# TrigonometryMechanical Vibrations - Linear Combinations

#### alane1994

##### Active member
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?

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

Last edited:

#### topsquark

##### Well-known member
MHB Math Helper
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?

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

##### Active member
Is the phase shift the $$\delta$$?

#### MarkFL

Staff member
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:

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

in the form:

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

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

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

Distributing the $R$, we have:

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

Comparison of this with the desired form, we find:

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

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

$$\displaystyle \omega_0=3$$

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

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

Dividing the second equation by the first, we find:

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

and so we may state:

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

#### alane1994

##### Active member
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 #### MarkFL

Staff member
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 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:

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

MHB Math Helper

#### MarkFL

Staff member
Yup!

-Dan
I believe the phase shift would actually be:

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

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

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

#### topsquark

##### Well-known member
MHB Math Helper
I believe the phase shift would actually be:

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

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

$$\displaystyle u(t)=R\cos\left(\omega_0\left(t-\frac{\delta}{\omega_0} \right) \right)$$
(Ahem!) That's one on MHF and now one on MHB. I'm going to bed.

Thanks for the catch.

-Dan