Oscillations - NEET Physics Notes, Formulas & Concepts

Oscillations — NEET Physics Essentials Concept Summary cover page

1. Chapter Introduction

Oscillations describe repeated motion about an equilibrium position. Simple harmonic motion (SHM) is the ideal model in which restoring acceleration is proportional to displacement and directed towards equilibrium. Oscillations connects position, velocity, acceleration, phase, energy, system parameters, damping and resonance.

2. Chapter in One View

Central idea Locate equilibrium first. If restoring acceleration is proportional to displacement and opposite in direction, the motion is SHM.

Identify whether motion is periodic, oscillatory, or specifically SHM.
Choose the equation and phase from initial position and direction.
Connect displacement, velocity and acceleration; use phase for timing.
Read time graphs and state graphs using axes, slopes and extrema.
Use energy relations for position and energy questions.
Build effective spring stiffness or identify the correct pendulum/system model.
When conditions change, locate the new equilibrium before finding the new amplitude.
Distinguish free damped motion from driven steady-state motion and resonance.

3. Must-Know Concepts

Concept	Core idea / recognition
Periodic motion	Repeats after equal time intervals; it need not be to-and-fro motion.
Oscillatory motion	Repeated motion about equilibrium; it need not be sinusoidal.
SHM condition	\(a=-\omega^2 x\); restoring acceleration is linear in displacement and opposite in sign.
Amplitude A	Maximum magnitude of displacement from equilibrium.
Angular frequency ω	\(\omega=2\pi f=\frac{2\pi}{T}\); units rad s⁻¹.
Phase	For \(x=A\cos(\omega t+\phi)\), phase is ωt + φ and identifies the state in a cycle.
Equilibrium	In SHM, speed is maximum and acceleration is zero.
Extreme positions	Velocity is zero; acceleration magnitude is maximum toward equilibrium.
Ideal SHM energy	Total mechanical energy is constant; kinetic and potential energies exchange.
Damping	Energy is dissipated and amplitude generally decreases.
Forced oscillation	A periodic external force drives the system; after transients, response occurs at the driving frequency.
Resonance	Response amplitude peaks near a resonant driving frequency; damping limits the peak.

4. Formula + Meaning

4.1 SHM definition and kinematics

a=-\omega^2x\qquad F=-kx\quad\text{(linear spring)}

Restoring acceleration/force is proportional to displacement and opposite in direction. Use x measured from equilibrium.

x(t)=A\cos(\omega t+\phi)

v(t)=-A\omega\sin(\omega t+\phi)\qquad a(t)=-\omega^2x

v^2=\omega^2(A^2-x^2)

Use the initial position and velocity direction to determine φ. The squared-velocity relation gives speed magnitude only; the sign of velocity must be determined separately.

4.2 Time period and phase

T=\frac{2\pi}{\omega}\qquad f=\frac{1}{T}\qquad \omega=2\pi f

Motion segment	Time
Extreme to opposite extreme	T/2
Extreme to equilibrium	T/4
Equilibrium to adjacent extreme	T/4
Same position and same direction again	T

Exam trap Equal distances do not generally take equal times. Speed is greatest near equilibrium and zero at the extremes.

4.3 Energy

U=\frac12m\omega^2x^2=\frac12kx^2

K=\frac12m\omega^2(A^2-x^2)

E=K+U=\frac12m\omega^2A^2=\frac12kA^2

For ideal SHM, total mechanical energy is constant. U is maximum at the extremes; K is maximum at equilibrium. Do not apply conserved mechanical energy when damping or external driving transfers energy.

4.4 Spring–mass systems

T=2\pi\sqrt{\frac{m}{k_{\rm eff}}}\qquad \omega=\sqrt{\frac{k_{\rm eff}}{m}}

x_{\rm eq}=\frac{mg}{k}\quad\text{(vertical spring extension)}

\text{Series: }\frac1{k_{\rm eff}}=\frac1{k_1}+\frac1{k_2}+\cdots

\text{Parallel: }k_{\rm eff}=k_1+k_2+\cdots

Gravity shifts the equilibrium of an ideal vertical spring but does not change its period. For two springs acting on the same block, check geometry before using the parallel formula.

k\propto\frac1\ell

If a uniform spring is cut into n equal pieces, each piece has stiffness nk; with the same mass its period is T/√n.

4.5 Two free masses and reduced mass

\mu=\frac{m_1m_2}{m_1+m_2}\qquad T=2\pi\sqrt{\frac{\mu}{k}}

Use reduced mass for relative oscillation of two free masses joined by an ideal spring. Do not use it for a mass attached to a fixed wall.

4.6 Simple pendulum

T=2\pi\sqrt{\frac{L}{g_{\rm eff}}}\qquad \omega=\sqrt{\frac{g_{\rm eff}}{L}}

Assumptions: small angular displacement (sin θ ≈ θ), ideal string, negligible damping, and uniform effective gravity.

v^2=2gL(\cos\theta-\cos\theta_0)

\text{At lowest point: }\mathcal{T}=mg+\frac{mv^2}{L}

For a bob released from rest at θ₀, the energy relation gives speed at angle θ. The tension relation follows from radial force balance.

4.7 Effective gravity

Situation	Effective gravity
Lift accelerating upward with magnitude a	g_eff = g + a
Lift accelerating downward, a < g	g_eff = g − a
Free fall	g_eff = 0; ordinary pendulum formula is not applicable
Support accelerating horizontally	Find the effective-gravity vector and new equilibrium direction first

Find the effective-gravity vector and the new equilibrium direction first, then apply the small-oscillation model.

4.8 Other oscillating systems

\text{Floating body: }\omega=\sqrt{\frac{\rho gS}{m}}

Assumes a small vertical displacement and constant cross-sectional area S at the liquid surface.

\text{Uniform U-tube: }\omega=\sqrt{\frac{2g}{L}}\qquad T=2\pi\sqrt{\frac{L}{2g}}

L is the total length of moving liquid; assume incompressible liquid and negligible viscosity.

\text{Torsional oscillator: }\tau=-C\theta\qquad T=2\pi\sqrt{\frac{I}{C}}

C is torsional constant and I is moment of inertia about the oscillation axis.

4.9 Damped oscillations

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

For linear viscous damping, the damping force is proportional to velocity and opposes the motion. The equation describes free damped oscillations.

In the linear viscous-damping model, the damping force opposes velocity. Energy is dissipated, so the amplitude envelope decays. For underdamping, motion continues while its amplitude decreases.

Graph 5. Damped motion: oscillations remain, but their envelope decreases with time.

Notice the shrinking peaks. Distinguish damped motion from ideal SHM and do not assume constant mechanical energy.

4.10 Forced oscillations and resonance

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

A(\omega)=\frac{F_0}{\sqrt{(k-m\omega^2)^2+b^2\omega^2}}\qquad \omega_0=\sqrt{\frac{k}{m}}

\tan\delta=\frac{b\omega}{k-m\omega^2}\quad\text{(select the phase quadrant consistent with the response)}

Here, m is the oscillating mass, b is the damping constant, k is the spring constant, F₀ is the driving-force amplitude, and ω is the driving angular frequency. The natural angular frequency is ω₀ = √(k/m).

After the transient response dies away, steady-state motion has the driving frequency ω, not necessarily the natural frequency. δ is the phase lag of displacement behind the driving force under the cosine convention.

Driving regime	Response / phase idea
\(\omega\ll\omega_0\)	Displacement approximately in phase with driving force.
\(\omega\approx\omega_0\)	Amplitude becomes large; damping limits the peak.
\(\omega\gg\omega_0\)	Displacement tends toward opposition in phase to driving force.
Greater damping	Lower, broader amplitude peak.

Natural and resonant frequency For a lightly damped oscillator, amplitude becomes maximum when driving frequency is close to ω₀. Damping limits the maximum amplitude; greater damping produces a lower and broader resonance peak. Resonance does not mean infinite amplitude in a real damped system.

Graph 4. Steady-state amplitude versus driving frequency for different damping strengths.

5. Formula Selection Guide

See this situation	Think this model	Use this / check
a proportional to −x	SHM definition	\(a=-\omega^2 x\); coefficient gives ω².
Initial x and direction	Initial phase	Use x and v equations; check both signs.
Speed at known x	SHM state relation	\(v^2=\omega^2(A^2-x^2)\); choose sign separately.
Time to reach a position	Phase/timing	Solve phase and select branch consistent with direction.
Mass and spring stiffness	Spring–mass oscillator	\(T=2\pi\sqrt{\frac{m}{k_{\rm eff}}}\).
Two free masses joined by spring	Relative oscillation	Use reduced mass \(\mu=\frac{m_1m_2}{m_1+m_2}\).
Small-angle pendulum	Pendulum model	\(T=2\pi\sqrt{\frac{L}{g_{\rm eff}}}\).
Amplitude decreases with time	Damping	Energy dissipated; mechanical energy decreases.
Periodic external force	Forced oscillation	Steady-state frequency is \(\omega\).
Amplitude plotted against ω	Resonance curve	Peak location/height depends on damping.
Phase of driven response	Driven oscillator	Use \(\tan\delta\) and select the correct quadrant.
A graph is supplied	Graph reasoning	Read axes, units, slopes, intercepts and extrema first.

6. When to Use / When Not to Use

SHM relations

Use when restoring acceleration is linear in displacement and directed toward equilibrium.
Do not use blindly when motion is merely periodic, force is nonlinear over the stated range, or x is measured from the wrong origin.

Ideal energy relations

Use when mechanical energy is conserved for the ideal oscillator.
Do not use blindly when damping dissipates energy or external driving adds/removes energy.

Simple pendulum

Use when angular displacement is small and effective gravity is known.
Do not use blindly for large amplitudes, free fall, or an accelerating support without correcting effective gravity.

Forced oscillations

Use the driven-oscillator amplitude formula for steady-state response of a linear oscillator under sinusoidal driving.
Do not confuse transient response with steady state, driving frequency ω with natural frequency ω₀, or resonance with zero damping.

Changed conditions Find the new equilibrium → determine effective stiffness/gravity → calculate the new frequency → apply the position and velocity at the instant of change to determine the new amplitude.

7. Common Misconceptions

Wrong idea	Correct physics	Why it matters
Every periodic motion is SHM.	SHM requires \(a=-\omega^2 x\).	Avoids using SHM equations on the wrong model.
At equilibrium both v and a are zero.	In SHM, speed is maximum and acceleration is zero.	Correct state/graph reading.
At an extreme acceleration is zero.	At an extreme, \|a\| = ω²A.	Separates velocity and acceleration.
v² relation gives signed velocity.	It gives speed magnitude only.	Use direction/phase to select sign.
Gravity changes vertical spring period.	It shifts equilibrium only for an ideal linear spring.	Use displacement from the new equilibrium.
Driven motion always occurs at natural frequency.	Steady-state motion occurs at driving frequency ω.	Correct frequency selection.
Resonance always gives infinite amplitude.	Damping limits real response amplitude.	Avoids idealized overclaim.
Mechanical energy stays constant in damping.	Damping dissipates mechanical energy.	Select the right energy model.

8. Exam Traps

Using a formula before identifying the model.
Measuring spring displacement from natural length instead of equilibrium.
Confusing speed with signed velocity.
Choosing the wrong phase branch when finding time.
Assuming equal distances take equal times.
Reading graph slopes without checking axes.
Using reduced mass for a mass attached to a fixed wall.
Applying small-angle pendulum relations at large amplitude.
Confusing natural, driving and resonant frequencies.
Treating resonance peak as infinite or assuming damping has no effect.
Using ideal energy conservation when damping or driving transfers energy.

9. See This → Think This

See this	Think this
a proportional to −x	SHM; coefficient is −ω².
At \(x=0\)	Speed maximum, acceleration zero.
At \(x=\pm A\)	Velocity zero, acceleration magnitude maximum.
Same x, opposite directions	Same speed magnitude, opposite velocity signs.
Slope of x–t graph	Velocity.
a–x straight line	Slope \(=-\omega^2\).
\(U/E=1/4\)	\(\|x\|=A/2\).
Vertical spring	Gravity shifts equilibrium; ideal period unchanged.
Damped amplitude envelope	Energy dissipated.
Periodic external force	Steady-state response follows driving frequency.
Tall, sharp resonance peak	Weak damping.
Lift accelerates	Find effective gravity and equilibrium first.

10. Key Graphs / Graph Reasoning

10.1 Time graphs: x, v and a

Graph 1. Normalized displacement, velocity and acceleration for x = A cos(ωt).

x and a are opposite in phase; v is shifted by one quarter-cycle. At t = 0, x = +A, v = 0 and a = −ω²A.

Exam use: Initial conditions, signs, phase differences, and identifying which quantity is plotted.

10.2 Acceleration versus displacement

Graph 2. Acceleration against displacement for SHM.

The graph is a straight line through the origin with negative slope. For physical, unnormalized axes, slope = −ω², so \(\omega=\sqrt{-\text{slope}}\).

10.3 Energy versus displacement

Energy distribution in ideal SHM: potential, kinetic and total energy versus displacement.

Potential energy is zero at equilibrium and maximum at the extremes; kinetic energy does the opposite; total energy is constant.

Exam use: Energy ratios and turning-point identification.

10.4 Damped motion

See Graph 5 in Section 4.9. Its envelope shows why constant-amplitude SHM equations and conserved mechanical energy cannot be applied to damped motion.

10.5 Resonance response

See Graph 4 in Section 4.10. Compare peak height and width; do not assume the damped peak occurs exactly at ω₀.

11. Quick Checks

What identifies SHM? \(a=-\omega^2 x\), with constant positive ω².
Where is speed maximum? At equilibrium, x = 0.
Where is acceleration magnitude maximum? At \(x=\pm A\).
Time from an extreme to equilibrium? T/4.
If U = E/4, what is |x|? A/2.
Does gravity change an ideal vertical spring period? No; it shifts equilibrium.
Which mass is used for two free masses joined by a spring? Reduced mass.
What frequency appears in steady-state forced motion? Driving frequency ω.
What happens to the resonance peak when damping increases? It becomes lower and broader.
Does \(v^2=\omega^2(A^2-x^2)\) give velocity sign? No; only speed magnitude.
What should be checked before using the pendulum formula? Small-angle condition and effective gravity.
What happens to mechanical energy during damping? It is dissipated.

12. Mini Exam Reasoning

Case A — Identify angular frequency

Clue: a = −25x, with x measured from equilibrium. Model: SHM. Compare with \(a=-\omega^2 x\): \(\omega=5\,\text{rad s}^{-1}\). Therefore \(T=2\pi/5\,\text{s}\). Check units: the coefficient has units s⁻².

Case B — Energy ratio

Clue: \(U=E/4\). Use U/E = x²/A². Therefore \(x=\pm A/2\). Both signs are possible unless direction/history is given.

Case C — Interpret a resonance curve

Clue: two curves have different peak heights; one is taller and sharper. The taller, sharper peak corresponds to weaker damping. Compare curves at the same driving frequency, not just their peak values.

Case D — Pendulum in an upward-accelerating lift

Clue: lift acceleration a upward. Effective gravity: \(g_{\rm eff}=g+a\). Small-angle period:

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

Bob mass does not enter the ideal period.

13. One-Minute Recall

SHM: \(a=-\omega^2 x\); \(x=A\cos(\omega t+\phi)\), \(v=-A\omega\sin(\omega t+\phi)\), and \(v^2=\omega^2(A^2-x^2)\). The squared relation gives speed, not velocity sign.
\(\omega=2\pi/T=2\pi f\). At equilibrium, speed is maximum and acceleration is zero; at extremes, speed is zero and |a| is maximum.
Ideal SHM energy: U = ½mω²x², K = ½mω²(A² − x²), E = ½mω²A².
Damping dissipates energy and reduces amplitude.
Spring–mass: \(T=2\pi\sqrt{\frac{m}{k_{\rm eff}}}\); vertical gravity shifts equilibrium only. For two free masses, use \(\mu=\frac{m_1m_2}{m_1+m_2}\).
Small-angle pendulum: \(T=2\pi\sqrt{\frac{L}{g_{\rm eff}}}\); establish effective gravity first.
Forced steady state follows driving frequency ω. Damping controls resonance height and width; distinguish natural, driving and resonant frequencies.

14. Scope and Final Recognition Sequence

This website summary is designed for rapid concept recall and formula selection. Detailed derivations and extended solved problems belong in the complete Concept Book.

Final principle Recognize the motion → choose the model → select the formula → check the condition → solve → verify the result.

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