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📐 Chapter 03 · Formulas & Dimensional Analysis

Complete Formula Bank for SHM

Every SHM formula, its dimensional verification, SI units, and the critical condition when it applies. Use the search bar to find any formula instantly.

🔬 CBSE Focus

CBSE expects: standard formulas + their dimensions + 1–2 formula-based numericals. Derivation of pendulum time period may be asked (3 marks).

🔬 NEET Focus

NEET: identify formula, substitute, compute. Speed is key. All formulas here are MCQ-ready. Focus on T expressions, v at x, energy ratios.

🔬 JEE Main Focus

JEE Main: numerical value questions require exact answers. Don't approximate π or √. Keep answers as fractions/surds where possible.

🔬 JEE Advanced Focus

JEE Advanced: dimensional analysis is used to verify answers in derivation-type questions. Know which quantities are dimensionless (phase, strain).

Interactive SHM Calculators

🔧 Spring-Mass Period Calculator

T = 2π√(m/k)

Mass (m) in kg

Spring Constant (k) in N/m

🕰 Simple Pendulum Period Calculator

T = 2π√(L/g)

Length (L) in m

g (m/s²) — default 9.8

⚡ Velocity at Position x

v = ω√(A² − x²)

Amplitude (A) in m

Position (x) in m

Angular Frequency (ω) in rad/s

⚡ Energy at Position x

KE + PE = ½mω²A²

Mass (m) kg

Amplitude (A) m

Position (x) m

ω (rad/s)

🔍 Searchable Formula Bank

Type any keyword — formula name, quantity, or unit — to filter instantly.

Displacement & Kinematics

Displacement (general)

x(t) = A cos(ωt + φ₀)

A = amplitude, ω = angular frequency, φ₀ = initial phase

Core

Displacement (sine form)

x(t) = A sin(ωt + φ₀)

Use when x=0 at t=0. Both forms are equivalent with different φ₀.

Core

Velocity (time)

v(t) = −Aω sin(ωt + φ₀)

Derivative of x(t). Maximum magnitude = Aω.

Kinematics

Velocity (position)

v = ω√(A² − x²)

v is max at x=0, zero at x=±A

Kinematics

Maximum velocity

v_max = Aω = A(2π/T)

Occurs at equilibrium position (x = 0)

NEET Key

Acceleration (time)

a(t) = −Aω² cos(ωt + φ₀) = −ω²x

Always directed toward equilibrium. a ∝ −x is the SHM condition.

Kinematics

Maximum acceleration

a_max = Aω²

Occurs at extreme positions (x = ±A)

NEET Key

Time Period & Frequency

Time Period (general)

T = 2π/ω

T in seconds, ω in rad/s

Core

Frequency

f = 1/T = ω/(2π)

f in Hz (cycles per second)

Core

Angular frequency

ω = 2πf = 2π/T = √(k/m)

Unit: rad/s. Not the same as frequency!

Core

Spring-mass time period

T = 2π√(m/k)

k = spring constant (N/m), m = mass (kg). Independent of g and A.

Core

Simple pendulum time period

T = 2π√(L/g)

Valid only for small angles (θ < 5°). Independent of mass m and amplitude.

Core

Pendulum — effective gravity

T = 2π√(L/g_eff) where g_eff = g ± a

+ for upward acceleration, − for downward. In free fall, T → ∞.

JEE Key

Springs in series

1/k_eff = 1/k₁ + 1/k₂ → k_eff = k₁k₂/(k₁+k₂)

Weaker combined spring. T increases.

System

Springs in parallel

k_eff = k₁ + k₂

Stiffer combined spring. T decreases.

System

Spring cut to n equal parts

k_new = n × k_original

Shorter spring → larger k. New T = T_original/√n

NEET Key

Energy in SHM

Total energy

E = ½mω²A² = ½kA²

Constant. Proportional to A². E ∝ ω²A²

Core

Kinetic energy at x

KE = ½mω²(A² − x²) = ½mv²

Max at x=0: KE_max = ½mω²A²

Core

Potential energy at x

PE = ½mω²x² = ½kx²

Min (=0) at x=0, Max at x=±A

Core

Position where KE = PE

x = ±A/√2 = ±0.707A

At this position, each = E/2 = ¼mω²A²

NEET Key

Energy frequency

Energy oscillates at 2f (twice the SHM frequency)

Both KE and PE complete 2 cycles per SHM period.

JEE Key

Force & Phase

Restoring force

F = −kx = −mω²x

Always directed toward equilibrium. Defining property of SHM.

Core

Phase difference: v leads x by

π/2 (90°)

When x is at max, v = 0. When x = 0, v is at max.

Phase

Phase difference: a leads v by

π/2 (90°)

Equivalently, acceleration is π (180°) out of phase with displacement.

Phase

Dimensional Analysis Verification

🔬 Exam Insight

CBSE boards often ask: "Verify the dimensional consistency of T = 2π√(L/g)." JEE uses dimensional analysis to check derived formulas. Master this process.

Verify: T = 2π√(m/k)

We need to show RHS has dimensions of [T] = [time] = T¹

[m] = M (mass)
[k] = Force/Length = MLT⁻²/L = MT⁻² (spring constant)
[m/k] = M / MT⁻² = T²
[√(m/k)] = T (seconds)
2π is dimensionless, so [T] = T ✓

Dimensional Formula of k[k] = MT⁻²

🎯 Memory Tip

Remember [k] = MT⁻². This is also the dimensional formula for surface tension — a fact that helps identify k in complex problems.

Verify: T = 2π√(L/g)

[L] = L (length)
[g] = acceleration = LT⁻²
[L/g] = L / LT⁻² = T²
[√(L/g)] = T ✓
2π dimensionless. RHS has dimensions of time. ✓

🔬 Derivation Check

This also tells you: Only L and g can affect T for a pendulum. Mass m and amplitude A do NOT have dimensions of time when combined with any expression, confirming T is independent of m and A.

Verify: E = ½mω²A²

[m] = M
[ω] = rad/s = T⁻¹ (radians are dimensionless)
[ω²] = T⁻²
[A²] = L² (amplitude squared)
[mω²A²] = M · T⁻² · L² = ML²T⁻² = [Energy] = [Joule] ✓

Dimensional Formula of Energy[E] = ML²T⁻²

Verify: ω = √(k/m)

[k] = MT⁻²
[m] = M
[k/m] = MT⁻² / M = T⁻²
[√(k/m)] = T⁻¹ = rad/s ✓

🧠 Thinking Step

ω has dimension T⁻¹. So does angular velocity in circular motion. This dimensional equivalence is why circular motion phasors work for SHM — they share the same angular frequency concept.

SI Units & Dimensional Formulae — Quick Reference

Quantity	Symbol	SI Unit	Dimensional Formula	Notes
Displacement	x	m	L	Can be negative
Amplitude	A	m	L	Always positive
Time Period	T	s	T	—
Frequency	f or ν	Hz = s⁻¹	T⁻¹	—
Angular Frequency	ω	rad/s	T⁻¹	ω = 2πf
Phase	φ	rad	Dimensionless	M⁰L⁰T⁰
Spring Constant	k	N/m	MT⁻²	Same as surface tension
Velocity in SHM	v	m/s	LT⁻¹	Max at x=0
Acceleration	a	m/s²	LT⁻²	Max at x=±A
Total Energy	E	J	ML²T⁻²	Constant in ideal SHM
Force constant	k_eff	N/m	MT⁻²	Effective k for combined springs
Restoring Force	F	N	MLT⁻²	F = −kx

❌ Common Mistake Alert

Students write the unit of ω as "Hz". Wrong. ω is in rad/s, NOT Hz. Frequency f is in Hz. ω = 2πf. If ω = 10 rad/s, then f = 10/2π ≈ 1.59 Hz. These are NOT equal.

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