Simple Harmonic Motion & Harmonic Oscillator

1. Simple Harmonic Motion and Harmonic Oscillator

The projection of uniform circular motion along the diameter of a circle is called S.H.M.

A particle is said to execute S.H.M. if the restoring force is directly proportional to the displacement and always directed towards the fixed point (or mean position).

Let us consider a mass $m$ executing S.H.M. Let $F$ be the restoring force and $x$ be its displacement. Then,

F \propto x$$ $$F = -kx

(The negative sign indicates that the restoring force opposes the displacement.)

From Newton's law of motion,

ma = -kx$$ $$m\frac{d^{2}x}{dt^{2}} = -kx$$ $$\frac{d^{2}x}{dt^{2}} = -\frac{k}{m}x$$ $$\frac{d^{2}x}{dt^{2}} + \frac{k}{m}x = 0$$ $$\frac{d^{2}x}{dt^{2}} + \omega_0^2 x = 0 \tag{1}\label{eq:1}

where $\omega_0^2 = \dfrac{k}{m}$.

Equation \eqref{eq:1} is the differential equation of S.H.M. / Simple Harmonic Oscillator.

Solution of the Differential Equation

Let the trial solution of equation \eqref{eq:1} be

x = Ae^{\alpha t} \tag{a}

Differentiating with respect to $t$:

\frac{dx}{dt} = A\alpha e^{\alpha t}, \qquad \frac{d^{2}x}{dt^{2}} = A\alpha^2 e^{\alpha t}

Substituting into equation \eqref{eq:1}:

A\alpha^2 e^{\alpha t} + \omega_0^2 A e^{\alpha t} = 0$$ $$Ae^{\alpha t}\left(\alpha^2 + \omega_0^2\right) = 0

Since $Ae^{\alpha t} \neq 0$,

\alpha^2 + \omega_0^2 = 0 \quad\Rightarrow\quad \alpha^2 = -\omega_0^2 \quad\Rightarrow\quad \alpha = \pm i\omega_0

Therefore, the possible solutions of equation (a) are:

x = A_1 e^{i\omega_0 t} \qquad \text{and} \qquad x = A_2 e^{-i\omega_0 t}

Any linear combination of these solutions is also a solution of equation \eqref{eq:1}. The general solution is:

x = A_1 e^{i\omega_0 t} + A_2 e^{-i\omega_0 t} \tag{2}\label{eq:2}

where $A_1$ and $A_2$ are constants. Using Euler's formula,

x = A_1\{\cos\omega_0 t + i\sin\omega_0 t\} + A_2\{\cos\omega_0 t - i\sin\omega_0 t\}$$ $$x = (A_1+A_2)\cos\omega_0 t + i(A_1-A_2)\sin\omega_0 t$$ $$x = A_3\cos\omega_0 t + A_4 \sin \omega_0 t \tag{3}\label{eq:3}

where $A_3 = A_1+A_2$ and $A_4 = i(A_1-A_2)$.

Let $A_3 = a\sin\phi$ and $A_4 = a\cos\phi$. Then,

x = a\{\cos\omega_0t\sin\phi + \sin\omega_0t\cos\phi\}$$ $$x = a\sin(\omega_0 t + \phi) \tag{4}\label{eq:4}

where $a$ is the amplitude and $\phi$ is the phase angle.

If $\phi = \dfrac{\pi}{2}+\delta$, then

x = a\sin\left(\omega_0 t + \frac{\pi}{2}+\delta\right)$$ $$x = a\cos(\omega_0 t + \delta) \tag{5}\label{eq:5}

Equations \eqref{eq:2}, \eqref{eq:3}, \eqref{eq:4}, and \eqref{eq:5} are all general solutions of the simple harmonic oscillator.

Time Period (T)

Time taken to complete one vibration is called the time period. We know,

\omega_0 = \frac{2\pi}{T}$$ $$T = \frac{2\pi}{\omega_0} = \frac{2\pi}{\sqrt{k/m}} = 2\pi\sqrt{\frac{m}{k}}

Velocity of the Oscillator

v = \frac{dx}{dt} = \frac{d}{dt}\left[a\sin(\omega_0 t+\phi)\right] = a\omega_0\cos(\omega_0 t + \phi)$$ $$v = a\omega_0\sqrt{1-\sin^2(\omega_0 t+\phi)}$$ $$v = \omega_0\sqrt{a^2-x^2}

Acceleration

\text{Acceleration} = \frac{dv}{dt} = \frac{d}{dt}\left[a\omega_0\cos(\omega_0 t+\phi)\right] = a\omega_0 \cdot \omega_0\{-\sin(\omega_0 t+\phi)\}$$ $$a_{\text{acc}} = -\omega_0^2 x

2. Energy of Harmonic Oscillator

Total Energy ($E$) = Kinetic Energy (K.E.) + Potential Energy (P.E.)

E = \frac{1}{2}mv^2 + \frac{1}{2}kx^2 = \frac{1}{2}m\left(\omega_0\sqrt{a^2-x^2}\right)^2 + \frac{1}{2}m\omega_0^2 x^2$$ $$E = \frac{1}{2}m\omega_0^2(a^2-x^2) + \frac{1}{2}m\omega_0^2 x^2$$ $$E = \frac{1}{2}m\omega_0^2 a^2 = \text{constant}

Thus, the total energy of a harmonic oscillator is independent of displacement and is constant (conserved).

Average Kinetic Energy over One Time Period

\overline{K.E.} = \int_0^T \frac{K.E.}{T}\,dt = \int_0^T \frac{\frac{1}{2}mv^2}{T}\,dt = \frac{m}{2T}\int_0^T v^2\,dt$$ $$\overline{K.E.} = \frac{m}{2T}\int_0^T a^2\omega_0^2\cos^2(\omega_0 t+\phi)\,dt = \frac{ma^2\omega_0^2}{2T}\int_0^T \cos^2(\omega_0 t+\phi)\,dt$$ $$= \frac{ma^2\omega_0^2}{2T}\int_0^T \frac{1+\cos2(\omega_0 t+\phi)}{2}\,dt = \frac{ma^2\omega_0^2}{4T}\left[\int_0^T dt + \int_0^T \cos2(\omega_0 t+\phi)\,dt\right]

The second integral vanishes over a complete period, so

\overline{K.E.} = \frac{ma^2\omega_0^2}{4T}\cdot T = \frac{1}{4}ma^2\omega_0^2

Average Potential Energy over One Time Period

\overline{P.E.} = \int_0^T \frac{P.E.}{T}\,dt, \qquad F\propto x,\ F=-kx,\ dW = -F(-dx) = kx\,dx$$ $$W = \int dW = \int kx\,dx = \frac{1}{2}kx^2$$ $$\overline{P.E.} = \int_0^T \frac{\frac12 kx^2}{T}\,dt = \frac{k}{2T}\int_0^T x^2\,dt = \frac{k}{2T}\int_0^T a^2\sin^2(\omega_0 t+\phi)\,dt$$ $$= \frac{ka^2}{2T}\int_0^T\frac{1-\cos2(\omega_0 t+\phi)}{2}\,dt = \frac{ka^2}{4T}\left[\int_0^T dt - \int_0^T\cos2(\omega_0 t+\phi)\,dt\right]$$ $$\overline{P.E.} = \frac{ka^2T}{4T} = \frac{1}{4}ka^2 = \frac{1}{4}m\omega_0^2 a^2

Average Total Energy

\overline{E} = \overline{K.E.} + \overline{P.E.} = \frac{1}{4}m\omega_0^2 a^2 + \frac{1}{4}m\omega_0^2 a^2$$ $$\overline{E} = \frac{1}{2}m\omega_0^2 a^2

3. Simple Pendulum

A simple pendulum is a heavy point mass suspended from a rigid support with light, inextensible, and flexible string such that it can oscillate freely.

Let $m$ = mass of the bob, $\ell$ = effective length of the pendulum. When the bob is displaced to position $A$ through a small angle $\theta$, the component $mg\cos\theta$ balances the tension on the string and $mg\sin\theta$ provides the restoring force.

\text{Restoring force } (F) = -mg\sin\theta

For small $\theta$, $\sin\theta \approx \theta$, so $F = -mg\theta$.

From Newton's second law:

ma = -mg\theta \quad\Rightarrow\quad \frac{d^{2}x}{dt^{2}} = -g\theta

Since $\theta = x/\ell$, we get $x = \theta\ell$, so $\frac{dx}{dt} = \ell\frac{d\theta}{dt}$ and $\frac{d^{2}x}{dt^{2}}=\ell\frac{d^{2}\theta}{dt^{2}}$.

\frac{d^{2}x}{dt^{2}} = -\frac{g}{\ell}x \quad\Rightarrow\quad \frac{d^{2}x}{dt^{2}}+\frac{g}{\ell}x=0, \qquad \omega_0^2 = \frac{g}{\ell}

Equivalently, in terms of $\theta$:

\frac{d^{2}\theta}{dt^{2}} + \frac{g}{\ell}\theta = 0 \tag{6}\label{eq:pendulum}

Comparing with $\frac{d^{2}\theta}{dt^{2}}+\omega_0^2\theta=0$, we get $\omega_0 = \sqrt{g/\ell}$, so

\frac{2\pi}{T} = \sqrt{\frac{g}{\ell}} \quad\Rightarrow\quad T = 2\pi\sqrt{\frac{\ell}{g}}

The solution of equation \eqref{eq:pendulum} is $\theta = \theta_0\sin(\omega_0 t + \phi)$.

Limitations of the Simple Pendulum

A heavy point-mass bob is impossible to realize.
A weightless string is impossible to realize.
Moment of inertia should be taken into account, which is not done in the simple pendulum.
The centre of gravity is considered as the centre of oscillation in a simple pendulum, which is a wrong assumption -- the centre of oscillation always lies at some distance away from the centre of gravity.

4. Compound Pendulum

A compound pendulum is a rigid body of any shape capable of oscillation in a vertical plane about a horizontal axis not passing through its centre of gravity.

Let $XY$ = horizontal axis through $S$, where $S$ = point of suspension, $C$ = centre of gravity, $\ell$ = distance between the point of suspension and the centre of gravity, and $mg$ = weight of the compound pendulum.

When the pendulum is shifted through a small angle $\theta$ and then released, it oscillates about axis $XY$. Since deflecting torque = restoring torque,

I\alpha = -mg\cdot \ell\sin\theta, \qquad (\tau = F\times r)

For small $\theta$: $I\frac{d^{2}\theta}{dt^{2}} = -mg\ell\theta$

\frac{d^{2}\theta}{dt^{2}}+\frac{mg\ell}{I}\theta = 0 \tag{7}\label{eq:compound}

which is the second order differential equation of an angular harmonic oscillator, whose solution is $\theta = \theta_0\sin(\omega t+\phi)$.

Comparing equation \eqref{eq:compound} with $\frac{d^{2}\theta}{dt^{2}}+\omega^2\theta=0$, we get

\omega^2 = \frac{mg\ell}{I}, \qquad \omega = \sqrt{\frac{mg\ell}{I}}, \qquad \frac{2\pi}{T} = \sqrt{\frac{mg\ell}{I}}$$ $$T = 2\pi\sqrt{\frac{I}{mg\ell}}

Using the theorem of parallel axes, $I = I_{cm}+m\ell^2 = mk^2+m\ell^2$ (where $k$ is the radius of gyration):

T = 2\pi\sqrt{\frac{mk^2+m\ell^2}{mg\ell}} = 2\pi\sqrt{\frac{k^2+\ell^2}{g\ell}}

Interchangeability of Point of Suspension and Centre of Oscillation

We have

T = 2\pi\sqrt{\frac{k^2/\ell+\ell}{g}}

Let $\ell' = k^2/\ell$, so $T = 2\pi\sqrt{\dfrac{\ell'+\ell}{g}}$. Let $L=\ell'+\ell$, so

T = 2\pi\sqrt{\frac{L}{g}}

When the axis passes through $S$ (point of suspension):

T_1 = 2\pi\sqrt{\frac{k^2/\ell+\ell}{g}} = 2\pi\sqrt{\frac{\ell'+\ell}{g}} = 2\pi\sqrt{\frac{L}{g}}

When the axis passes through $O$ (centre of oscillation), with new distance $\ell'$:

T_2 = 2\pi\sqrt{\frac{k^2/\ell'+\ell'}{g}} = 2\pi\sqrt{\frac{\ell+\ell'}{g}} = 2\pi\sqrt{\frac{L}{g}}

Hence $T_1 = T_2$. So, the point of suspension and the centre of oscillation are interchangeable.

Maximum and Minimum Time Period of a Compound Pendulum

We know,

T = 2\pi\sqrt{\frac{k^2/\ell + \ell}{g}} \tag{8}\label{eq:cpT}

When $\ell=0$, the axis passes through the centre of gravity:

T = 2\pi\sqrt{\frac{k^2/0+0}{g}} \quad\Rightarrow\quad T_{\max} = \infty

As the time period is infinite, it cannot behave as a pendulum in this case.

Squaring equation \eqref{eq:cpT}:

T^2 = 4\pi^2\left(\frac{k^2}{\ell}+\ell\right)\frac{1}{g} = \frac{4\pi^2 k^2}{\ell g}+\frac{4\pi^2\ell}{g}

Differentiating with respect to $\ell$:

2T\frac{dT}{d\ell} = \frac{4\pi^2 k^2}{g}\left(-\frac{1}{\ell^2}\right)+\frac{4\pi^2}{g}

For minimum, $\frac{dT}{d\ell}=0$:

0 = -\frac{4\pi^2k^2}{g\ell^2}+\frac{4\pi^2}{g} \quad\Rightarrow\quad \frac{4\pi^2}{g} = \frac{4\pi^2k^2}{g\ell^2}$$ $$1 = \frac{k^2}{\ell^2} \quad\Rightarrow\quad k^2=\ell^2 \quad\Rightarrow\quad k=\pm\ell$$ $$T_{\min} = 2\pi\sqrt{\frac{\ell^2/\ell+\ell}{g}} = 2\pi\sqrt{\frac{2\ell}{g}}

Determination of Acceleration Due to Gravity

We have

T = 2\pi\sqrt{\frac{k^2/\ell+\ell}{g}}

Squaring,

T^2 = \frac{4\pi^2(k^2+\ell^2)}{\ell g}$$ $$T^2 = \frac{4\pi^2k^2}{g}\cdot\frac{1}{\ell}+\frac{4\pi^2}{g}\cdot\ell$$ $$T^2\ell = \frac{4\pi^2k^2}{g}+\frac{4\pi^2}{g}\ell^2

Let $T^2\ell = y$, $\dfrac{4\pi^2}{g}=m$ (slope), $\ell^2=x$, $\dfrac{4\pi^2k^2}{g}=c$ (intercept). This is of the form $y=mx+c$, a straight line.

Plotting $T^2\ell$ against $\ell^2$ gives a straight line of slope $m = a/b$ where $a,b$ are the graph's rise and run. Then,

\frac{4\pi^2}{g} = \frac{a}{b} \quad\Rightarrow\quad g = \frac{4\pi^2}{\text{slope of graph}}

5. Torsional Pendulum

A rigid body capable of executing angular harmonic motion in a horizontal plane about a vertical axis passing through the centre of gravity of the body is called a torsional pendulum.

It consists of a disc suspended by a wire which is attached to the centre of gravity of the disc. Let $O$ = centre of the disc, $A$ = initial equilibrium position of the disc. When the disc is rotated from $A$ to $B$ in the horizontal plane, the wire is twisted and a restoring torque tends to bring the disc back to the equilibrium position.

In this case, the restoring torque is directly proportional to the angle of twist:

\tau \propto \theta, \qquad \tau = -C\theta

where $C$ = torsional constant (depends on the material of the wire); the negative sign indicates the restoring torque opposes the angle of twist.

I\alpha = -C\theta \quad\Rightarrow\quad \alpha = -\frac{C}{I}\theta \quad\Rightarrow\quad \frac{d^{2}\theta}{dt^{2}} = -\frac{C}{I}\theta$$ $$\frac{d^{2}\theta}{dt^{2}}+\frac{C}{I}\theta = 0 \tag{9}\label{eq:torsional}

The solution of equation \eqref{eq:torsional} is $\theta = \theta_0\sin(\omega t+\phi)$. Comparing with $\frac{d^{2}\theta}{dt^{2}}+\omega^2\theta=0$:

\omega^2 = \frac{C}{I}, \qquad \omega = \sqrt{\frac{C}{I}}, \qquad \frac{2\pi}{T} = \sqrt{\frac{C}{I}}$$ $$T = 2\pi\sqrt{\frac{I}{C}}

6. Spring and Mass System

It consists of a mass placed on a horizontal frictionless surface. One end of the spring is connected to the mass and the other end is connected to a rigid support (fixed point).

When some force is applied to stretch or compress the spring, the restoring force tends to regain the equilibrium position, due to which the mass executes simple harmonic motion.

From Hooke's law, restoring force $\propto$ displacement:

F \propto x, \qquad F = -kx$$ $$ma = -kx \qquad (k = \text{spring constant})$$ $$a = -\frac{k}{m}x \quad\Rightarrow\quad \frac{d^{2}x}{dt^{2}} = -\frac{k}{m}x$$ $$\frac{d^{2}x}{dt^{2}}+\frac{k}{m}x = 0 \tag{10}\label{eq:spring}

which is the differential equation of S.H.M., and its solution is $x = a\sin(\omega t+\phi)$.

Comparing equation \eqref{eq:spring} with $\frac{d^{2}x}{dt^{2}}+\omega^2 x=0$:

\omega^2 = \frac{k}{m}, \qquad \omega=\sqrt{\frac{k}{m}}, \qquad \frac{2\pi}{T}=\sqrt{\frac{k}{m}}$$ $$T = 2\pi\sqrt{\frac{m}{k}}

7. Helmholtz Resonator

It is a device which is used to analyse the complex note (frequency) of sound. It consists of a large vessel of glass or metal, spherical or cylindrical in shape. It has a narrow neck ($N$) through which it communicates with outside air and receives the complex note. It has a narrow aperture ($O$) cut opposite to $N$, which is to be plugged into the ear.

A spherical resonator analyses only the frequency equal to its natural frequency, whereas a cylindrical resonator is used to detect different frequencies by sliding one part over another -- its frequency can be changed.

Working Principle

The air in the neck acts as an air plug. It performs oscillatory motion like a piston inside a cylinder.

Theory

Stress = Pressure (for a gas), Let $\ell$ = length of neck, $A$ = area of cross-section of the neck, $S$ = density of air.

Mass of the air plug in the neck $= \text{volume}\times\text{density} = A\ell S$.

If the air plug is forced inward through a small distance $x$, the decrease in volume is

\delta V = -Ax

(the negative sign indicates a decrease in volume). If $V$ = total initial volume of air, the bulk modulus of air is given by

k = \frac{\delta p}{-\delta V/V} \quad\Rightarrow\quad \delta p = -k\frac{\delta V}{V}

Force acting on the air plug = pressure $\times$ area:

ma = \delta p \times A = -\frac{k\delta V}{V}A = -\frac{kA^2x}{V}$$ $$m\frac{d^{2}x}{dt^{2}} = -\frac{kA^2 x}{V} \quad\Rightarrow\quad \frac{d^{2}x}{dt^{2}} = -\frac{kA^2x}{Vm} = -\frac{kA^2x}{V\cdot A\ell S} = -\frac{kAx}{V\ell S}$$ $$\frac{d^{2}x}{dt^{2}}+\frac{kA}{V\ell S}x = 0 \tag{11}\label{eq:helm}

which is the differential equation of S.H.M. Comparing with $\frac{d^{2}x}{dt^{2}}+\omega_0^2 x=0$:

\omega_0^2 = \frac{kA}{V\ell S}, \qquad \omega_0 = \sqrt{\frac{kA}{V\ell S}}, \qquad \frac{2\pi}{T} = \sqrt{\frac{kA}{V\ell S}}$$ $$T = 2\pi\sqrt{\frac{V\ell S}{kA}} = \frac{2\pi}{v}\sqrt{\frac{V\ell}{A}}, \qquad \left(v=\sqrt{\frac{k}{S}} \text{ is the velocity of sound}\right)

and frequency

f = \frac{1}{T} = \frac{v}{2\pi}\sqrt{\frac{A}{V\ell}}

8. N-Coupled Oscillator (Two Masses Connected by a Spring / Two-Body Oscillator)

Consider a two-body oscillator consisting of two masses $m_1$ and $m_2$ connected by a horizontal spring of force constant $k$, placed on a frictionless horizontal surface such that it can oscillate freely along the length of the spring.

Let $\ell$ = length of spring, $x_1$ and $x_2$ = positions of the two ends of the spring from a fixed origin. Let the spring be extended through a distance $x$ and then released.

\text{Extension } (x) = (x_2-x_1)-\ell

For extension, $x>0$; for compression, $x<0 and="" at="" compressed="" has="" if="" is="" its="" length.="" length="" natural="" negative="" normal="" p="" positive="" spring="" stretched="" the="" x="" zero="">

The force exerted by the spring on mass $m_1$ is

m_1\frac{d^{2}x_1}{dt^{2}} = kx \tag{12}\label{eq:m1}

The force exerted by the spring on mass $m_2$ is

m_2\frac{d^{2}x_2}{dt^{2}} = -kx \tag{13}\label{eq:m2}

Computing $m_1\times\text{\eqref{eq:m2}} - m_2\times\text{\eqref{eq:m1}}$:

m_1m_2\frac{d^{2}x_2}{dt^{2}}-m_1m_2\frac{d^{2}x_1}{dt^{2}} = -m_1kx-m_2kx$$ $$m_1m_2\frac{d^2}{dt^2}(x_2-x_1) = -kx(m_1+m_2)$$ $$\frac{m_1m_2}{m_1+m_2}\frac{d^{2}}{dt^{2}}(x_2-x_1)+kx = 0

Since $x=(x_2-x_1)-\ell$ and $\ell$ is constant, $\dfrac{d^2x}{dt^2}=\dfrac{d^2}{dt^2}(x_2-x_1)$. Let the reduced mass $\mu = \dfrac{m_1m_2}{m_1+m_2}$. Then,

\mu\frac{d^{2}x}{dt^{2}}+kx = 0 \quad\Rightarrow\quad \frac{d^{2}x}{dt^{2}}+\frac{k}{\mu}x=0 \tag{14}\label{eq:coupled}

which is the differential equation of S.H.M., whose solution is $x=a\sin(\omega_0 t+\phi)$. Comparing equation \eqref{eq:coupled} with $\frac{d^{2}x}{dt^{2}}+\omega_0^2 x=0$:

\omega_0^2 = \frac{k}{\mu}, \qquad \omega_0 = \sqrt{\frac{k}{\mu}}, \qquad \frac{2\pi}{T}=\sqrt{\frac{k}{\mu}}$$ $$T = 2\pi\sqrt{\frac{\mu}{k}}, \qquad f = \frac{1}{T}=\frac{1}{2\pi}\sqrt{\frac{k}{\mu}}

9. Damped Harmonic Oscillator

When a body oscillates in the presence of some dissipating force, the amplitude of vibration continuously decreases. Such an oscillation is called damped oscillation, and such an oscillator is called a damped harmonic oscillator.

Damping (or resisting) force is directly proportional to the velocity, acting in the opposite direction:

$$F_1 = -bv$$

where $b$ is the damping constant (negative sign indicates the force opposes the velocity).

The linear restoring force acting on the body is $F_2 = -kx$.

Total resultant force on the oscillating body:

F = F_1+F_2 = -bv-kx$$ $$ma = -kx-bv \quad\Rightarrow\quad m\frac{d^{2}x}{dt^{2}} = -kx-b\frac{dx}{dt}$$ $$\frac{d^{2}x}{dt^{2}} = -\frac{k}{m}x-\frac{b}{m}\frac{dx}{dt}$$ $$\frac{d^{2}x}{dt^{2}}+\frac{b}{m}\frac{dx}{dt}+\frac{k}{m}x = 0 \tag{15}\label{eq:damp1}

Let $\dfrac{k}{m}=\omega_0^2$ and $\dfrac{b}{m}=\dfrac{1}{\tau}$, where $\tau$ is the relaxation time. Then equation \eqref{eq:damp1} becomes

\frac{d^{2}x}{dt^{2}}+\frac{1}{\tau}\frac{dx}{dt}+\omega_0^2 x = 0 \tag{16}\label{eq:damp2}

Let $x=e^{\alpha t}$ be the trial solution of equation \eqref{eq:damp2}. Then

\frac{dx}{dt}=\alpha e^{\alpha t}, \qquad \frac{d^{2}x}{dt^{2}}=\alpha^2e^{\alpha t}

Substituting,

\alpha^2 e^{\alpha t}+\frac{1}{\tau}\alpha e^{\alpha t}+\omega_0^2 e^{\alpha t}=0$$ $$e^{\alpha t}\left(\alpha^2+\frac{1}{\tau}\alpha+\omega_0^2\right)=0

Since $e^{\alpha t}\neq0$,

\alpha^2+\frac{1}{\tau}\alpha+\omega_0^2=0$$ $$\alpha = \frac{-\frac{1}{\tau}\pm\sqrt{\frac{1}{\tau^2}-4\omega_0^2}}{2} = -\frac{1}{2\tau}\pm\sqrt{\left(\frac{1}{2\tau}\right)^2-\omega_0^2}

Let $\beta = \sqrt{\left(\dfrac{1}{2\tau}\right)^2-\omega_0^2}$. Then $\alpha = -\dfrac{1}{2\tau}\pm\beta$, giving two roots:

\alpha_1 = -\frac{1}{2\tau}+\beta, \qquad \alpha_2 = -\frac{1}{2\tau}-\beta

The general solution of equation \eqref{eq:damp2} is

x = A_1e^{\alpha_1t}+A_2e^{\alpha_2t} = A_1e^{\left(-\frac{1}{2\tau}+\beta\right)t}+A_2e^{\left(-\frac{1}{2\tau}-\beta\right)t}$$ $$x = e^{-t/2\tau}\left\{A_1e^{\beta t}+A_2e^{-\beta t}\right\} \tag{17}\label{eq:damp_gen}

which is the equation of the damped harmonic oscillator. Depending on the value of $\beta$, there are three cases:

Case I: Real Damping (Overdamped)

When $\beta$ is real, the damping is very high and the term $\sqrt{(1/2\tau)^2-\omega_0^2}$ is real and positive. Therefore, the displacement decays exponentially without changing direction -- there is no oscillation. This is the case of damped motion, also known as aperiodic motion or dead-beat motion.

Case II: Critically Damped

The damping in which $\dfrac{1}{2\tau}=\omega_0$, i.e., $\beta=0$, gives the critically damped condition.

From equation \eqref{eq:damp_gen}, with $A_1 = \dfrac{x_0}{2}\left(1+\dfrac{1}{2\beta\tau}\right)$ and $A_2 = \dfrac{x_0}{2}\left(1-\dfrac{1}{2\beta\tau}\right)$, and expanding binomially while neglecting higher powers of $\beta$:

x = \frac{x_0}{2}e^{-t/2\tau}\left\{\left(1+\frac{1}{2\beta\tau}\right)(1+\beta t)+\left(1-\frac{1}{2\beta\tau}\right)(1-\beta t)\right\}$$ $$x = \frac{x_0}{2}e^{-t/2\tau}\left(2+\frac{2t}{2\tau}\right)$$ $$x = x_0e^{-t/2\tau}\left(1+\frac{t}{2\tau}\right)

Here, the second term decays less rapidly than the first term ($x_0e^{-t/2\tau}$), so the system returns to its equilibrium position in the shortest possible time.

Case III: Underdamped

When damping is small, $\dfrac{1}{2\tau}<\omega_0$, and $\beta=\sqrt{\dfrac{1}{4\tau^2}-\omega_0^2}$ is an imaginary quantity. Let $\beta=i\omega$, where $\omega = \sqrt{\omega_0^2-\dfrac{1}{4\tau^2}}$.

The displacement becomes

x = \frac{x_0}{2}e^{-t/2\tau}\left\{\left(1+\frac{1}{2\beta\tau}\right)e^{\beta t}+\left(1-\frac{1}{2\beta\tau}\right)e^{-\beta t}\right\}

Substituting $\beta=i\omega$ and expanding using $e^{\pm i\omega t}=\cos\omega t \pm i\sin\omega t$:

x = \frac{x_0}{2}e^{-t/2\tau}\left(2\cos\omega t+\frac{2}{2\omega\tau}i\sin\omega t\right)$$ $$x = x_0e^{-t/2\tau}\left(\cos\omega t+\frac{\sin\omega t}{2\omega\tau}\right) \tag{18}\label{eq:underdamped1}

Let $x_0=a\sin\phi$ and $\dfrac{x_0}{2\omega\tau}=a\cos\phi$. Then equation \eqref{eq:underdamped1} becomes

x = e^{-t/2\tau}\,a\{\sin\phi\cos\omega t+\cos\phi\sin\omega t\}$$ $$x = a\,e^{-t/2\tau}\sin(\omega t+\phi) \tag{19}\label{eq:underdamped_final}

Squaring and adding $x_0=a\sin\phi$ and $\dfrac{x_0}{2\omega\tau}=a\cos\phi$:

x_0^2+\frac{x_0^2}{4\omega^2\tau^2} = a^2 \quad\Rightarrow\quad x_0^2\left(1+\frac{1}{4\omega^2\tau^2}\right)=a^2$$ $$a = x_0\left(1+\frac{1}{4\omega^2\tau^2}\right)^{1/2}

Also,

2\omega\tau = \tan\phi \quad\Rightarrow\quad \phi = \tan^{-1}(2\omega\tau)

Thus, equation \eqref{eq:underdamped_final} represents oscillatory motion of decreasing amplitude, with frequency

f = \frac{\omega}{2\pi} = \frac{1}{2\pi}\sqrt{\omega_0^2-\frac{1}{4\tau^2}} = \frac{1}{2\pi}\sqrt{\frac{k}{m}-\frac{b^2}{4m^2}}

If there is no damping, $b=0$ and $\omega_0=\omega$, so $f=\dfrac{1}{2\pi}\sqrt{k/m}$. When damping is negligible, the damped frequency of the body approaches its natural frequency. The amplitude of oscillation, $ae^{-t/2\tau}$, decreases exponentially with time, resulting in lower power dissipation.

10. Power Dissipation

The displacement is given by $x = a\,e^{-t/2\tau}\sin(\omega t+\phi)$. Then,

v = \frac{dx}{dt} = a\omega\,e^{-t/2\tau}\cos(\omega t+\phi)-\frac{a}{2\tau}e^{-t/2\tau}\sin(\omega t+\phi)

Now,

K.E. = \frac{1}{2}mv^2 = \frac{1}{2}ma^2e^{-t/\tau}\left\{\omega\cos(\omega t+\phi)-\frac{1}{2\tau}\sin(\omega t+\phi)\right\}^2$$ $$=\frac{1}{2}ma^2e^{-t/\tau}\left\{\omega^2\cos^2(\omega t+\phi)-2\omega\cos(\omega t+\phi)\cdot\frac{1}{2\tau}\sin(\omega t+\phi)+\frac{1}{4\tau^2}\sin^2(\omega t+\phi)\right\}

The average kinetic energy over one period, using $\int_0^T\cos^2(\omega t+\phi)\,dt/T = \int_0^T\sin^2(\omega t+\phi)\,dt/T = \tfrac12$ and $\int_0^T \sin2(\omega t+\phi)\,dt=0$:

\overline{K.E.} = \frac{1}{4}ma^2e^{-t/\tau}\left(\omega^2+\frac{1}{4\tau^2}\right)

Similarly,

P.E. = \frac{1}{2}kx^2 = \frac{1}{2}m\omega_0^2a^2e^{-t/\tau}\sin^2(\omega t+\phi)$$ $$\overline{P.E.} = \frac{1}{4}m\omega_0^2a^2e^{-t/\tau}

Total average energy:

\overline{E} = \overline{K.E.}+\overline{P.E.} = \frac{1}{4}ma^2e^{-t/\tau}\left(\omega^2+\omega_0^2+\frac{1}{4\tau^2}\right)

For small damping, $\tau$ is very large so $\dfrac{1}{4\tau^2}$ can be neglected, and $\omega\approx\omega_0$. Then,

\overline{E} = \frac{1}{4}ma^2e^{-t/\tau}\cdot2\omega_0^2$$ $$\overline{E} = \frac{1}{2}m\omega_0^2a^2e^{-t/\tau}

Average power dissipated:

\overline{P} = -\frac{d\overline{E}}{dt} \quad \text{(negative sign indicates energy/power loss)}$$ $$\overline{P} = -\frac{d}{dt}\left\{\frac{1}{2}m\omega_0^2a^2e^{-t/\tau}\right\} = -\frac{1}{2}m\omega_0^2a^2\left(-\frac{1}{\tau}\right)e^{-t/\tau}$$ $$\overline{P} = \frac{1}{2\tau}m\omega_0^2a^2e^{-t/\tau}

Quality Factor (Q)

Q = 2\pi\times\frac{\text{average energy per cycle}}{\text{average energy dissipated per cycle}} = 2\pi\times\frac{\text{average energy per cycle}}{T\times\text{average power dissipated}}$$ $$Q = 2\pi\cdot\frac{\frac{1}{2}m\omega_0^2a^2e^{-t/\tau}}{T\times\frac{1}{2\tau}m\omega_0^2a^2e^{-t/\tau}} = \frac{2\pi\tau}{T}$$ $$Q = \omega\tau

For low damping, $\omega\approx\omega_0$, so

Q = \omega_0\tau

11. Driven (Forced) Harmonic Oscillator

If a particle is acted upon by an external periodic force $F_0\sin pt$ along with damping forces, the resulting oscillation is known as a forced or driven harmonic oscillator.

In this case, the differential equation of motion is:

\frac{d^{2}x}{dt^{2}}+\frac{b}{m}\frac{dx}{dt}+\frac{k}{m}x = \frac{F_0}{m}\sin pt \tag{20}\label{eq:driven1}

Let $\dfrac{b}{m}=\dfrac{1}{\tau}$, $\dfrac{k}{m}=\omega_0^2$, and $\dfrac{F_0}{m}=f_0$. Then equation \eqref{eq:driven1} becomes

\frac{d^{2}x}{dt^{2}}+\frac{1}{\tau}\frac{dx}{dt}+\omega_0^2x = f_0\sin pt \tag{21}\label{eq:driven2}

The (steady-state) trial solution of equation \eqref{eq:driven2} is

x = a\sin(pt-\phi) \tag{A}

Differentiating with respect to $t$:

\frac{dx}{dt} = ap\cos(pt-\phi), \qquad \frac{d^{2}x}{dt^{2}} = -ap^2\sin(pt-\phi)

Substituting into equation \eqref{eq:driven2}:

-ap^2\sin(pt-\phi)+\frac{ap}{\tau}\cos(pt-\phi)+\omega_0^2 a\sin(pt-\phi) = f_0\sin pt$$ $$(\omega_0^2-p^2)a\sin(pt-\phi)+\frac{ap}{\tau}\cos(pt-\phi) = f_0\sin pt

Expanding $\sin(pt-\phi)$ and $\cos(pt-\phi)$:

(\omega_0^2-p^2)\{a\sin pt\cos\phi-a\cos pt\sin\phi\}+\frac{ap}{\tau}\{\cos pt\cos\phi+\sin pt\sin\phi\}=f_0\sin pt$$ $$\left\{(\omega_0^2-p^2)\cos\phi+\frac{p}{\tau}\sin\phi\right\}a\sin pt + \left\{-(\omega_0^2-p^2)\sin\phi+\frac{p}{\tau}\cos\phi\right\}a\cos pt = f_0\sin pt

Comparing coefficients of $\sin pt$ and $\cos pt$:

\left\{(\omega_0^2-p^2)\cos\phi+\frac{p}{\tau}\sin\phi\right\}a = f_0 \tag{22}\label{eq:coefA}$$ $$-(\omega_0^2-p^2)\sin\phi+\frac{p}{\tau}\cos\phi = 0 \tag{23}\label{eq:coefB}

From equation \eqref{eq:coefB}:

(\omega_0^2-p^2)\sin\phi = \frac{p}{\tau}\cos\phi \quad\Rightarrow\quad \tan\phi = \frac{p/\tau}{\omega_0^2-p^2}

Letting $y=\sqrt{(\omega_0^2-p^2)^2+p^2/\tau^2}$, from the right-triangle relation:

\cos\phi = \frac{\omega_0^2-p^2}{y}, \qquad \sin\phi = \frac{p/\tau}{y}

Substituting into equation \eqref{eq:coefA}:

a = \frac{f_0}{y\cos^2\phi+y\sin^2\phi} = \frac{f_0}{y}$$ $$a = \frac{f_0}{\sqrt{(\omega_0^2-p^2)^2+\dfrac{p^2}{\tau^2}}} \tag{24}\label{eq:amplitude}

Substituting the values of $a$ and $\phi$ into (A):

x = \frac{f_0}{\sqrt{(\omega_0^2-p^2)^2+\dfrac{p^2}{\tau^2}}}\;\sin\left\{pt-\tan^{-1}\left(\frac{p/\tau}{\omega_0^2-p^2}\right)\right\} \tag{25}\label{eq:driven_sol}

This is not the complete solution. To make it complete, we add the complementary function $a_0e^{-Kt}\sin(\omega t+\theta)$. Therefore, the complete solution of equation \eqref{eq:driven2} is:

x = a_0e^{-Kt}\sin(\omega t+\theta)+a\sin(pt-\phi)

Here, the first term represents the initial damped oscillation of frequency $\dfrac{\omega}{2\pi}$, and the second term represents the forced/driven oscillation of frequency $\dfrac{p}{2\pi}$. The first term dies out quickly, and the second term remains effective. Therefore, we are left with

x = a\sin(pt-\phi)

as the (steady-state) solution of motion of a forced or driven harmonic oscillator.

Resonance -- Condition for Maximum Amplitude

For the maximum amplitude of oscillation, the denominator of equation \eqref{eq:amplitude} must have a minimum value. So, applying the condition of minima:

\frac{d}{dp}\left\{(\omega_0^2-p^2)^2+\frac{p^2}{\tau^2}\right\} = 0$$ $$2(\omega_0^2-p^2)(-2p)+\frac{2p}{\tau^2}=0$$ $$-2(\omega_0^2-p^2)+\frac{1}{\tau^2}=0 \quad\Rightarrow\quad (\omega_0^2-p^2)=\frac{1}{2\tau^2}$$ $$p^2 = \omega_0^2-\frac{1}{2\tau^2}$$ $$p = \sqrt{\omega_0^2-\frac{1}{2\tau^2}}

Let $\dfrac{1}{2\tau}=K$. Then $p=\sqrt{\omega_0^2-2K^2}$.

The amplitude of oscillation is maximum when the driving frequency is $\dfrac{p_R}{2\pi}=\dfrac{\sqrt{\omega_0^2-2K^2}}{2\pi}$. Here, $p_R$ is also known as the resonant frequency. This frequency is smaller than both the natural frequency ($\omega_0/2\pi$) and the damped frequency ($\omega/2\pi$).

Maximum Amplitude at Resonance

From equation \eqref{eq:amplitude}:

a = \frac{f_0}{\{(2K^2)^2+p^2(2K)^2\}^{1/2}}

At resonance, $p\to p_R$ and $a\to a_{\max}$:

a_{\max} = \frac{f_0}{\{4K^4+p_R^2\cdot4K^2\}^{1/2}} = \frac{f_0}{[4K^2(K^2+p_R^2)]^{1/2}} = \frac{f_0}{2K(K^2+p_R^2)^{1/2}}$$ $$a_{\max} = \frac{f_0\tau}{(K^2+p_R^2)^{1/2}}

Case I: At low damping. $K$ can be neglected, so

a_{\max} = \frac{f_0\tau}{p_R} = \frac{f_0\tau}{\omega_0} \qquad \left(\because p^2=\omega_0^2-2K^2\right)

Case II: At low damping and negligible driving frequency ($p\to0$). $a = \dfrac{f_0}{\omega_0^2}$. Then,

\frac{a_{\max}}{a} = \frac{f_0\tau/\omega_0}{f_0/\omega_0^2} = \omega_0\tau = Q \quad \text{(Quality factor)}

which is the quality factor of the forced harmonic oscillator, defined as the ratio of the maximum amplitude at resonance to the amplitude.

Case III: At low damping and high driving frequency. $\omega_0$ may be neglected, so

a = \frac{f_0}{p^2}

As $p$ increases from zero, the amplitude curve rises to a peak near resonance and then falls off; the peak becomes sharper and taller as damping decreases (zero damping $\to$ light damping $\to$ medium damping $\to$ high damping, in decreasing order of sharpness).

12. Power Absorption

When the oscillator settles into steady-state oscillation, the average power absorbed is equal to the average power dissipated.

If $F=F_0\sin pt$ is the driving force and $dx$ is the displacement in time $dt$, then $dE=F\,dx$:

\frac{dE}{dt} = F\frac{dx}{dt} = F\cdot\frac{d}{dt}\{a\sin(pt-\phi)\}$$ $$\frac{dE}{dt} = aFp\cos(pt-\phi) = \frac{f_0}{\sqrt{(\omega_0^2-p^2)^2+4K^2p^2}}\cdot F\cdot p\cos(pt-\phi)

Since $F = F_0\sin pt = mf_0\sin pt$:

\frac{dE}{dt} = \frac{f_0}{\sqrt{(\omega_0^2-p^2)^2+4K^2p^2}}\cdot mf_0\sin pt\cdot p\cos(pt-\phi)$$ $$P = \frac{mf_0^2p\sin pt\cos(pt-\phi)}{\sqrt{(\omega_0^2-p^2)^2+4K^2p^2}}

Now, the average power is given by (using $\overline{\sin pt\cos(pt-\phi)}=\tfrac12\sin\phi$):

\overline{P} = \frac{mf_0^2p}{\sqrt{(\omega_0^2-p^2)^2+4K^2p^2}}\cdot\frac{1}{2}\sin\phi

Using $\sin\phi = \dfrac{2Kp}{f_0}$ (from the phase relations):

\overline{P} = \frac{mf_0^2p}{\sqrt{(\omega_0^2-p^2)^2+4K^2p^2}}\cdot K\cdot\frac{f_0}{\sqrt{(\omega_0^2-p^2)^2+4K^2p^2}}\cdot\frac{p}{f_0}$$ $$\overline{P} = \frac{mKf_0^2p^2}{(\omega_0^2-p^2)^2+4K^2p^2}

Letting $V_0^2 = \dfrac{f_0^2p^2}{(\omega_0^2-p^2)^2+4K^2p^2}$:

\overline{P} = mKV_0^2 = m\cdot\frac{1}{2\tau}V_0^2$$ $$\overline{P} = \frac{1}{2}rV_0^2, \qquad \text{where } r=\frac{m}{\tau}

And the maximum power (at $p=\omega_0$):

P_{\max} = \frac{1}{4}\cdot\frac{mf_0^2}{K}

Harmonic Oscillator