Oscillations & Simple Harmonic Motion
Build the mental model first. Every SHM formula is a consequence of one thing: F = −kx. Understand this deeply and everything else follows logically.
CBSE focuses on: definition of SHM, equations of motion, energy expressions, and derivation of T for pendulum and spring-mass. Expect 1–2 conceptual + 1 numerical question.
NEET tests: SHM identification (which function is SHM?), time period formulas, velocity at given displacement, energy at given position. All single-correct MCQ — speed matters more than depth.
JEE Main adds: modified pendulums (accelerating frame), spring combinations, damping concept. Numerical value questions require precise calculation — avoid approximation errors.
JEE Advanced tests: superposition of SHMs, SHM in non-inertial frames, coupled oscillators, SHM with varying restoring force, energy diagrams. Multiple-correct and paragraph-type questions are common.
1. Periodic & Oscillatory Motion
Periodic Motion
Any motion that repeats itself at regular time intervals is periodic.
Examples: Earth revolving around Sun, motion of clock hands, vibration of a guitar string.
All oscillatory motion is periodic, but not all periodic motion is oscillatory. Circular motion is periodic but not oscillatory (no back-and-forth about an equilibrium).
Oscillatory (Vibratory) Motion
Motion where the object moves back and forth about a fixed equilibrium position.
Key feature: The body always has a restoring tendency — when displaced, it returns toward equilibrium.
Ask: Is there a position the body always "wants" to return to? If yes → oscillatory. Does it repeat in time? If yes → also periodic. Most oscillations in physics satisfy both.
Key Definitions You Must Know Cold
2. Defining Simple Harmonic Motion
The Core Definition
SHM is a special type of oscillatory motion where the restoring force is directly proportional to displacement from equilibrium and directed opposite to it.
Here: k = spring/force constant (N/m), x = displacement from equilibrium, negative sign = force is always toward equilibrium (restoring in nature).
This single equation F = −kx is the entire physics. If you can show that the net force on a system is proportional to displacement and opposite in direction, then that system performs SHM. Everything else — T, v, energy — is derived from this.
In JEE problems, always verify SHM by checking if F ∝ −x. If the restoring force is not linearly proportional (e.g., F = −kx²), the motion is NOT SHM but may still be oscillatory.
Derivation: F = ma → Differential Equation of SHM
From Newton's second law: F = ma = m(d²x/dt²)
Since F = −kx:
- Write F = −kx (force law)
- Apply Newton's 2nd Law: m(d²x/dt²) = −kx
- Rearrange: d²x/dt² = −(k/m)x
- Define ω² = k/m, so d²x/dt² = −ω²x
- This is the standard SHM differential equation
JEE Advanced may give you an expression like d²x/dt² = −4x and ask for time period. Directly: ω² = 4 → ω = 2 rad/s → T = 2π/ω = π seconds. Don't overthink it.
General Solution: Equation of Displacement
The general solution to d²x/dt² + ω²x = 0 is:
Or equivalently: x(t) = A sin(ωt + φ₀)
Which form to use?
- If x = A at t = 0 → use cosine with φ₀ = 0
- If x = 0 at t = 0 (moving in +x direction) → use sine with φ₀ = 0
- If x = 0 at t = 0 (moving in −x direction) → use −sine or sine with φ₀ = π
This is where most students lose marks: Velocity is zero at x = ±A (extreme positions), NOT at x = 0. Acceleration is zero at x = 0 (equilibrium), NOT at x = ±A. Internalize this.
SHM as Projection of Uniform Circular Motion
Consider a particle moving in a circle of radius A with uniform angular speed ω. The projection of this circular motion on any diameter is SHM.
This view is powerful because it immediately tells you:
- Amplitude A = radius of reference circle
- Angular frequency ω = angular speed of circular motion
- Phase = angle subtended at center at t = 0
When JEE gives you two SHMs and asks for the resultant — draw the phasors (circular representation). Addition of SHMs with same frequency = phasor addition. This is tested in JEE Advanced regularly.
3. Standard SHM Systems
A mass m attached to a spring (spring constant k) on a frictionless surface.
T depends on m and k, NOT on amplitude A or g. This is why SHM time period doesn't change with amplitude.
If the spring is cut in half, its spring constant DOUBLES (k' = 2k). New T = 2π√(m/2k) = T/√2.
Mass hangs from a spring vertically. At equilibrium: mg = kx₀ (static extension x₀).
When the mass oscillates about the new equilibrium position, the equation of motion gives the same SHM equation. Gravity doesn't affect T — it only shifts the equilibrium position down by x₀ = mg/k.
This is a trap question in NEET/JEE Main. The spring-mass vertical system has the same T = 2π√(m/k) because gravity only shifts equilibrium. Many students write T = 2π√(m+g/k) — this is wrong.
A bob of mass m attached to a massless inextensible string of length L, oscillating with small angle (θ < 5°).
- Resolve forces: restoring = −mg sinθ
- For small angles: sinθ ≈ θ = x/L
- F = −(mg/L)x → This is SHM with k_eff = mg/L
- T = 2π√(m/k_eff) = 2π√(mL/mg) = 2π√(L/g)
T doesn't depend on mass m or amplitude A (for small oscillations). Only L and g matter. On the Moon (g = 1.6 m/s²), the pendulum beats slower — T increases.
- Pendulum in accelerating elevator (upward with a): T = 2π√(L/(g+a))
- Pendulum in free fall: T → ∞ (weightlessness, no restoring force)
- Pendulum on incline (angle α): T = 2π√(L/(g cosα))
Series (one after another):
In series: k_eff < k₁, k₂ → weaker spring → longer T.
Parallel (side by side):
In parallel: k_eff > k₁, k₂ → stiffer system → shorter T.
Series = weaker = bigger T. Parallel = stronger = smaller T. Remember this direction and you won't confuse the formulas in an exam.
4. Energy in Simple Harmonic Motion
In SHM, energy continuously converts between kinetic and potential forms, but total mechanical energy remains constant (no friction).
When does KE = PE?
½mω²(A²−x²) = ½mω²x² → A² − x² = x² → x² = A²/2 → x = ±A/√2
KE = PE at x = ±A/√2 ≈ ±0.707A. This is a very frequently asked result in NEET and JEE Main.
If amplitude doubles (A → 2A), total energy quadruples (E → 4E). If frequency doubles (ω → 2ω), and A stays same, E → 4E (since E ∝ ω²A²). JEE Main loves numerical questions on this.
5. Velocity as a Function of Position
From energy conservation (or by differentiating x = A cosωt):
- At x = 0 (equilibrium): v = ωA = v_max
- At x = ±A (extreme): v = 0
- At x = A/2: v = ω√(A² − A²/4) = ω√(3A²/4) = (√3/2)ωA
In exams, if asked for velocity at x = A/2, don't substitute and compute from scratch every time. Memorize: v = (√3/2)v_max. At x = A/√2: v = v_max/√2. These shortcuts save time in MCQ.