Mechanical Vibrations - Linear Combinations

[alane1994]

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?

 

[topsquark]

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?

Looks good. One note: When I graphed these (easiest way to check) I needed several more digits for the phase shift to make it work right.

-Dan

 

[alane1994]

Is the phase shift the \(\delta\)?

 

[MarkFL]

I have moved this topic to our Trigonometry sub-forum since the problem, while it comes from an application of a second order linear ODE, involves only trigonometry. I have also edited the title.

We want to express the solution:

$$u(t)=4\cos(3t)-2\sin(3t)$$

in the form:

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

I would use the angle-difference identity for cosine to write:

$$u(t)=R\left(\cos(\omega_0 t)\cos(\delta)+\sin(\omega_0 t)\sin(\delta) \right)$$

Distributing the $R$, we have:

$$u(t)=R\cos(\omega_0 t)\cos(\delta)+R\sin(\omega_0 t)\sin(\delta)$$

Comparison of this with the desired form, we find:

$$R\cos(\delta)=4$$

$$R\sin(\delta)=-2$$

$$\omega_0=3$$

Squaring the first two equations, and adding, we get:

$$R^2=20\implies\,R=2\sqrt{5}$$

Dividing the second equation by the first, we find:

$$\tan(\delta)=-\frac{1}{2}\implies\delta=-\tan^{-1}\left(\frac{1}{2} \right)$$

and so we may state:

$$u(t)=2\sqrt{5}\cos\left(3t+\tan^{-1}\left(\frac{1}{2} \right) \right)$$

 

[alane1994]

Ok, so you do just plug them back into the friendly equation from earlier in the problem!
I need to stop over-thinking things, it just seemed too easy for a course of this level :P

 

[MarkFL]

Ok, so you do just plug them back into the friendly equation from earlier in the problem!
I need to stop over-thinking things, it just seemed too easy for a course of this level :P

Yes, you did everything correctly, the only thing I would have done further is reduce the argument of the inverse tangent function and avoided using a decimal approximation for the resulting angle $\delta$.

Technically, I should have written:

$$ u(t)=2\sqrt{5}\cos\left(3t-\left(-\tan^{-1}\left(\frac{1}{2} \right) \right) \right)$$

 

[topsquark]

Is the phase shift the \(\delta\)?

Yup!

-Dan

 

[MarkFL]

Yup!

-Dan

I believe the phase shift would actually be:

$$\frac{\delta}{\omega_0}$$

which can be seen by writing the solution in the form:

$$u(t)=R\cos\left(\omega_0\left(t-\frac{\delta}{\omega_0} \right) \right)$$

 

[topsquark]

I believe the phase shift would actually be:

$$\frac{\delta}{\omega_0}$$

which can be seen by writing the solution in the form:

$$u(t)=R\cos\left(\omega_0\left(t-\frac{\delta}{\omega_0} \right) \right)$$

(Ahem!) That's one on MHF and now one on MHB. I'm going to bed.

Thanks for the catch.

-Dan