Deriving x(t) into A sin (\omega*t + \varphi)

[krnhseya]

Homework Statement

Derive...

x(t)=(exp^(-\zeta\omegat))*(a1(exp^(i\omegasqrt(1-\zeta^2)*t)))+a2(exp^(-i\omegasqrt(1-\zeta^2)*t))))

into

x(t)=(exp^(-\zeta\omegat))*(A sin (\omega*t + \varphi))

Homework Equations

n/a

The Attempt at a Solution

I've managed to get x(t) = (exp^(-\zeta\omegat))*((a1(cos\omega*t) + i sin (\omega*t))+(a2(cos\omega*t) - i sin (\omega*t))) then when i simplify things...sin terms cancel out and i end up geting...

exponential term * (a1+a2) * (2cos \omega*t)

 

[quantumdude]

Hi,

Is the \zeta a constant here?

I've managed to get x(t) = (exp^(-\zeta\omegat))*((a1(cos\omega*t) + i sin (\omega*t))+(a2(cos\omega*t) - i sin (\omega*t))) then when i simplify things...sin terms cancel out and i end up geting...

exponential term * (a1+a2) * (2cos \omega*t)

No, the sine terms do not cancel out. You have the following.

x(t)=e^{-\zeta\omega t}\left[(a_1+a_2)\cos\left(\omega\sqrt{1-\zeta^2}t\right)+(a_1-a_2)\sin\left(\omega\sqrt{1-\zeta^2}t\right)\right].

The only way the sine term will vanish is if a_1=a_2, and you did not say that that is the case in your problem statement.

 

[krnhseya]

Hi,

Is the \zeta a constant here?



No, the sine terms do not cancel out. You have the following.

x(t)=e^{-\zeta\omega t}\left[(a_1+a_2)\cos\left(\omega\sqrt{1-\zeta^2}t\right)+(a_1-a_2)\sin\left(\omega\sqrt{1-\zeta^2}t\right)\right].

The only way the sine term will vanish is if a_1=a_2, and you did not say that that is the case in your problem statement.

Hello. It is a constant.
Yeah, I figured out that I made a mistake.

that's as far as i can go but i need to combine that into sin like I've posted above.

thanks for the reply!

And, there should be a bracket infront of "sqrt" so exponential part it outside the whole thing.

[edit] I think I got it...Thank you very much! :)