Damped Free Forced Vibration

[shorty]

Something i am doing is not adding up. I don't understand the part with the external force. here's the question:

Show that the solution for the damped free forced vibration given by is when , where

something along the way i'm doing wrong, because not for heck can i get that fraction with the sin theta part.

Help Please?

 

[Chris L T521]

Something i am doing is not adding up. I don't understand the part with the external force. here's the question:

Show that the solution for the damped free forced vibration given by View attachment 155 is View attachment 156 when View attachment 157, where View attachment 158

something along the way i'm doing wrong, because not for heck can i get that fraction with the sin theta part.

Help Please?

If you solve the homogeneous equation $my^{\prime\prime}+ky=0$, you get $y_c = c_1\cos(\omega t)+c_2\sin(\omega t)$, where $\omega = \sqrt{\dfrac{k}{m}}$.

Now, for the particular solution, we can proceed by the method of undetermined coefficients where $y_p = (A+Bt)\cos(\omega t) + (C+Dt)\sin(\omega t)$. This then satisfies the differential equation $my_p^{\prime\prime}+ky_p=\cos(\theta t)$.

Solve for $A$, $B$, $C$, and $D$; you should then get the particular solution. $y=y_c+y_p$ should then be the desired final result.

I hope this helps!

 

[CaptainBlack]

Something i am doing is not adding up. I don't understand the part with the external force. here's the question:

Show that the solution for the damped free forced vibration given by View attachment 155 is View attachment 156 when View attachment 157, where View attachment 158

something along the way i'm doing wrong, because not for heck can i get that fraction with the sin theta part.

Help Please?

This is a "show" question, which means you only need demonstrate that the given solution is indeed a solution of the DE. Presumably you can solve the homogeneous part, so you need only show that \(f(t)=\frac{t \sin(\omega t)}{2\omega}\) is a solution of:

\(my''+ky=\cos(\omega t) \)

You might note at this point that as \(\Theta=\omega \) you have resonance and the amplitude of the oscilation becomes arbitarily large as \(t\) increases without bound.

CB

 

[shorty]

If you solve the homogeneous equation $my^{\prime\prime}+ky=0$, you get $y_c = c_1\cos(\omega t)+c_2\sin(\omega t)$, where $\omega = \sqrt{\dfrac{k}{m}}$.

Now, for the particular solution, we can proceed by the method of undetermined coefficients where $y_p = (A+Bt)\cos(\omega t) + (C+Dt)\sin(\omega t)$. This then satisfies the differential equation $my_p^{\prime\prime}+ky_p=\cos(\theta t)$.

Solve for $A$, $B$, $C$, and $D$; you should then get the particular solution. $y=y_c+y_p$ should then be the desired final result.

I hope this helps!

This was the easy part. I got that far but when I did my substitution for the coefficients into the equation my particular solution wasn't adding up..

 

[ifeg]

This is what i am unsure about. How do i go about showing that \(f(t)=\frac{t \sin(\omega t)}{2\omega}\) is a solution of:

\(my''+ky=\cos(\omega t) \)? i don't understand the resonance and the amplitude aspects..

This is a "show" question, which means you only need demonstrate that the given solution is indeed a solution of the DE. Presumably you can solve the homogeneous part, so you need only show that \(f(t)=\frac{t \sin(\omega t)}{2\omega}\) is a solution of:

\(my''+ky=\cos(\omega t) \)

You might note at this point that as \(\Theta=\omega \) you have resonance and the amplitude of the oscilation becomes arbitarily large as \(t\) increases without bound.

CB

 

[Mr Fantastic]

This is what i am unsure about. How do i go about showing that \(f(t)=\frac{t \sin(\omega t)}{2\omega}\) is a solution of:

\(my''+ky=\cos(\omega t) \)? i don't understand the resonance and the amplitude aspects..

Calculate the first and second derivatives of y. Substitute them into the left hand side of the given differential equation. Simplify the result. The simplified result will equal the right hand side, provided $\omega$ is equal to a particular expression involving m and k. QED.

 

[ifeg]

Thank you for the tip. i tried it though and it didn't quite work out. but i will keep playing with it and see.

Calculate the first and second derivatives of y. Substitute them into the left hand side of the given differential equation. Simplify the result. The simplified result will equal the right hand side, provided $\omega$ is equal to a particular expression involving m and k. QED.

 

[CaptainBlack]

Thank you for the tip. i tried it though and it didn't quite work out. but i will keep playing with it and see.

I can assure you it does, if you post what you have tried here we will see if we can help you fing the errors etc...

CB