 # What is the equation of harmonic oscillation?

Restoring force is directly proportional to displacement from equilibrium restoring force is the driving force of harmonic oscillations.

In LC circuit the restoring force is due to electrostatic repulsive force between electrons which tends to distribute electrons equally on plates of capacitor, to make there is no net charge, but on the other hand inductor tends to oppose this redistribution of electrons (i.e) opposes the increase of current at any instant of time, the voltage across the inductor is

$$\begin{array}{l}{{V}_{L}}=-L\frac{di}{dt}=-L\frac{{{d}^{2}}q}{d{{t}^{2}}}\end{array}$$

The minus sign indicates that voltage opposes the increases of current. But applying kirchoff’s law to the given LC circuit, the potential drop across the inductor must be equal to potential drop across the capacitor.

VL = VC

$$\begin{array}{l}L\frac{{{d}^{2}}q}{d{{t}^{2}}}=\frac{q}{c}\end{array}$$

$$\begin{array}{l}\frac{{{d}^{2}}q}{d{{t}^{2}}}=-{{\omega }^{2}}q\end{array}$$

$$\begin{array}{l}\omega =\frac{1}{\sqrt{LC}}\end{array}$$

$$\begin{array}{l}\frac{{{d}^{2}}q}{d{{t}^{2}}}+{{\omega }^{2}}q=0\end{array}$$
…………….(1)

The above equation 1 is known as differential equation of harmonic oscillation. Which resembles the mechanical harmonic oscillation as

$$\begin{array}{l}\frac{{{d}^{2}}x}{d{{t}^{2}}}+{{\omega }^{2}}x=0\end{array}$$
…………….(1a)

In electrical circuit, (i.e) LC circuit charges oscillates harmonically with an angular frequency of

$$\begin{array}{l}\omega =\frac{1}{\sqrt{LC}}\end{array}$$

and

$$\begin{array}{l}T=2\pi \sqrt{LC}\end{array}$$

The general solution of equation (1a) becomes

\begin{array}{l}\begin{matrix} x(t)=Acos (\omega t+\phi ) \begin{align} & similarly,for,equation,1 \ & q(t)={{q}_{0}}cos (\omega t+\phi ) \end{align} \end{matrix}\end{array}
_______(2)

.Equation 2 gives at any instant charge as a function of time. Where

$$\begin{array}{l}{{q}_{0}}- is\,maximum\,charge\end{array}$$

$$\begin{array}{l}\phi – phase\, of\, the\, electron\, oscillation\end{array}$$

If we differentiate the equation 2 with respect to time, we can get current as function of time at any instant of time

$$\begin{array}{l}\begin{matrix} i=\frac{dq}{dt}=-\omega {{q}_{0}}sin (\omega t+\phi ) \ i=-{{i}_{0}}sin (\omega t+\phi ),…… \end{matrix}\end{array}$$
__________(3)

Where

io = ꞷqo = is maximum current

qo = cvo and

$$\begin{array}{l}\omega =\frac{1}{\sqrt{LC}}\end{array}$$
then

$$\begin{array}{l}{{i}_{o}}={{v}_{o}}\sqrt{\frac{C}{L}}\end{array}$$

From the above discussion we conclude in LC circuit charges oscillating harmonically.

Alternate method:

In LC circuit there is no damping factor (i.e) Resistor (R). So there is Energy loss at any instant total energy stored in the circuit (UTotal) is equal to sum of energy stored across the capacitor and inductor.

Total energy in the circuit (UTotal)

$$\begin{array}{l}{{U}_{total}}=\frac{1}{2}L{{i}^{2}}+\frac{{{phi }^{2}}}{2c}\end{array}$$
________(4)

Since,

There is No Damping factor, Energy remains constant

$$\begin{array}{l}\frac{d{{U}_{Total}}}{dt}=0\end{array}$$

$$\begin{array}{l}\frac{d}{dt}\left( \frac{1}{2}L{{i}^{2}}+\frac{{{\phi }^{2}}}{2c} \right)=0\end{array}$$

Where

• L and c = constants
• i and q = are variables
$$\begin{array}{l}Li\frac{di}{dt}+\frac{1}{c}q\frac{dq}{dt}=0\end{array}$$
__________(5)

$$\begin{array}{l}\frac{dq}{dt}=i,and,\frac{di}{dt}=\frac{{{d}^{2}}q}{d{{t}^{2}}}\end{array}$$

$$\begin{array}{l}\frac{{{d}^{2}}q}{d{{t}^{2}}}+\frac{q}{LC}=0\end{array}$$
_________(6)

We know,

$$\begin{array}{l}\omega =\frac{1}{\sqrt{Lc}}=\sqrt{\frac{K}{m}}\end{array}$$

$$\begin{array}{l}\frac{{{d}^{2}}q}{d{{t}^{2}}}+{{\omega }^{2}}q=0\end{array}$$
________(7)

This equation 7 is similar to the linear differential equation of a simple harmonic oscillator composed of a spring and mass.

$$\begin{array}{l}\frac{{{d}^{2}}x}{d{{t}^{2}}}+{{\omega }^{2}}x=0\end{array}$$
_________(8)

General solution of this equation becomes

x = A cos (ꞷt + ϕ)

• A – maximum displacement (i.e) Amplitude
• ϕ – is phase constant

by comparing equation (7) and (8)

$$\begin{array}{l}q={{q}_{0}}cos (\omega t+\phi )\end{array}$$
______________(9)

• q0 = is maximum charge
• ϕ – is phase constant

then current at any instant of time is

$$\begin{array}{l}i=\frac{dq}{dt}=-\omega {{q}_{0}}sin (\omega t+\phi )\end{array}$$
__________(10)

By substituting initial condition in equation 9 and 10

At time t = 0, ϕ = 0

• q = q0 cos ꞷt and
• i = i0 sin ꞷt

Energy of Harmonic oscillator:

From the discussion we saw the LC circuit resembles a spring – mass system both of them has a characteristic frequency.

To understanding LC circuit oscillation Let us assume that,

Case (I): at time (t = 0)

Charge across the capacitor is q and current in the inductor (i) = 0. Now energy stored in the circuit will be equal to electrostatic energy stored across the capacitor

$$\begin{array}{l}{{E}_{c}}=\frac{{{Q}^{2}}}{2c}…….(i)\end{array}$$

Energy stored in the inductor is zero because current passing through the inductor is zero.

Case (II) : at time (t = t)

After time (t) capacitor starts discharging through the inductor current (i) will be established across the inductor

$$\begin{array}{l}i=\frac{dq}{dt}\end{array}$$
. i starts increases, when i increases q will start decreasing. When q attains zero capacitor is completely discharged, at this instant current will reach maximum. Now energy stored in the circuit will be completely magnetic energy because q = 0 and I = is maximum.

$$\begin{array}{l}{{E}_{L}}=\frac{1}{2}L{{i}^{2}},…………(ii)\end{array}$$

Whole of the energy stored across inductor only.

Case (III):

The large current in the inductor starts transporting charges to the capacitor plates and capacitor gets charged again after attaining maximum charge it will starts discharging again but now current flow in to the opposite direction. Eventually the current returns to its initial value and the process keep on continues.

The energy exchange occurs between the capacitor and inductor in the form electrostatic energy and magnetic energy respectively but total energy of the system is conserved.

Energy = constant

Here there is no energy Loss, because there is no damping factor (i.e) resistor component so energy is conserved

$$\begin{array}{l}{{E}_{Total}}={{E}_{C}}+{{E}_{L}}\end{array}$$
$$\begin{array}{l}=\frac{{{q}^{2}}}{2c}+\frac{1}{2}L{{i}^{2}}=constan t,…………..(iii)\end{array}$$

Since,

• q = qo cos ꞷ t
• i = – io sin ꞷ t = – ꞷ qo sin ꞷ t
$$\begin{array}{l}{{E}_{Total}}=\frac{q_{0}^{2}{{cos }^{2}}\omega t}{2c}+\frac{1}{2}L{{\omega }^{2}}q_{0}^{2}{{sin }^{2}}\omega t\end{array}$$

$$\begin{array}{l}=\frac{q_{0}^{2}{{sin }^{2}}\omega t}{2c}+\frac{q_{0}^{2}cos {{t}^{2}}\omega t}{2c}\end{array}$$

=

$$\begin{array}{l}=\frac{q_{0}^{2}}{2c}[{{sin }^{2}}\omega t+{{cos }^{2}}\omega t]\end{array}$$

$$\begin{array}{l}{{E}_{Total}}=\frac{q_{0}^{2}}{2c}…………..(iv)\end{array}$$

Total energy of the system is constant and it is equal to initial energy.

Graphical representation of charge (q) and current (i) as function of time

 Spring mass system LC oscillator $$\begin{array}{l}\frac{{{d}^{2}}x}{d{{t}^{2}}}+{{\omega }^{2}}x=0\end{array}$$ $$\begin{array}{l}\frac{{{d}^{2}}q}{d{{t}^{2}}}+{{\omega }^{2}}q=0\end{array}$$ $$\begin{array}{l}\frac{{{d}^{2}}x}{d{{t}^{2}}}+\frac{k}{m}x=0\end{array}$$ $$\begin{array}{l}\frac{{{d}^{2}}q}{d{{t}^{2}}}+\frac{1}{LC}=0\end{array}$$ x = A cos (ꞷt + ϕ) q = q0 cos (ꞷt + ϕ) $$\begin{array}{l}V=\frac{dx}{dt}=-\omega Asin (\omega t+\phi )\end{array}$$ $$\begin{array}{l}i=\frac{dq}{dt}=-\omega {{q}_{0}}sin (\omega t+\phi )\end{array}$$ $$\begin{array}{l}a=\frac{dv}{dt}=-{{\omega }^{2}}Acos (\omega t+\phi )\end{array}$$ $$\begin{array}{l}\frac{di}{dt}=\frac{{{V}_{L}}}{L}=-\omega _{{}}^{2}{{q}_{0}}cos (\omega t+\phi )\end{array}$$ $$\begin{array}{l}\frac{1}{c}\end{array}$$ K L M $$\begin{array}{l}\omega =\frac{1}{\sqrt{LC}}\end{array}$$ $$\begin{array}{l}\omega =\sqrt{\frac{K}{m}}\end{array}$$ $$\begin{array}{l}{{E}_{c}}={{U}_{c}}=\frac{{{q}^{2}}}{2c}\end{array}$$ $$\begin{array}{l}PE=U=\frac{1}{2}K{{x}^{2}}\end{array}$$ $$\begin{array}{l}{{E}_{L}}={{U}_{L}}=\frac{1}{2}L{{i}^{2}}\end{array}$$ $$\begin{array}{l}KE=\frac{1}{2}m{{v}^{2}}\end{array}$$ (0) (1)