-φ)]-w_0^2*A*i*e^[i(wt-φ)] = F_0/m * (-i*e^(iwt))

I think that's all the question is looking for.

 

[RJLiberator]

Hey guys, sorry to bump this with an extra post.
But only 2 hours until class starts and was interested in if my answer was looking good, as it is a hard problem for me to understand fully.

 

[ehild]

Hey guys, sorry to bump this with an extra post.
But only 2 hours until class starts and was interested in if my answer was looking good, as it is a hard problem for me to understand fully.

If you take the imaginary part of the solution, the driving function is Fo sin(wt)

 

[RJLiberator]

I'm not sure what you mean by that.

The problem states:

if the forcing is F_0*sin(wt) instead of F_0*cos(wt).

So, isn't that already assumed.

 

[ehild]

You solve the differential equation for the case F=F₀e^iwt, and you get the steady-state solution as z=Ae^iwt. In principle, the amplitude can be complex, but you get that A=F₀/(w₀²-w²), which is real. Taking the imaginary part, the real-life solution is

y=F₀/(w₀²-w²)sin(wt). sine function instead of cosine.

 

[RJLiberator]

Hm.
Few concerns:

The forcing part of it is F_0*sin(wt)

so we have
m*a = -kx + F_0*sin(wt)

If we are to convert this to complex numbers we have
m*d^2z/dt^2 +kz = F_0 * (-i*e^(iwt))

Why do you say z = Ae^(iwt) and not what I think should be (-i*e^(iwt)) as my part seems to provide a real part of sin(wt) instead of an imaginary part.

 

[ehild]

Hm.
Few concerns:

The forcing part of it is F_0*sin(wt)

so we have
m*a = -kx + F_0*sin(wt)

If we are to convert this to complex numbers we have
m*d^2z/dt^2 +kz = F_0 * (-i*e^(iwt))

Why do you say z = Ae^(iwt) and not what I think should be (-i*e^(iwt)) as my part seems to provide a real part of sin(wt) instead of an imaginary part.

F_0*sin(wt) is not the same as F_0 * (-i*e^(iwt)), but F_0*sin(wt)=Im(F_0 e^(iwt))

 

[RJLiberator]

Okay, thatttt makes a lot of sense to me.

So using z = Ae^[i(wt-Φ)] is the correct choice, but instead of Re(z) =x we let Im(z) = x.

Solution would be then:
(iw)^2*A*e^[i(wt-Φ)]+w_0^2*A*e^[i(wt-Φ)] = F_0/m * e^(iwt)

With Im(z) = x.

Would that be the correct way to display the solution (I guess here I am getting lost on what the problem actually wants.)
It seems like this would be the steady-state motion of a forced oscillator. But this is still the same equation as with cos(wt) just where cos(wt) is with Re(z).

Or, do I input for z = sin(wt) and solve?

 

[ehild]

Okay, thatttt makes a lot of sense to me.

So using z = Ae^[i(wt-Φ)] is the correct choice, but instead of Re(z) =x we let Im(z) = x.

Solution would be then:
(iw)^2*A*e^[i(wt-Φ)]+w_0^2*A*e^[i(wt-Φ)] = F_0/m * e^(iwt)

Solve this equation for A, and then give the solution as A*sin(wt-Φ)
But you get the same solution if you just input x=Asin(wt-Φ). Φ will be 0 or pi.

 

[RJLiberator]

A = (F_0/m) * (1/(w^2-w_0^2) * sin(Φ)

:D ?

 

[ehild]

A = (F_0/m) * (1/(w^2-w_0^2) * sin(Φ)

:D ?

And what is Φ?