Real and Complex representations of an oscillation equation

[e_mts]

Homework Statement

For each case find the complex representation for all three quantities – displacement, velocity, and acceleration – in the form: $Q(t) = Qe^{−i(ωt+φ)}$

Relevant Equations

$e ^ {i \omega t} = Cos(\omega t) + iSin(\omega t)$

More in the attached picture

I've been trying to continue my education by self-teaching during quarantine (since I can't really go to college right now) with the MIT Opencourseware courses. I landed on one section that's got me stuck for a while which is the second part of this problem (I managed to finish the first part without any issues):

For problem d, I tried both working with $Q(t)=Qe^{-i(\omega t + \phi)}$ to make it look like d but I got as far as this (by setting $\phi=0$): $$Q(t)=QCos(\omega t) - QiSin(\omega t)$$ which can't be simplified to look like d (as far as I can tell) because $\beta$ is a real constant (might be wrong here).

The other thing I tried to do was working with d to make it look like $Q(t)=Qe^{-i(\omega t + \phi)}$ but that leaves me with: $$\alpha \frac{e^{i\omega t}+e^{-i\omega t}}{2}-\beta \frac{e^{i\omega t}-e^{-i\omega t}}{2i}$$ after trying to simplify that I landed on: $$\frac{1}{2i}(e^{i\omega t}(\alpha i - \beta)+e^{-i\omega t}(\alpha i +\beta))$$ which doesn't look at all like what I need to get.

I'd like to get some help for this problem because I'm not even sure about what to study to start figuring out how to figure it out. It would be very appreciated, thank you.

 

[Master1022]

Hi,

For (d), perhaps you can reformulate/condense $x(t)$ as a double angle cos and use that to represent the solution as a complex phasor.

Hope that makes sense as a starting point. If not, let me know and I can try to explain better

[EDIT]:
As you stated above, $Qe^{i (\omega t + \phi)} = Qcos(\omega t + \phi) + i Qsin(\omega t + \phi)$ and therefore we have the real part of the solution and want to represent it as $Qcos(\omega t + \phi)$ such that we can use the complex exponential form.

 

[vela]

Which course is this problem from?

 

[e_mts]

Which course is this problem from?

This is from Course 8.03 Physics III: Vibrations and Waves, the problem is from https://ocw.mit.edu/courses/physics/8-03sc-physics-iii-vibrations-and-waves-fall-2016/part-i-mechanical-vibrations-and-waves/lecture-1/MIT8_03SCF16_ProblemSet1.pdf.

 

[e_mts]

Hi,

For (d), perhaps you can reformulate/condense $x(t)$ as a double angle cos and use that to represent the solution as a complex phasor.

Hope that makes sense as a starting point. If not, let me know and I can try to explain better

[EDIT]:
As you stated above, $Qe^{i (\omega t + \phi)} = Qcos(\omega t + \phi) + i Qsin(\omega t + \phi)$ and therefore we have the real part of the solution and want to represent it as $Qcos(\omega t + \phi)$ such that we can use the complex exponential form.

Thank you, I will try this as soon as I can. I was too focused on the euler form to think about that.

 

[vela]

You want to use $A \cos(\omega t + \phi) = \text{real part of }~Ae^{i(\omega t + \phi)}$.

For part (d), you want to first figure out what $A$ and $\phi$ are in terms of $\alpha$ and $\beta$. Try using the cosine angle-addition formula to expand $A \cos(\omega t + \phi)$ and compare it to the expression given in (d).

 

[e_mts]

You want to use $A \cos(\omega t + \phi) = \text{real part of }~Ae^{i(\omega t + \phi)}$.

For part (d), you want to first figure out what $A$ and $\phi$ are in terms of $\alpha$ and $\beta$. Try using the cosine angle-addition formula to expand $A \cos(\omega t + \phi)$ and compare it to the expression given in (d).

Admittedly, I seem to be missing a math tool needed to understand that then. I haven't taken the complex variable functions class at school but I'm willing to learn what I have to to be able to finish this problem. Would you be able to point me to the right direction for what I need to study, I don't really understand when you talk about "the real part" or "the complex part" of the function.

EDIT: Just to throw this out there, would I need to then work through the equation until I find both the real and imaginary parts and then only work with the real part?

 

[vela]

Let $z=x+iy$ be a complex number. Here $x$ and $y$ are real. The real part of $z$ is $x$.

 

[e_mts]

Here's what I got for (d), using $Q(t)=Qe^{-i(\omega t+\phi)}=QCos(\omega t+\phi)-iQSin(\omega t+\phi)$, taking the real part $QCos(\omega t+\phi)$ and working with that: $$QCos(\omega t+\phi)=Q[Cos(\omega t)Cos(\phi)-Sin(\omega t)Sin(\phi)]$$ Considering that $\alpha \neq \beta$ (I think that's implied), we can take any $\phi$ so that $Sin(\phi)\neq 0, Cos(\phi)\neq 0$ and $Sin(\phi)\neq Cos(\phi)$: $$QCos(\omega t+\phi)=aQCos(\omega t)-bQSin(\omega t)$$ Since both $aQ, bQ \in \mathbb{R}$ then we can say that $aQ=\alpha , bQ=\beta$ $$\implies Q(t)=\alpha Cos(\omega t) - \beta Sin(\omega t)$$ $$\therefore Q(t)=x(t)$$ Is that right? If it is, I think I can work through the rest of the problems since they basically just boil down to this.

 

[Master1022]

Here's what I got for (d), using $Q(t)=Qe^{-i(\omega t+\phi)}=QCos(\omega t+\phi)-iQSin(\omega t+\phi)$, taking the real part $QCos(\omega t+\phi)$ and working with that: $$QCos(\omega t+\phi)=Q[Cos(\omega t)Cos(\phi)-Sin(\omega t)Sin(\phi)]$$ Considering that $\alpha \neq \beta$ (I think that's implied), we can take any $\phi$ so that $Sin(\phi)\neq 0, Cos(\phi)\neq 0$ and $Sin(\phi)\neq Cos(\phi)$: $$QCos(\omega t+\phi)=aQCos(\omega t)-bQSin(\omega t)$$ Since both $aQ, bQ \in \mathbb{R}$ then we can say that $aQ=\alpha , bQ=\beta$ $$\implies Q(t)=\alpha Cos(\omega t) - \beta Sin(\omega t)$$ $$\therefore Q(t)=x(t)$$ Is that right? If it is, I think I can work through the rest of the problems since they basically just boil down to this.

You need to include the phase information. If you look at the expansion of the double angle: $$QCos(\omega t+\phi)=Q[Cos(\omega t)Cos(\phi)-Sin(\omega t)Sin(\phi)]$$ and compare it to: $$ x(t) = \alpha cos(\omega t) - \beta sin(\omega t) $$, then we can see that $\alpha = Qcos(\phi)$ and $\beta = Qsin(\phi)$. Given those two expressions (found by comparing coefficients of the sin and cos), can you now re-write $Qcos(\omega t + \phi)$ in terms of those variables (you need to re-write Q and $\phi$ in terms of $\alpha$ and $\beta$).

To start off, can we use the $\alpha$ and $\beta$ equations to find Q in terms of $\alpha$ and $\beta$?

I hope that makes sense

 

[Master1022]

To follow up on your post,
$$ x(t) = Re(Q(t)) $$ Remember that the goal is to condense the function of time into a complex phasor of the form $Q e^{-i(\omega t + \phi)}$. Now we need to condense your expression $Q(t) = \alpha cos(\omega t) - \beta sin(\omega t)$ into the singular cosine with a phase shift such that we can it as that phasor.

I hope I am making some sense. If not, please let me know and I can explain more. I am going into this response assuming that you have seen phasors (complex exponentials) - please let me know if I am wrong.

Overview (just to try and make sure I am clear) for (d):
- You have a function of time involving two sinusoids which we want to represent as a phasor
- We are told that $x(t)$ is the real component of the phasor which is $Qcos(\omega t + \phi)$
- Thus we need to represent our $x(t)$ in that above cosine form with the phase shift. Why? From there, we can directly write down the phasor as we will have expressions for Q and $\phi$

 

[e_mts]

To follow up on your post,
$$ x(t) = Re(Q(t)) $$ Remember that the goal is to condense the function of time into a complex phasor of the form $Q e^{-i(\omega t + \phi)}$. Now we need to condense your expression $Q(t) = \alpha cos(\omega t) - \beta sin(\omega t)$ into the singular cosine with a phase shift such that we can it as that phasor.

I hope I am making some sense. If not, please let me know and I can explain more. I am going into this response assuming that you have seen phasors (complex exponentials) - please let me know if I am wrong.

Overview (just to try and make sure I am clear) for (d):
- You have a function of time involving two sinusoids which we want to represent as a phasor
- We are told that $x(t)$ is the real component of the phasor which is $Qcos(\omega t + \phi)$
- Thus we need to represent our $x(t)$ in that above cosine form with the phase shift. Why? From there, we can directly write down the phasor as we will have expressions for Q and $\phi$

No, I'm sorry I haven't studied complex exponentials admittedly and am not too familiar with wave equations (which is why I'm studying this). Which math topic should I start studying in order to understand this?

 

[vela]

Considering that $\alpha \neq \beta$ (I think that's implied), we can take any $\phi$ so that $Sin(\phi)\neq 0, Cos(\phi)\neq 0$ and $Sin(\phi)\neq Cos(\phi)$

There's no need to assume $\alpha \ne \beta$ or that one of them can't be zero. I'm not sure why you're adding those assumptions.

$$QCos(\omega t+\phi)=aQCos(\omega t)-bQSin(\omega t)$$

You never defined what $a$ and $b$ are.

$$\therefore Q(t)=x(t)$$ Is that right?

No. $Q(t)$ is complex while $x(t)$ is purely real. They're not equal.

As @Master1022 says, you need to find an expression for $Q$ in terms of just $\alpha$ and $\beta$. Likewise for $\phi$.

 

[Master1022]

No, I'm sorry I haven't studied complex exponentials admittedly and am not too familiar with wave equations (which is why I'm studying this). Which math topic should I start studying in order to understand this?

Okay, not to worry. There isn't much math required for phasors (simply put: they are 'lines' used to represent a complex number) but you might want to look at:
- Complex numbers: properties, euler notation, argand diagram, etc.
- Calculus: differentiation/ integration
- Trigonometry identities: double angle and the use of 'harmonic form' (this is what we are trying to do above and there are some youtube videos and google articles to explain what I was talking about when representing $x(t)$ as the single cosine with a phase shift)
(I am just a fellow student, so don't take this list as being exhaustive)

I did have a quick glance at some of the later topics in the course which become increasingly involved with mathematics and thus may require exploration of other topics (e.g. vector calculus and/or differential equations) - I am sure you can tell when there is something that is unfamiliar in the notes/ lectures.

To get back to the original problem, I am not sure I am allowed to just give the answer here. If you look into the 'harmonic form' within trigonometry, you should be able to see how we can represent
$$\alpha cos(\omega t) - \beta sin(\omega t)$$ as $$Q cos(\omega t + \phi)$$
by using the two equations: $\alpha = Qcos(\phi)$ and $\beta = Qsin(\phi)$ to solve for $Q$ and $\phi$

 

[e_mts]

Thank you both @Master1022 and @vela for the patience and help, I'm feeling confident about this one: $$R\{Q(t)\}=QCos(\omega t+\phi)$$ $$QCos(\omega t+\phi)=QCos(\phi)Cos(\omega t)-QSin(\phi)Sin(\omega t)$$ And as @Master1022 mentioned, we need to compare this to: $$x(t)=\alpha Cos(\omega t) - \beta Sin(\omega t)$$ Which leaves us with: $$\alpha =QCos(\phi), \beta =QSin(\phi)$$ After a bit of manipulation we arrive to: $$\phi = tan^-1(\frac{\beta}{\alpha}), Q=\alpha \sqrt{\frac{\beta^2}{\alpha^2}+1}$$ Which we substitute then in (skipping a couple steps in $Cos(tan^-1(x))$: $$R\{Q(t)\}=\bigg(\alpha \sqrt{\frac{\beta^2}{\alpha^2}+1}\bigg)\Bigg(\frac{1}{\sqrt{\frac{\beta^2}{\alpha^2}+1}}\Bigg)Cos(\omega t)-\bigg(\alpha \sqrt{\frac{\beta^2}{\alpha^2}+1}\bigg)\Bigg(\frac{\beta}{\alpha \sqrt{\frac{\beta^2}{\alpha^2}+1}}\Bigg)Sin(\omega t)$$ $$\implies R\{Q(t)\}=\alpha Cos(\omega t)-\beta Sin(\omega t)$$ And, as @vela mentioned, it was a dumb mistake to say that $Q(t)=x(t)$ so: $$\therefore R\{Q(t)\}=x(t)$$ Is this procedure correct?

 

[Master1022]

Thank you both @Master1022 and @vela for the patience and help, I'm feeling confident about this one: $$R\{Q(t)\}=QCos(\omega t+\phi)$$ $$QCos(\omega t+\phi)=QCos(\phi)Cos(\omega t)-QSin(\phi)Sin(\omega t)$$ And as @Master1022 mentioned, we need to compare this to: $$x(t)=\alpha Cos(\omega t) - \beta Sin(\omega t)$$ Which leaves us with: $$\alpha =QCos(\phi), \beta =QSin(\phi)$$ After a bit of manipulation we arrive to: $$\phi = tan^-1(\frac{\beta}{\alpha}), Q=\alpha \sqrt{\frac{\beta^2}{\alpha^2}+1}$$

Yes this approach is correct. After you have found $Q$ and $\phi$ in terms of $\alpha$ and $\beta$, there is no need to resubstitute it back into the expression. Those extra lines after finding Q and $\phi$ just verify that your expressions are correct and yield $x(t)$ (I'm not saying its a bad thing to check the working, but it isn't a necessary step to show).

Now that you have found Q and $\phi$, we can write $Q(t)$ in terms of $\alpha$ and $\beta$.
$$Q(t) = Qe^{-i(\omega t + \phi)}$$
so you just have directly substitute the expressions for Q and $\phi$ into this.

I hope that makes sense. If not, please do let me know.

Spoiler

$$Q(t) = Qe^{-i(\omega t + \phi)} = \sqrt{\alpha^2 + \beta^2} e^{-i(\omega t + tan^{-1} (\frac{\beta}{\alpha}))}$$

You can now picture an arrow of length $\sqrt{\alpha^2 + \beta^2}$, centered at the origin of the argand diagram, rotating around the origin with angular velocity $\omega$ in the clock-wise direction (due to the -ve sign). The $\phi$ represents a phase difference between this signal and an arbitrary one that starts directly on the real-line (take note of the -ve sign here as well, so it will start off as being a negative angle). The real part of this arrow at any given point is the projection of the arrow onto the number line and this represents $x(t)$.