Another Driven Harmonic Oscillator problem

[Benzoate]

Homework Statement

A driven oscillator satisfies the equation

x'' + omega²=F0cos(omega(1+episilon)t]

where episilon is a positive constant. Show that the solution that satisfies the iniitial conditions x=0 and x'=0 when t=0 is

x= (F0*sin(.5episilon*omega * t) sin(omega(1+.5episilon)t)/(episilon(1+.5episilon)omega^2)

Homework Equations

The Attempt at a Solution

cos(omega*t)=cos omega(1+.5episilon)t-.5omega*t)
cos(omega(1+episilon)t)=cos(omega(1+.5episilon)t + .5omega*t)

let x=ce^iw_pt

x'= ciw_pe^iw_pt
x''= -c*w_pe^]iw_pt
c= F0/(-w_p²+omega^2)

Therefore
x= (F0/(-w_p²+omega^2))*ce^iw_pt

Am I heading in the right direction

Should I derived an equation for the complimentary solution?

 

[olgranpappy]

Homework Statement

A driven oscillator satisfies the equation

x'' + omega²=F0cos(omega(1+episilon)t]

where episilon is a positive constant. Show that the solution that satisfies the iniitial conditions x=0 and x'=0 when t=0 is

x= (F0*sin(.5episilon*omega * t) sin(omega(1+.5episilon)t)/(episilon(1+.5episilon)omega^2)

Homework Equations

The Attempt at a Solution

cos(omega*t)=cos omega(1+.5episilon)t-.5omega*t)
cos(omega(1+episilon)t)=cos(omega(1+.5episilon)t + .5omega*t)

let x=ce^iw_pt

x'= ciw_pe^iw_pt
x''= -c*w_pe^]iw_pt
c= F0/(-w_p²+omega^2)

Therefore
x= (F0/(-w_p²+omega^2))*ce^iw_pt

Am I heading in the right direction

Should I derived an equation for the complimentary solution?

You are given a solutions, and all you are asked to do it verify that what you are given is indeed a solution. You are not asked to solve the equation for x(t) you are only asked to check that the given solution is correct.

So... take what you are given and first check that it satisfies the boundary conditions. For very first, check that it satisfies x(0)=0. Can you verify this?

 

[Benzoate]

You are given a solutions, and all you are asked to do it verify that what you are given is indeed a solution. You are not asked to solve the equation for x(t) you are only asked to check that the given solution is correct.

So... take what you are given and first check that it satisfies the boundary conditions. For very first, check that it satisfies x(0)=0. Can you verify this?

my equation for my driven response looks like:

x_p=(F₀*e^iw_p*t

For complementary/transient motion should my equation be:

x_c=A*cos(w_p*t)+B*sin(w_p*t)=0

 

[Hootenanny]

Have you even read the previous post? You are not required to solve any equations, you need to simply check that the given solution satisfies the boundary value problem.

 

[Benzoate]

Have you even read the previous post? You are not required to solve any equations, you need to simply check that the given solution satisfies the boundary value problem.

But don't I need to find the equation for the driven response and the equation for the transient motion?

general solution

x=xc+xp= A*cos(wp*t)+B*sin(wp*t)+(F0*eiwp*t)/(w0^2-w^2)^2 and now that I have a general solution I can plug in my initial conditions that will helped me find A and B and I now that I have A and B, I can convert my general solution to the particular solution in the book?

 

[Hootenanny]

But don't I need to find the equation for the driven response and the equation for the transient motion?

general solution

x=xc+xp= A*cos(wp*t)+B*sin(wp*t)+(F0*eiwp*t)/(w0^2-w^2)^2 and now that I have a general solution I can plug in my initial conditions that will helped me find A and B and I now that I have A and B, I can convert my general solution to the particular solution in the book?

No, you don't need to find the general solution, you are given the particular solution! Of course you could go through all the trouble of solving the differential equation, then imposing the boundary conditions and finally showing that your solution is equivalent to that given. However, a much more straight forward method would be to simply substitute the given solution into the differential equation and check that the equation (and all boundary conditions) are satisfied.

Do you follow?

 

[Benzoate]

No, you don't need to find the general solution, you are given the particular solution! Of course you could go through all the trouble of solving the differential equation, then imposing the boundary conditions and finally showing that your solution is equivalent to that given. However, a much more straight forward method would be to simply substitute the given solution into the differential equation and check that the equation (and all boundary conditions) are satisfied.

Do you follow?

yes. Thanks.