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#### alane1994

##### Active member

- Oct 16, 2012

- 126

The title may be incorrect, I named this after the section of my book in which this is located.

My problem is as follows.

Determine \(\omega_0\), R, and \(\delta\) so as to write the given expression in the form

\(u=R\cos(\omega_0 t-\delta)\)

\(\color{blue}{u=4\cos(3t)-2\sin(3t)},~\text{My Problem}\)

I know that,

\(u(t)=A\cos(\omega_0 t)+B\sin(\omega_0 t)\)

\(\omega_0=\sqrt{\dfrac{k}{m}}\)

\(A=R\cos(\delta),~~B=R\sin(\delta)~~\Rightarrow~R=\sqrt{A^2+B^2},~\tan(\delta)=\dfrac{B}{A}\)

So that means that,

\(\omega_0=3\)

\(A=4\)

\(B=-2\)

\(R=2\sqrt{5}\)

\(\delta=\tan^{-1}(\dfrac{-2}{4})\approx-.463648\)

Now I am a little confused as to where to go from here. Any thoughts?

ADDITIONAL THOUGHTS:

Would I then just plug in the values into the desired format above?

My problem is as follows.

Determine \(\omega_0\), R, and \(\delta\) so as to write the given expression in the form

\(u=R\cos(\omega_0 t-\delta)\)

\(\color{blue}{u=4\cos(3t)-2\sin(3t)},~\text{My Problem}\)

I know that,

\(u(t)=A\cos(\omega_0 t)+B\sin(\omega_0 t)\)

\(\omega_0=\sqrt{\dfrac{k}{m}}\)

\(A=R\cos(\delta),~~B=R\sin(\delta)~~\Rightarrow~R=\sqrt{A^2+B^2},~\tan(\delta)=\dfrac{B}{A}\)

So that means that,

\(\omega_0=3\)

\(A=4\)

\(B=-2\)

\(R=2\sqrt{5}\)

\(\delta=\tan^{-1}(\dfrac{-2}{4})\approx-.463648\)

Now I am a little confused as to where to go from here. Any thoughts?

ADDITIONAL THOUGHTS:

Would I then just plug in the values into the desired format above?

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