What is the meaning of the intercept of a particular solution to damped oscillator?

[zenterix]

Homework Statement

Suppose we have a mass $m$ moving on a horizontal air track.

The mass is attached on both sides to two identical springs each with spring constant $k$ and relaxed length $l_0$.

The end of the first spring is fixed.

The end of the second spring is attached to an electrical motor that causes it to undergo harmonic motion with amplitude $d\sin{\omega_d t}$.

At $t=0$ the springs are relaxed and the mass is not moving.

At $t=0$ the motor is switched on.

Define the coordinate system such that the origin is at the position of the mass at $t=0$.

Relevant Equations

Assume there is a small air friction given by $b\dot{x}$.

The mass is under the effect of two forces $F_a$ and $F_b$.

Let's denote the right edge of the mass as $b$ and the left edge as $a$. Let's denote the right end of the right spring as $c$.

Denote the width of the mass by $w$.

The left hand side

$x$ is the position of the middle of the mass.

$\Delta x_a$ is the change in the position of $a$ and $\delta x$ is the change in the position of $c$.

Then the positions are

$$r_a=-w+\Delta x_a\tag{1}$$

$$r_b=w+\Delta x_a\tag{2}$$

$$r_c=w+l_0+\Delta x_c\tag{3}$$

Any time the right spring is in equilibrium we have $r_b-r_c=-l_0$.

$$(r_b-r_c)_{eq}=-l_0\tag{4}$$

In addition we have

$$\Delta x_a=x\tag{5}$$

$$\Delta x_c=d \sin{\omega_d t}\tag{6}$$

Thus

$$F_b=-k((r_b-r_c)-(r_b-r_c)_eq)\tag{7}$$
$$=k(\Delta x_c-\Delta x_a)\tag{8}$$

$$F_a=-k((r_a-r_{wall})-(r_a-r_{wall})_{eq}\tag{9}$$

$$=-k\Delta x_a\tag{10}$$

Equation of Motion
$$F_a+F_b+F_{drag}=m\ddot{x}\tag{11}$$

$$-k\Delta x_a+k(\Delta x_c-\Delta x_a)-b\dot{x}=m\ddot{x}\tag{12}$$

$$-kx+k(d\sin{\omega_d t}-x)-b\dot{x}=m\ddot{x}\tag{13}$$

$$\ddot{x}+\frac{b}{m}\dot{x}+\frac{2k}{m}x=\frac{k}{m}d\sin{\omega_d t}\tag{14}$$

$$\ddot{x}+\gamma\dot{x}+\omega_0^2x=A_0\sin{\omega_d t}\tag{15}$$

Solution
$$x(t)=\frac{A_0}{(\omega_0^2-\omega_d^2)^2+\gamma^2\omega_d^2}\sin{(\omega_d t-\beta)}\tag{16}$$

with

$$\beta=arg(\omega_0^2-\omega_d^2+i\gamma\omega_d)\tag{17}$$

One question I have is about writing the following

$$\tan{\beta}=\frac{\gamma\omega_d}{\omega_0^2-\omega_d^2}\tag{18}$$

If you try to plot this with Maple, you will get a result that isn't correct for the problem. It seems we lose information in going from (17) to (18).

What justifies writing (18) instead of (17)? If I recall correctly, I see (18) much more frequently than (17) in textbooks/notes/etc.

Numerical Example Parameters
Let's consider a concrete example.

Assume

$$m=1$$

$$k=1$$

$$d=1$$

$$\gamma=\frac{\omega_0}{4}$$

Then

$$A_0=\frac{kd}{m}=1$$

$$\omega_0=\sqrt{\frac{2k}{m}}=\sqrt{2}$$

$$\gamma=\frac{1}{4}$$

$$b=m\gamma=\frac{1}{4}$$

The steady-state solution is

$$x(t)=\frac{1}{(2-\omega_d^2)^2+\frac{\omega_d^2}{16}}\sin{(\omega_d t-\beta)}$$

The amplitude of the steady-state solution is

$$B=\frac{A_0}{(\omega_0^2-\omega_d^2)^2-\gamma^2\omega_d^2}=\frac{1}{(2-\omega_d^2)^2+\frac{\omega_d^2}{16}}$$

A plot of this function is

The phase lag of the steady-state solution is

$$\beta=arg(\omega_0^2-\omega_d^2+i\gamma\omega_d)=arg\left (2-\omega_d^2+i\frac{\omega_d^2}{16}\right )$$

The amplitude maximum does not occur at $\omega_0$. It occurs at

$$\omega_{d,max}=\sqrt{\frac{2\omega_0^2-\gamma^2}{2}}<\omega_0$$

My question is about the consideration of the initial conditions.

Suppose

$$\omega_d=\frac{\omega_0}{2}$$

Then, the steady-state solution is

$$x_p(t)=\frac{32}{73}\sin{(0.7t-0.34)}$$

The amplitude is $32/73$ and the phase lag is $-0.34\text{rad}$.

I am interested in what is happening at $t=0$ here.

What is the meaning of the negative value of $x_p(0)$ here?

I guess that for the initial conditions implicit above, the homogeneous solution is just zero. But what does this mean exactly?

I think what I am trying to understand is the following.

First of all, is there more than one possible particular solution?

I know that the homogeneous solution has two parameters so there is an infinite number of those solutions.

Given a particular solution, and given initial conditions, we obtain the two parameters to pin down a homogeneous solution such that the general solution describes the motion of our mass with the initial conditions.

If we were to have started with a different particular solution, we would get other values for the homogeneous parameters (given the initial conditions).

Are there other particular solutions?

 

[zenterix]

Here are the calculations to obtain the general solution

$$x_g(t)=x_h(t)+x_p(t)=e^{-\frac{\gamma t}{2}}A\cos{(\omega t-\phi)}+\frac{A_0}{(\omega_0^2-\omega_d^2)^2+\gamma^2\omega-d^2}\sin{(\omega_d t-\beta)}$$

The two unknowns here are $A$ and $\phi$, ie the amplitude and phase of the homogeneous oscillations.

Imposing $x(0)=0$ and $\dot{x}(0)=0$ we obtain

$$\tan{\phi}=\frac{\frac{\gamma}{2}d_1-d_2}{\omega d_1}$$

$$A=\frac{d_1}{\cos{\phi}}$$

where

$$d_1=\frac{A_0}{(\omega_0^2-\omega_d^2)^2+\gamma^2\omega-d^2}\sin{\beta}$$

$$d_2=\frac{A_0\omega_d}{(\omega_0^2-\omega_d^2)^2+\gamma^2\omega-d^2}\cos{\beta}$$

For the numerical example in the OP, a plot of $x_h$, $x_p$, and $x_g$ is

The green curve is the solution. It is composed of the blue particular solution and the red homogeneous solution.

We see that the homogeneous equation is chosen to make the initial conditions true.

 

[haruspex]

If you try to plot this with Maple, you will get a result that isn't correct for the problem. It seems we lose information in going from (17) to (18).

What exactly are you plotting? If it involves arctan, that introduces an ambiguity. $\tan(\alpha)=\tan(\alpha+\pi)$.

 

[zenterix]

What exactly are you plotting? If it involves arctan, that introduces an ambiguity. $\tan(\alpha)=\tan(\alpha+\pi)$.

Exactly, it introduces ambiguity.

The expression $arg(...)$ does not have this ambiguity. This is why I am now using this expression.

 

[haruspex]

Exactly, it introduces ambiguity.

The expression $arg(...)$ does not have this ambiguity. This is why I am now using this expression.

Ok, but you can also fix it by adding pi if the denominator in $\arctan(\frac yx)$ is negative. (I think that works.)

 

[haruspex]

is there more than one possible particular solution?

If p(t) is a particular solution and h(t) is the homogeneous solution then $p+\lambda h$ is the general solution if we consider $\lambda$ to be a free variable, but if we fix a value of it, $\lambda_0$, then $p+\lambda_0 h$ represents another particular solution, no?

 

[zenterix]

That's true. We have infinite particular solutions because we can add any homogeneous solution to any particular solution.

I don't know the specifics, but in terms of vector spaces, it seems the homogeneous solutions are a vector space and the set of particular solutions are not a vector space, but rather an affine subset.

The initial value problem in the OP has a unique solution. So even though we may start with a particular solution that is different from the one shown in the OP, it is different by a specific homogeneous solution. When we add a general homogeneous solution and determine the two unknown parameters to give the initial conditions, I imagine that the sum of this new homogeneous solution with the one added to the the particular solution will always give the same result which is the one I derived in the second post.