Oscillations – Simple Harmonic Motion, Damped & Forced Oscillations, Resonance

1. What is an Oscillation?

An oscillation is a repetitive variation of a physical quantity about an equilibrium position. When the restoring force is directly proportional to the displacement and opposite in direction, the motion is simple harmonic.

$$F=-kx\text{(Hooke’s law for a linear restoring force)}$$

The corresponding differential equation

$$m\frac{d^{2}x}{d{t}^2}+kx=0$$

has the solution

$$x(t)=A\cos (\omega t+\varphi )$$

where

• $A$ = amplitude (maximum displacement)
• $\omega =\sqrt{k/m}$ = angular frequency (rad s${}^{-1}$)
• $f=\omega /2\pi$ = frequency (Hz)
• $T=1/f=2\pi \sqrt{m/k}$ = period (s)

IMAGE: simple pendulum labelled diagram | Simple pendulum showing bob, string, and equilibrium line

IMAGE: mass spring system labelled diagram | Mass‑spring oscillator with spring constant $k$ and mass $m$

2. Types of Oscillatory Systems

flowchart TD
    A["Oscillations"] --> B["Free (undriven)"]
    A --> C["Damped"]
    A --> D["Forced"]
    B --> E["Simple Harmonic Motion"]
    C --> F["Underdamped"]
    C --> G["Critically Damped"]
    C --> H["Overdamped"]
    D --> I["Resonance"]

2.1 Free Simple Harmonic Motion

• No external forces act after the system is released.
• Energy is conserved: $E=\frac{1}{2}k{A}^2=\frac{1}{2}m{\omega }^2{A}^2$.

2.2 Damped Oscillations

Real systems lose energy to friction, air resistance, or internal material damping. The equation of motion becomes

$$m\frac{d^{2}x}{d{t}^2}+b\frac{dx}{dt}+kx=0$$

where $b$ is the damping coefficient. The solution depends on the damping ratio $\zeta =\frac{b}{2\sqrt{mk}}$.

• Underdamped motion still oscillates but with exponentially decreasing amplitude.
• Critically damped returns to equilibrium fastest without overshoot – essential for analog meters.
• Overdamped returns slowly, never overshooting.

2.3 Forced Oscillations and Resonance

When an external periodic force $F_d=F_0\cos ({\omega }_{d}t)$ drives the system, the equation is

$$m\frac{d^{2}x}{d{t}^2}+b\frac{dx}{dt}+kx=F_0\cos ({\omega }_{d}t)$$

The steady‑state amplitude

$$A({\omega }_d)=\frac{F_0/m}{\sqrt{({\omega }_0^2-{\omega }_d^2{)}^2+(2\zeta {\omega }_0{\omega }_d{)}^2}}$$

peaks when ${\omega }_d$ ≈ ${\omega }_0$ (natural frequency). This is resonance. The maximum amplitude

$$A_{\max }=\frac{F_0}{2m{\omega }_0\zeta }$$

occurs at

$${\omega }_r={\omega }_0\sqrt{1-2{\zeta }^2}(\text{for }\zeta <\frac{1}{\sqrt{2}})$$

IMAGE: oscilloscope forced oscillation labelled diagram | Oscilloscope trace showing resonance peak in a driven RLC circuit

3. Energy in Simple Harmonic Motion

For an ideal (undamped) oscillator:

• Kinetic Energy $K=\frac{1}{2}m{v}^2=\frac{1}{2}m{\omega }^2{A}^2{\sin }^2(\omega t+\varphi )$
• Potential Energy $U=\frac{1}{2}k{x}^2=\frac{1}{2}k{A}^2{\cos }^2(\omega t+\varphi )$

Total energy $E=K+U=\frac{1}{2}k{A}^2$ is constant.

In a damped system, energy decays as

$$E(t)=E_0{e}^{-2\zeta {\omega }_{0}t}$$

4. Worked Example – Damped Mass‑Spring System

Problem: A 0.5 kg mass is attached to a spring of constant $k=200{\text{N m}}^{-1}$. The system experiences a damping force $bv$ with $b=4{\text{kg s}}^{-1}$. Determine (a) the damping ratio, (b) the type of damping, (c) the displacement after 5 s if the initial amplitude is 0.1 m and the motion is released from rest.

Solution:

1. Natural angular frequency

$${\omega }_0=\sqrt{\frac{k}{m}}=\sqrt{\frac{200}{0.5}}=\sqrt{400}=20{\text{rad s}}^{-1}$$

2. Damping ratio

$$\zeta =\frac{b}{2\sqrt{mk}}=\frac{4}{2\sqrt{0.5\times 200}}=\frac{4}{2\sqrt{100}}=\frac{4}{20}=0.20$$

Since $\zeta <1$, the system is underdamped.

3. Damped angular frequency

$${\omega }_d={\omega }_0\sqrt{1-{\zeta }^2}=20\sqrt{1-0.04}=20\sqrt{0.96}=19.6{\text{rad s}}^{-1}$$

4. Displacement expression (released from rest ⇒ $\varphi =0$)

$$x(t)=A{e}^{-\zeta {\omega }_{0}t}\cos ({\omega }_{d}t)$$

Plug numbers:

$$x(5)=0.1e^{-0.20\times 20\times 5}\cos (19.6\times 5)=0.1e^{-20}\cos (98)$$

$e^{-20}\approx 2.06\times 10^{-9}$; $\cos (98\text{rad})$ ≈ 0.54.

$$x(5)\approx 0.1\times 2.06\times 10^{-9}\times 0.54\approx 1.1\times 10^{-10}\text{m}$$

The amplitude has essentially vanished after 5 s, illustrating rapid energy loss due to damping.

5. Resonance in an RLC Circuit (Forced EM Oscillation)

Consider a series RLC circuit driven by an AC source $V(t)=V_0\cos (\omega t)$. The differential equation for charge $q$ is

$$L\frac{d^{2}q}{d{t}^2}+R\frac{dq}{dt}+\frac{q}{C}=V_0\cos (\omega t)$$

The resonant frequency (ignoring resistance) is

$${\omega }_0=\frac{1}{\sqrt{LC}}$$

At resonance, the impedance is minimum ($Z=R$), and the current amplitude

$$I_{\max }=\frac{V_0}{R}$$

Derivation (brief)

1. Write phasor form: $Z=R+j(\omega L-1/(\omega C))$.
2. Magnitude $|Z|=\sqrt{R^2+(\omega L-1/(\omega C){)}^2}$.
3. Minimum $|Z|$ occurs when $\omega L=1/(\omega C)\Rightarrow \omega ={\omega }_0$.

Thus the maximum current is $I_{\max }=V_0/R$.

6. Applications and Advantages

7. In the Real World

• eSewa & Khalti payment gateways: When a user clicks “Pay”, the server sends a request that triggers a forced oscillation in the network traffic. The system’s load balancer is tuned to the resonant frequency of its request‑handling loop, ensuring maximum throughput without overload.
• Daraz order queue: Orders are processed by a damped queue system. The queue length behaves like an under‑damped oscillator; after a surge (e.g., flash sale) the system’s automatic scaling reduces the “amplitude” of pending orders exponentially, preventing endless backlog.
• NTC mobile towers: Antenna tuning circuits are LC resonant circuits. By adjusting the capacitor bank, the tower resonates at the carrier frequency (e.g., 900 MHz), maximizing radiated power and minimizing reflected power, which directly improves call quality.

Worked real‑world example: A Nepali bank offers a loan with an annual interest rate of 12 % compounded monthly. The monthly payment schedule can be modeled as a forced damped oscillator, where the principal acts as the “mass”, the interest accrual as the “restoring force”, and monthly repayments as the “damping”. Using the SHM formula for exponential decay, the remaining principal after $n$ months is

$$P_n=P_0{e}^{-n\ln (1+0.12/12)}=P_0(1+0.01{)}^{-n}$$

Thus after 12 months, $P_{12}=P_0\times (0.99{)}^{12}\approx 0.886P_0$, showing a 11.4 % reduction—exactly the effect of the monthly payment plus interest.

8. Comparison of Free, Damped, and Forced Oscillations

flowchart LR
    F["Free SHM"] -->|"No energy loss"| U["Constant amplitude"]
    D["Damped"] -->|"Energy loss"| V["Amplitude ↓ exponentially"]
    R["Forced"] -->|"External drive"| W["Steady‑state amplitude depends on driving frequency"]
    F -->|"If disturbed"| D
    D -->|"If driven"| R

9. Common Mistakes to Avoid

1. Confusing natural frequency ${\omega }_0$ with driving frequency ${\omega }_d$. The resonance condition involves equality of these two only when damping is small.
2. Using $T=2\pi \sqrt{L/g}$ for a pendulum of large amplitude. The formula is valid only for small angles (< 5°).
3. Neglecting the phase angle $\varphi$ in forced oscillations. At resonance the phase difference between force and displacement is $\pi /2$.
4. Assuming critical damping gives the fastest return for any system. It is fastest without overshoot; for some control systems an under‑damped response may be preferred for speed.

10. Sample Past‑Exam Question & Solution

Question: Derive an expression for the resonant frequency in a forced EM oscillation (series RLC circuit). Hence, find the maximum current when the source voltage is $V_0=10\text{V}$ and the resistance is $R=2\Omega$.

Solution Sketch:

1. Impedance $Z=\sqrt{R^2+(\omega L-1/\omega C{)}^2}$.
2. Resonance when $\omega L=1/(\omega C)\Rightarrow {\omega }_0=1/\sqrt{LC}$.
3. At resonance, $Z_{\min }=R$; current amplitude $I_{\max }=V_0/R=10/2=5\text{A}$.

Key points for exam: write the differential equation, identify the term that vanishes at resonance, and clearly state the final current expression.

Exam tip

• Memorise the three standard forms of the displacement equation (undamped, under‑damped, critically damped). Write them down quickly with the symbols ${\omega }_0,\zeta ,\lambda$.
• Derivation shortcut: For resonance in any linear system, set the reactive part of the impedance (or equivalent) to zero; this instantly gives ${\omega }_0=1/\sqrt{LC}$ for RLC or ${\omega }_0=\sqrt{k/m}$ for mechanical.
• Units check: Always keep ${\text{rad s}}^{-1}$ for angular frequency; convert to Hz only at the final step.
• Worked example recall: The damped mass‑spring problem above demonstrates the use of $\zeta$ and ${\omega }_d$. Replicate the steps for any numerical question.
• Past paper pattern: Questions often ask for “percentage of maximum amplitude after n time constants”. Remember that for under‑damped motion the amplitude after one time constant $\tau =1/(\zeta {\omega }_0)$ is $A(\tau )=A_0{e}^{-1}\approx 0.37A_0$.