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The following differential equation represents the motion of a body of mass $m$ and displacement $x$ from the mean position, that is attached to a spring of force constant $a$ and viscous damping coefficient $b$ : $$\bbox[5px,border:1px solid black] { m\frac{d^2x}{dt^2}=-ax-b\frac{dx}{dt}}$$

On rearrangement of the above equation we get the following: $$\bbox[5px,border:1px solid black] { \frac{d^2x}{dt^2}+\frac bm\cdot\frac{dx}{dt}+\frac amx=0 }$$ Defining $k\triangleq\frac{b}{2m}$ and $\omega^2=\frac am$, we have: $$\bbox[5px,border:1px solid black] { \frac{d^2x}{dt^2}+2k\frac{dx}{dt}+\omega^2x=0 }$$

Hence, we solve it as:

            Image 1]

In the situation of the undamped oscillation (shown in the image below):

  • What is the significance of the term $i(A-B)\sin(\omega't)$?
  • And also is the line highlighted with green justified? Why or why not?

               Image 2

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closed as off-topic by Aaron Stevens, Kyle Kanos, ZeroTheHero, Jon Custer, user191954 Nov 16 '18 at 17:03

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  • $\begingroup$ Term with iota? Did you mean "phi" ($\phi$)? $\endgroup$ – Vinicius ACP Nov 3 '18 at 8:50
  • $\begingroup$ No i(A-B)sin(w't) $\endgroup$ – Abhishek Ghosh Nov 3 '18 at 8:51
  • $\begingroup$ Also, I'm voting to move this to Mathematics SE, since this is not a physics question, even though it has applications in physics. $\endgroup$ – Aaron Stevens Nov 3 '18 at 11:03
  • $\begingroup$ The green part along with the purple part are just assignment of two new variables (C and $\phi$) instead of the old variables (A and B). Note that the new variables can be complex. The final line in the image presents the solution using these new variables. The advantage of the new presentation is that you can easily relate it to initial conditions, say $x(t=0)$ and $v(t=0)$. Initial conditions, being real, will enforce C and $\phi$ to be real as well. $\endgroup$ – npojo Nov 3 '18 at 14:40
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We have a 2nd order Linear Homogeneous Ordinary Differential Equation (LH-ODE): $$\frac{d^2x}{dt^2}+2k\frac{dx}{dt}+\omega^2x=0 \tag{I} $$

Its general solution is, as pointed in the question: $$x(t)=e^{-kt}\cdot(Ae^{i\omega't}+Be^{-i\omega't})$$


What the significance of the term $i(A-B)\sin(\omega't)$ ?

Using Euler's formula $(e^{i\theta}=\cos\theta+i\sin\theta)$, we have: \begin{cases} e^{i\omega't}=\cos(\omega't)+i\sin(\omega't) \\ e^{-i\omega't}=\cos(-\omega't)+i\sin(-\omega't)=\cos(\omega't)-i\sin(\omega't) \end{cases}

Therefore: \begin{align} x(t) & = e^{-kt}\cdot\big[A\cos(\omega't)+iA\sin(\omega't)+B\cos(\omega't)-iB\sin(\omega't)\big]\\ & =e^{-kt}\cdot\big[(A+B)\cos\omega't+i(A-B)\sin\omega't\big]\\ & =e^{-kt}\cdot(A+B)\cos\omega't+i\cdot e^{-kt}(A-B)\sin\omega't \tag{II} \end{align}

Note that the equation $\text{(II)}$ above is still the general solution of $\text{(I)}$. Now let's remember the linearity of the solutions and the superposition principle valid for any LH-ODE:

If $x_1(t)$ and $x_2(t)$ are particular solutions, then $x_1(t)+x_2(t)$ is also a particular solution;

If $x(t)$ is a particular solution, then $u\cdot x(t)$, where $u\in\mathbb{C}$ is an arbitrary constant, is also a particular solution;

The general solution in given by: $x(t)=u\cdot x_1(t)+v\cdot x_2(t)$, where $u,v\in\mathbb{C}$ are arbitrary constants.

So, comparing the above general solution with $\text{(II)}$: \begin{cases} x_1(t)=e^{-kt}\cdot\cos(\omega't) & \text{and}\space\space\space\space u=A+B \\ x_2(t)=e^{-kt}\cdot\sin(\omega't) & \text{and}\space\space\space\space v=i\cdot(A-B) \end{cases} Therefore, $\space i\cdot(A-B)\sin\omega't\space$ has no meaning, but $\space i\cdot (A-B)e^{-kt}\sin\omega't\space$ has: is one of the particular solutions of $\text{(I)}$.


What is the justification for the use of $\space i\cdot(A-B)=C\cos\phi$ ?

First, let's find the values of $A$ and $B$ : $$\begin{align} \begin{cases} A+B=C\sin\phi\\ i\cdot(A-B)=C\cos\phi \end{cases} \iff \begin{cases} A=\frac C2\left(\sin\phi+i\cos\phi\right)\\ B=\frac C2\left(\sin\phi-i\cos\phi\right)\\ \end{cases} \,\end{align}$$ We can see that $A=B^{\space*}$, where $*$ denotes complex conjugation. Moreover, $\frac C2\left(\sin\phi+i\cos\phi\right)$ and $\frac C2\left(\sin\phi-i\cos\phi\right)$ are the polar forms of $A$ and $B$. If the values of $A$ and $B$ are substituted in $\text{(II)}$, we have: \begin{align} x(t) &= e^{-kt}\cdot\big[C\space\sin\phi\space\cos(\omega't)+C\space\cos\phi\space\sin(\omega't)\big]\\ &=C e^{-kt}\cdot\sin(\omega't+\phi) \end{align} So, the question now becomes "Why were the constants $A$ and $B$ chosen in this way?"

For three reasons, which I'll list below:

  • In a physical problem, $x(t)$ must be real (We can't have the position $x(t)=7+3i$ meters, for example) and the choice of $A=B^{\space*}$ ensures that $x(t)$ is real. This becomes more evident if we take the rectangular form of $A$ and $B$ and substitute them in $\text{(II)}$:

\begin{equation} \left. \begin{aligned} A &\triangleq a+ib\\ B &\triangleq a-ib\ \end{aligned} \right\} \implies \begin{aligned}[t] x(t) &= e^{-kt}\cdot\big[(a+ib+a-ib)\cos\omega't+i(a+ib-a+ib)\sin\omega't\big]\\ &= e^{-kt}\cdot\big[(2a)\cos\omega't+(-2b)\sin\omega't\big]\\ &\therefore\space \forall t, \space x(t)\in\mathbb{R} \end{aligned} \end{equation}

  • The choice of $\sin\phi+i\cos\phi$ instead of the "more natural" way $\cos\phi+i\sin\phi$ is arbitrary, because if we use $\phi=\frac{\pi}{2}-\theta\space$ (without loss of generality), it's possible to see that: $$\sin\phi+i\cos\phi=\sin\left(\frac{\pi}{2}-\theta\right)+\cos\left(\frac{\pi}{2}-\theta\right)=\cos\theta+i\sin\theta$$ So if we make the substitutions in $\text{(II)}$:

    – If we choose $A=B^{\space*}=\frac C2\left(\sin\phi+i\cos\phi\right)$, then we have $x(t)=C e^{-kt}\cdot\sin(\omega't+\phi)$

    – If we choose $A=B^{\space*}=\frac C2\left(\cos\phi+i\sin\phi\right)$, then we have $x(t)=C e^{-kt}\cdot\cos(\omega't+\phi)$

    (And both forms are valid to represent the position $x(t)$ in the underdamped harmonic motion)

  • The choice of $\frac C2$ and not just $C$ is for convenience:

    – If we choose $A=B^{\space*}=\frac C2\left(\sin\phi+i\cos\phi\right)$, then we have $x(t)=C e^{-kt}\cdot\sin(\omega't+\phi)$

    – If we choose $A=B^{\space*}=C\left(\sin\phi+i\cos\phi\right)$, then we have $x(t)=2 C e^{-kt}\cdot\sin(\omega't+\phi)$

    (And both forms are valid to represent the position $x(t)$ in the underdamped harmonic motion, but no one uses the latter, because if we define $D\triangleq2C$ it's easy to see that the two forms are equivalent)

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  • $\begingroup$ I don't have atleast 15 reputation so I can't upvote $\endgroup$ – Abhishek Ghosh Nov 3 '18 at 18:54
  • $\begingroup$ But I wish I could $\endgroup$ – Abhishek Ghosh Nov 3 '18 at 18:55
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This differential equation is linear with constant coefficients. Mathematically, the simplest way to solve such equations is to find the solutions which are complex numbers, even when the coefficients of the equation are real.

The reason is that if the auxiliary polynomial (used to find the values of $\alpha$ in the notation of the OP's images) is of degree $n$, it always has $n$ complex roots, which lead to different solutions of the differential equation.

If the coefficients of the differential equation are all real, the roots are either real or are in complex conjugate pairs, as in the OP's example when the motion is underdamped, so you can combine the two complex solutions for a complex pair of roots to get one solution which is real and another which is pure imaginary (i.e. the real part is zero).

Since the equation is linear, you can multiply any solution by a constant (either real or complex), so multiplying the pure imaginary solution by $i$ (or more usually, by $-i$ since $-i^2 = 1$) gives a second real solution.

It is often "neater" to do the math using the complex solutions until the final step of interpreting the results, which physically must be real of course, since $Ce^{i\omega t}$ where $C$ is a complex constant is more compact than $A \sin(\omega t) + B \cos(\omega t)$ or $A\sin(\omega t + \phi)$.

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  • $\begingroup$ you mean to say that in the above situation exp(-kt).(A+B)cos(w't) is a purely real solution while exp(-kt).(i(A-B)sin(w't)) is the purely imaginary solution ... The differential equation being linear we can multiply exp(-kt)(i(A-B)sin(w't)) with '-i' to get another purely real solution Did I get you right?? $\endgroup$ – Abhishek Ghosh Nov 3 '18 at 13:58

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