Simple Harmonic Motion

Simple harmonic motion refers to the periodic sinusoidal oscillation of an object or quantity. Simple harmonic motion is executed by any quantity obeying the Differential Equation

$$\ddot x+{\omega_0}^2x = 0,$$ (1)

where $\ddot x$ denotes the second Derivative of $x$ with respect to $t$, and ${\omega_0}$ is the angular frequency of oscillation. This Ordinary Differential Equation has an irregular Singularity at $\infty$. The general solution is

$$x = A\sin({\omega_0}t)+B\cos({\omega_0}t)$$ (2)

$$= C\cos({\omega_0}t+\phi),$$ (3)

where the two constants $A$ and $B$ (or $C$ and $\phi$) are determined from the initial conditions.

Many physical systems undergoing small displacements, including any objects obeying Hooke's law, exhibit simple harmonic motion. This equation arises, for example, in the analysis of the flow of current in an electronic CL circuit (which contains a capacitor and an inductor ). If a damping force such as Friction is present, an additional term $\beta\dot x$ must be added to the Differential Equation and motion dies out over time.

Adding a damping force proportional to $\dot x$, the first derivative of $x$ with respect to time, the equation of motion for damped simple harmonic motion is

$$\ddot x+\beta\dot x+{\omega_0}^2 x=0,$$ (4)

where $\beta$ is the damping constant. This equation arises, for example, in the analysis of the flow of current in an electronic CLR circuit, (which contains a capacitor, an inductor, and a resistor ). This Ordinary Differential Equation can be solved by looking for trial solutions of the form $x=e^{rt}$. Plugging this into (4) gives

$$(r^2+\beta r+{\omega_0}^2)e^{rt}=0$$ (5)

$$r^2+\beta r+{\omega_0}^2=0.$$ (6)

This is a Quadratic Equation with solutions

$$r = {\textstyle{1\over 2}}(-\beta\pm\sqrt{\beta^2-4{\omega_0}^2}\,).$$ (7)

There are therefore three solution regimes depending on the Sign of the quantity inside the Square Root,

$$\alpha\equiv\beta^2-4{\omega_0}^2.$$ (8)

The three regimes are

1. $\alpha>0$ is Positive: overdamped,

2. $\alpha=0$ is Zero: critically damped,

3. $\alpha<0$ is Negative: underdamped.

If a periodic (sinusoidal) forcing term is added at angular frequency $\omega$, the same three solution regimes are again obtained. Surprisingly, the resulting motion is still periodic (after an initial transient response, corresponding to the solution to the unforced case, has died out), but it has an amplitude different from the forcing amplitude.

The ``particular'' solution $x_p(t)$ to the forced second-order nonhomogeneous Ordinary Differential Equation

$$\ddot x+p(t)\dot x+q(t)x=A\cos(\omega t)$$ (9)

due to forcing is given by the equation

$$x_p(t)=-x_1(t)\int{x_2(t)g(t)\over W(t)}\,dt+x_2(t)\int {x_1(t)g(t)\over W(t)}\,dt,$$ (10)

where $x_1$ and $x_2$ are the homogeneous solutions to the unforced equation

$$\ddot x+p(t)\dot x+q(t)x=0$$ (11)

and $W(t)$ is the Wronskian of these two functions. Once the sinusoidal case of forcing is solved, it can be generalized to any periodic function by expressing the periodic function in a Fourier Series.

Critical damping is a special case of damped simple harmonic motion in which

$$\alpha\equiv\beta^2-4{\omega_0}^2=0,$$ (12)

so

$$\beta=2{\omega_0}.$$ (13)

The above plot shows an underdamped simple harmonic oscillator with $\omega=0.3$, $\beta=0.15$. The solid curve is for $(A,B)=(1,0)$, the dot-dashed for (0, 1), and the dotted for (1/2, 1/2). In this case, $\alpha=0$ so the solutions of the form $x=e^{rt}$ satisfy

$$r_{\pm}={\textstyle{1\over 2}}(-\beta) = -{\textstyle{1\over 2}}\beta = -{\omega_0}.$$ (14)

One of the solutions is therefore

$$x_1=e^{-{\omega_0}t}.$$ (15)

In order to find the other linearly independent solution, we can make use of the identity

$$x_2(t)=x_1(t)\int {e^{-\int p(t)\,dt}\over [x_1(t)]^2}\,dt.$$ (16)

Since we have $p(t)=2{\omega_0}$, $e^{-\int p(t)\,dt}$ simplifies to $e^{-2{\omega_0}t}$. Equation (16) therefore becomes

$$x_2(t) = e^{-{\omega_0}t} \int{e^{-2{\omega_0}t}\over [e^{-{... ...0}t}]^2}\, dt = e^{-{\omega_0}t} \int dt = t e^{-{\omega_0}t}.$$ (17)

The general solution is therefore

$$x=(A+Bt)e^{-{\omega_0}t}.$$ (18)

In terms of the constants $A$ and $B$, the initial values are

$$x(0) = A$$ (19)

$$x'(0) = B-A\omega,$$ (20)

so

$$A = x(0)$$ (21)

$$B = x'(0)+{\omega_0}x(0).$$ (22)

For sinusoidally forced simple harmonic motion with critical damping, the equation of motion is

$$\ddot x+2{\omega_0}\dot x+{\omega_0}^2x=A\cos(\omega t),$$ (23)

and the Wronskian is

$$W(t) \equiv x_1\dot x_2-\dot x_1 x_2$$

$$= e^{-{\omega_0}t}(e^{-{\omega_0}t}-{\omega_0}t e^{-{\omega_0}t})+{\omega_0}e^{-{\omega_0}t} t e^{-{\omega_0}t}$$

$$= e^{-2{\omega_0}t}(1-{\omega_0}t+{\omega_0}t) = e^{-2{\omega_0}t}.$$ (24)

Plugging this into the equation for the particular solution gives

$$x_p(t) = -e^{-{\omega_0}t}\int{te^{-{\omega_0}t}A\cos(\omega t)\over e^{-2... ...^{-{\omega_0}t}\int{e^{-{\omega_0}t}A\cos(\omega t)\over e^{-2{\omega_0}t}}\,dt$$

$$= Ae^{-{\omega_0}t}\left[{-\int te^{{\omega_0}t}\cos(\omega t)\,dt+t\int e^{{\omega_0}t}\cos(\omega t)\,dt}\right]$$

$$= Ae^{-{\omega_0}t}\left\{{-{e^{{\omega_0}t}\over (\omega^2+{\omega_0}^2)^2} [(\omega^2+t\omega^2{\omega_0}-{\omega_0}^2+t{\omega_0}^3)}\right.$$

$$\phantom{=} \times\cos(\omega t)+\omega(t\omega^2-2{\omega_0}+t{\omega_0}^2)\sin(\omega t)]$$

$$\phantom{=} \left.{\mathop{+}t {e^{{\omega_0}t}\over \omega^2+{\omega_0}^2}[{\omega_0}\cos(\omega t)+\omega \sin(\omega t)}\right\}$$

$$= {A\over(\omega^2+{\omega_0}^2)^2}[({\omega_0}^2-\omega^2)\cos(\omega t)+2\omega{\omega_0}\sin(\omega t)].$$ (25)

In order to put this in the desired form, note that we want to equate

$$C\cos\theta+S\sin\theta = Q\cos(\theta+\delta)$$

$$= Q(\cos\theta\cos\delta-\sin\theta\sin\delta).$$ (26)

This means

$$C \equiv Q\cos\delta = {\omega_0}^2-\omega^2$$ (27)

$$S \equiv -Q\sin\delta=2\omega{\omega_0},$$ (28)

so

$$Q = \sqrt{C^2+S^2}$$ (29)

$$\delta = \tan^{-1}\left({-{S\over C}}\right).$$ (30)

Plugging in,

$$Q = \sqrt{{\omega_0}^4-2{\omega_0}^2\omega^2+\omega^4+4{\omega_0}^2\omega^2}$$

$$= \sqrt{{\omega_0}^4+2{\omega_0}^2\omega^2+\omega^4} = {\omega_0}^2+\omega^2.$$ (31)

$$\delta = \tan^{-1}\left({2\omega{\omega_0}\over \omega^2-{\omega_0}^2}\right).$$ (32)

The solution in the requested form is therefore

$$x_p = {A\over (\omega^2+{\omega_0}^2)^2} ({\omega_0}^2+\omega^2) \cos(\omega t+\delta)$$

$$= {A\over \omega^2+{\omega_0}^2} \cos(\omega t+\delta),$$ (33)

where $\delta$ is defined by (32).

Overdamped simple harmonic motion occurs when

$$\beta^2-4{\omega_0}^2>0,$$ (34)

so

$$\alpha\equiv\beta^2-4{\omega_0}^2>0.$$ (35)

The above plot shows an overdamped simple harmonic oscillator with $\omega=0.3$, $\beta=0.075$. The solid curve is for $(A,B)=(1,0)$, the dot-dashed for (0, 1), and the dotted for (1/2, 1/2). The solutions are

$$x_1 = e^{r_-t}$$ (36)

$$x_2 = e^{r_+t},$$ (37)

where

$$r_\pm \equiv {\textstyle{1\over 2}}(-\beta\pm\sqrt{\beta^2-4{\omega_0}^2}\,).$$ (38)

The general solution is therefore

$$x = Ae^{r_-t}+Be^{r_+t},$$ (39)

where $A$ and $B$ are constants. The initial values are

$$x(0) = A+B$$ (40)

$$x'(0) = Ar_-+Br_+,$$ (41)

so

$$A = x(0)+{r_+x(0)-x'(0)\over r_--r_+}$$ (42)

$$B = -{r_+x(0)-x'(0)\over r_--r_+}.$$ (43)

For a cosinusoidally forced overdamped oscillator with forcing function $g(t)=C\cos(\omega t)$, the particular solutions are

$$y_1(t) = e^{r_1t}$$ (44)

$$y_2(t) = e^{r_2t},$$ (45)

where

$$r_1 \equiv {\textstyle{1\over 2}}(-\beta+\sqrt{\beta ^2-4{\omega_0}^2}\,)$$ (46)

$$r_2 \equiv {\textstyle{1\over 2}}(-\beta-\sqrt{\beta ^2-4{\omega_0}^2}\,).$$ (47)

These give the identities

$$r_1+r_2 = -\beta$$ (48)

$$r_1-r_2 = \sqrt{\beta^2-4{\omega_0}^2}$$ (49)

and

$${\omega_0}^2 = {\textstyle{1\over 4}}[\beta-(r_1-r_2)^2] = {\textstyle{1\over 4}}[(r_1+r_2)^2-(r_1-r_2)^2]$$

$$= {\textstyle{1\over 4}}[2r_1r_2+2r_1r_2] = r_1r_2.$$ (50)

The Wronskian is

$$W(t) \equiv y_1y_2'-y_1'y_2 = e^{r_1t}r_2e^{r_2t}-r_1e^{r_1t}e^{r_2t}$$

$$= (r_2-r_1)e^{(r_1+r_2)t}.$$ (51)

The particular solution is

$$y_p = -y_1 v_1+y_2 v_2,$$ (52)

where

$$v_1 \equiv \int {y_2 g(t)\over W(t)} = {C\over r_2-r_1} {\omega\sin (\omega t)-r_2\cos (\omega t)\over e^{r_2t}({r_2}^2+\omega^2)}$$ (53)

$$v_2 \equiv \int {y_2 g(t)\over W(t)} = {C\over r_2-r_1} {\omega\sin(\omega t)-r_1\cos (\omega t)\over e^{r_1t}({r_2}^2+\omega^2)}.$$ (54)

Therefore,

$$y_p = C {\cos (\omega t)(r_1r_2-\omega^2)-\sin (\omega t)\omega(r_1+r_2)\over ({r_1}^2+\omega^2)({r_2}^2+\omega^2)}$$

$$= \displaystylg C {({\omega_0}^2-\omega^2)\cos (\omega t)+\beta\omega\sin (\omega t)\over \omega^2\beta^2+(\omega^2-{\omega_0}^2)}$$

$$= {C\over\omega^2\beta ^2+(\omega^2-{\omega_0}^2)^2} \sqrt{(\omega^2-{\omega_0}^2)^2+\beta^2\omega^2}\cos(\omega t+\delta)$$

$$= {C\over\sqrt{\beta^2\omega^2+(\omega^2-{\omega_0}^2)^2}} \cos(\omega t+\delta),$$ (55)

where

$$\delta=\tan^{-1}\left({\beta\omega\over \omega^2-{\omega_0}^2}\right).$$ (56)

Underdamped simple harmonic motion occurs when

$$\beta^2-4{\omega_0}^2<0,$$ (57)

so

$$\alpha\equiv\beta^2-4{\omega_0}^2<0.$$ (58)

The above plot shows an underdamped simple harmonic oscillator with $\omega=0.3$, $\beta=0.4$. The solid curve is for $(A,B)=(1,0)$, the dot-dashed for (0, 1), and the dotted for (1/2, 1/2). Define

$$\gamma\equiv\sqrt{-\alpha}={\textstyle{1\over 2}}\sqrt{4{\omega_0}^2-\beta^2},$$ (59)

then solutions satisfy

$$r_{\pm}=-{\textstyle{1\over 2}}\beta\pm i\gamma,$$ (60)

where

$$r_\pm \equiv {\textstyle{1\over 2}}(-\beta\pm\sqrt{\beta^2-4{\omega_0}^2}\,),$$ (61)

and are of the form

$$x=e^{-(\beta/2\pm i\gamma)t}.$$ (62)

Using the Euler Formula

$$e^{ix}=\cos x+i\sin x,$$ (63)

this can be rewritten

$$x = e^{-(\beta/2)t} \left[{\cos\left({\gamma t}\right)\pm i \sin\left({\gamma t}\right)}\right].$$ (64)

We are interested in the real solutions. Since we are dealing here with a linear homogeneous ODE, linear sums of Linearly Independent solutions are also solutions. Since we have a sum of such solutions in (64), it follows that the Imaginary and Real Parts separately satisfy the ODE and are therefore the solutions we seek. The constant in front of the sine term is arbitrary, so we can identify the solutions as

$$x_1 = e^{-(\beta/2)t}\cos(\gamma t)$$ (65)

$$x_2 = e^{-(\beta/2)t}\sin(\gamma t),$$ (66)

so the general solution is

$$x= e^{-(\beta/2)t}[A\cos(\gamma t)+B\sin(\gamma t)].$$ (67)

The initial values are

$$x(0) = A$$ (68)

$$x'(0) = -{\textstyle{1\over 2}}\beta A+B,\gamma$$ (69)

so $A$ and $B$ can be expressed in terms of the initial conditions by

$$A = x(0)$$ (70)

$$B = {\beta x(0)\over 2\gamma}+{x'(0)\over\gamma}.$$ (71)

For a cosinusoidally forced underdamped oscillator with forcing function $g(t)=C\cos(\omega t)$, use

$$\gamma \equiv {\textstyle{1\over 2}}\sqrt{4{\omega_0}^2-\beta^2}$$ (72)

$$\alpha \equiv {\textstyle{1\over 2}}\beta$$ (73)

to obtain

$$4{\omega_0}^2-\beta^2 = 4\gamma^2$$ (74)

$${\omega_0}^2 = \gamma^2+{\textstyle{1\over 4}}\beta^2 = \gamma^2+\alpha^2$$ (75)

$$\beta = 2\alpha.$$ (76)

The particular solutions are

$$y_1(t) = e^{-\alpha t}\cos(\gamma t)$$ (77)

$$y_2(t) = e^{-\alpha t}\sin(\gamma t).$$ (78)

The Wronskian is

$$W(t) \equiv y_1y_2'-y_1'y_2$$

$$= e^{-\alpha t}\cos(\gamma t)[-\alpha e^{-\alpha t}\sin(\gamma t)+e^{-\alpha t}\gamma \cos(\gamma t)]$$

$$\phantom{=} \mathop{-} e^{-\alpha t}\sin(\gamma t)[-\alpha e^{-\alpha t} \cos(\gamma t)-e^{-\alpha t}\gamma \sin(\gamma t)]$$

$$= e^{-2\alpha t}\left\{{\alpha [-\sin (\gamma t)\cos (\gamma t)+\sin (\gamma t) \cos (\gamma t)]+\gamma[\cos ^2(\gamma t)+\sin ^2(\gamma t)]}\right\}$$

$$= \gamma e^{-2\alpha t}.$$ (79)

The particular solution is given by

$$y_p = -y_1 v_1+y_2 v_2,$$ (80)

where

$$v_1 \equiv \int {y_2 g(t)\over W(t)} = {C\over \gamma} \int e^{\alpha t}\cos (\gamma t)\cos(\omega t)\,dt$$ (81)

$$v_2 \equiv \int {y_2 g(t)\over W(t)} = {C\over \gamma} \int e^{\alpha t}\cos (\gamma t)\cos(\omega t)\,dt.$$ (82)

Using computer algebra to perform the algebra, the particular solution is

$$y_p(t) = C {(\alpha^2+\gamma^2-\omega^2)\cos (\omega t)+2\alpha \omega\sin (\omega t) \over [\alpha^2+(\gamma-\omega)^2][\alpha^2+(\gamma+\omega)^2]}$$

$$= C {({\omega_0}^2-\omega^2)\cos(\omega t)+\beta\omega\sin(\omega t)\over(\alpha^2+\gamma^2+\omega^2)^2-4\gamma^2\omega^2}$$

$$= C {({\omega_0}^2-\omega^2)\cos(\omega t)+\beta\omega\sin(\omega t)\over({\omega_0}^2+\omega^2)^2-4{1\over 4}(4{\omega_0}^2-\beta^2)\omega^2}$$

$$= C {({\omega_0}^2-\omega^2)\cos(\omega t)+\beta\omega\sin(\omega t)\over ({\omega_0}^2-\omega^2)^2-\omega^2(4{\omega_0}^2-\beta^2)}$$

$$= {C\sqrt{({\omega_0}^2-\omega^2)^2+\beta^2\omega^2}\over ({\omega_0}^2-\omega^2)^2-\omega^2(4{\omega_0}^2-\beta ^2)}\cos(\omega t+\delta)$$

$$= C {\sqrt{({\omega_0}^2-\omega^2)^2+\beta^2\omega^2}\over({\omega_0}^2-\omega^2)^2-\omega^2(4{\omega_0}^2-\beta^2)}\cos(\omega t+\delta),$$ (83)

where

$$\delta=\tan^{-1}\left({\beta\omega\over \omega^2-{\omega_0}^2}\right).$$ (84)

If the forcing function is sinusoidal instead of cosinusoidal, then

$$\delta'=\delta-{\textstyle{1\over 2}}\pi = \tan^{-1}x-{\textstyle{1\over 2}}\pi = \tan^{-1}\left({-{1\over x}}\right),$$ (85)

so

$$\delta' = \tan^{-1}\left({{\omega_0}^2-\omega^2\over \beta \omega}\right).$$ (86)