Oscillatory motion underlies sound waves, vibrations in aircraft structures, AC electrical systems, and radar. The principles of SHM — period, frequency, amplitude, restoring force, and energy — are fundamental to understanding all wave phenomena. Expect 3–5 direct questions from this chapter, especially on SHM equations, spring-mass systems, and simple pendulum.
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Section 13.2
Periodic & Oscillatory Motion
Periodic motion: Any motion that repeats itself at regular intervals of time. The smallest such interval is the period T (SI unit: second).
Oscillatory motion: Periodic to-and-fro motion about a mean equilibrium position. Every oscillatory motion is periodic, but not every periodic motion is oscillatory (e.g., circular motion is periodic but not oscillatory).
ν = 1/T ω = 2π/T = 2πν
ν = frequency (Hz = s⁻¹) | T = period (s) | ω = angular frequency (rad/s)
Simple Harmonic Motion (SHM) is the simplest form — restoring force is directly proportional to displacement and directed towards the mean position.
📌 Periodic Functions — Key Facts
sin ωt, cos ωt — periodic with period 2π/ω (SHM)
sin ωt + cos ωt = √2 sin(ωt + π/4) — periodic with T = 2π/ω (SHM)
e^(−ωt) — NOT periodic (decreases monotonically to zero)
log(ωt) — NOT periodic (increases without bound)
sin²ωt = ½ − ½cos(2ωt) — periodic with T = π/ω (NOT SHM, centre at ½ not 0)
🎯 Practice MCQs — Periodic & Oscillatory Motion
AAI ATC Style
1 A human heart beats 75 times per minute. Its time period is:
A 0.5 s
B 0.8 s
C 1.25 s
D 75 s
✅ ν = 75/60 = 1.25 Hz. T = 1/ν = 1/1.25 = 0.8 s. Frequency = beats per second = 75/60 = 1.25 Hz. Period = time for one complete beat = 0.8 s.
2 Which of the following functions represents a non-periodic motion?
A sin ωt + cos ωt
B sin ωt + cos 2ωt + sin 4ωt
C e^(−ωt)
D sin²ωt
✅ e^(−ωt) decreases monotonically from 1 to 0 as t increases — it never repeats its value. The other three options are all periodic: sin ωt + cos ωt has T = 2π/ω; the sum of three trig functions has T = 2π/ω; and sin²ωt has T = π/ω.
3 The angular frequency of a particle performing SHM with time period T = 0.5 s is:
A 0.5 rad/s
B 2π rad/s
C 4π rad/s
D π rad/s
✅ ω = 2π/T = 2π/0.5 = 4π rad/s. Angular frequency = 2π × frequency = 2π/T. For T = 0.5 s, ν = 2 Hz, so ω = 2π × 2 = 4π rad/s.
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Section 13.3 & 13.4
Simple Harmonic Motion
In SHM, displacement varies sinusoidally with time:
SHM can be viewed as the projection of uniform circular motion on a diameter. A particle moving in a circle of radius A with angular speed ω → its x-projection executes SHM with amplitude A.
Quantity
Symbol
Formula
Range
Amplitude
A
Max displacement from mean
Always +ve
Angular frequency
ω
2π/T = 2πν
rad/s
Phase
ωt + φ
Time-dependent argument
Any value
Phase constant
φ
Phase at t = 0
Depends on initial conditions
Period
T
2π/ω
Always +ve, in seconds
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Insert Diagram: x vs t graph for SHM — sinusoidal wave with amplitude A, period T
Recommended: Fig 13.5 from NCERT — Displacement as a continuous function of time in SHM
🎯 Practice MCQs — SHM Basics
AAI ATC Style
1 A body oscillates with SHM: x = 5 cos(2πt + π/4) m. At t = 1.5 s, the displacement is:
✅ a = −10x. For SHM: a ∝ −x (acceleration proportional to displacement, directed towards mean position — negative sign essential). Option (A) has positive sign (wrong direction). Options (B) and (D) are proportional to x² and x³ (not linear in x), so they are not SHM.
3 The function sin ωt − cos ωt represents:
A SHM with period 2π/ω and amplitude √2
B Periodic but not SHM, period π/ω
C Non-periodic motion
D SHM with period π/ω and amplitude 1
✅ sin ωt − cos ωt = √2 sin(ωt − π/4). This is SHM with amplitude √2 and period 2π/ω. Phase constant = −π/4. The conversion uses: A sin θ + B cos θ = √(A²+B²) sin(θ + φ) where tan φ = B/A.
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Section 13.5
Velocity & Acceleration in SHM
For x(t) = A cos(ωt + φ):
v(t) = −ωA sin(ωt + φ)
Velocity amplitude = ωA | Maximum velocity at x = 0 (mean position) | Zero at x = ±A (extremes)
a(t) = −ω²A cos(ωt + φ) = −ω²x(t)
Acceleration amplitude = ω²A | Maximum at x = ±A (extremes) | Zero at x = 0 (mean position)
Velocity leads displacement by phase π/2. Acceleration leads displacement by phase π (opposite direction).
🔑 Phase Relations in SHM: Displacement → Velocity (phase leads by π/2) → Acceleration (phase leads by another π/2 = total π from displacement). At mean position: velocity is max, acceleration is zero. At extreme: velocity is zero, acceleration is maximum (directed back towards mean).
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Insert Diagram: x, v, a vs time graphs — showing phase relationships (v leads x by π/2, a leads x by π)
Recommended: Fig 13.13 from NCERT — Displacement, velocity and acceleration in SHM
🎯 Practice MCQs — Velocity & Acceleration in SHM
AAI ATC Style
1 A body executes SHM: x = 5 cos(2πt + π/4) m. At t = 1.5 s, the speed of the body is approximately:
3 For the same spring-mass system (k = 1200 N/m, m = 3 kg, A = 2 cm), the maximum speed of the mass is:
A 0.2 m/s
B 0.3 m/s
C 0.4 m/s
D 0.8 m/s
✅ v_max = ωA = 20 × 0.02 = 0.4 m/s. Maximum speed occurs at the mean position (x = 0). v_max = ωA. Frequency = ω/2π = 20/2π ≈ 3.18 Hz.
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Section 13.6
Force Law for SHM — Spring-Mass System
The restoring force in SHM is always proportional to displacement and directed towards the mean position:
F = −kx → ω = √(k/m) → T = 2π√(m/k)
k = spring/force constant (N/m) | m = mass (kg) | T = period (s) | ω = angular frequency (rad/s)
The negative sign means force is always directed towards the mean position (restoring force). This is Hooke's Law applied to oscillations.
For two identical springs (constant k each) attached to both sides of a mass: Effective k = 2k → T = 2π√(m/2k).
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Insert Diagram: Block between two springs — showing restoring forces F₁ and F₂ when displaced by x
Recommended: Fig 13.14 & 13.15 from NCERT — Block with two springs
🎯 Practice MCQs — Force Law & Spring-Mass System
AAI ATC Style
1 A spring balance reads 0–50 kg over 20 cm. A body suspended from it oscillates with T = 0.6 s. The weight of the body is: (g = 10 m/s²)
A 100 N
B 200 N
C 219.8 N ≈ 220 N
D 500 N
✅ k = F/x = 50×10/0.2 = 2500 N/m. T = 2π√(m/k) → 0.6 = 2π√(m/2500) → m = (0.6/2π)² × 2500 = 22.8 kg. W = mg = 22.8 × 10 ≈ 228 N ≈ 220 N. Spring constant = max load × g / scale length.
2 Two identical springs (each with spring constant k) are attached to both sides of a mass m. The period of oscillation is:
A 2π√(m/k)
B 2π√(m/2k)
C 2π√(2m/k)
D π√(m/k)
✅ When mass is displaced by x, both springs exert restoring force: F = −kx − kx = −2kx. Effective spring constant = 2k. T = 2π√(m/2k). The period is reduced (T smaller) compared to a single spring, because the effective restoring force is doubled.
3 A piston in a locomotive moves with SHM with angular frequency 200 rad/min and stroke (= 2A) = 1.0 m. The maximum speed is:
A 100 m/min
B 50 m/min
C 100 m/min
D 200 m/min
✅ A = stroke/2 = 1.0/2 = 0.5 m. v_max = ωA = 200 × 0.5 = 100 m/min. Maximum speed of SHM occurs at the mean position. v_max = ωA. Here ω = 200 rad/min, A = 0.5 m.
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Section 13.7
Energy in Simple Harmonic Motion
KE = ½mω²A²sin²(ωt+φ) = ½k(A²−x²)
Maximum KE at x = 0 (mean position) = ½kA² | Zero KE at x = ±A (extremes)
PE = ½kx² = ½kA²cos²(ωt+φ)
Maximum PE at x = ±A (extremes) = ½kA² | Zero PE at x = 0 (mean position)
E = KE + PE = ½kA² = ½mω²A²
Total energy is CONSTANT — independent of time and position | Depends only on amplitude A
🔑 Energy Key Facts: Both KE and PE vary with period T/2 (twice the SHM frequency). Total energy E = ½kA² is constant — does NOT depend on time, position, velocity, or phase. KE and PE interchange energy between them. At x = 0: all KE. At x = ±A: all PE. At x = A/2: KE = (3/4)E, PE = (1/4)E.
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Insert Diagram: KE, PE, and Total energy vs displacement — showing KE+PE = constant E
Recommended: Fig 13.16(b) from NCERT — Energy vs displacement in SHM
🎯 Practice MCQs — Energy in SHM
AAI ATC Style
1 A block (m = 1 kg, k = 50 N/m) is pulled 10 cm from equilibrium and released. When the block is 5 cm from mean position, the KE is:
A 0.25 J (all energy)
B 0.0625 J
C 0.1875 J
D 0 J
✅ Total E = ½kA² = ½×50×0.01 = 0.25 J. PE at x=5cm = ½×50×0.0025 = 0.0625 J. KE = E − PE = 0.25 − 0.0625 = 0.1875 J. Or directly: KE = ½k(A²−x²) = ½×50×(0.01−0.0025) = 25×0.0075 = 0.1875 J.
2 If the amplitude of a SHM is doubled, the total energy becomes:
A Doubled
B Halved
C Four times
D Same
✅ E = ½kA² → E ∝ A². If A → 2A, then E → ½k(2A)² = 4×(½kA²) = 4E. Total energy becomes four times. This is why larger amplitude vibrations in aircraft structures carry significantly more energy.
3 At what displacement from the mean position is the KE equal to the PE in SHM?
A x = A
B x = A/2
C x = A/√2
D x = A√2
✅ KE = PE → ½k(A²−x²) = ½kx² → A²−x² = x² → A² = 2x² → x = A/√2. At this position, KE = PE = E/2 = ¼kA². This occurs when x = 0.707A, i.e., about 70.7% of amplitude from mean position.
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Section 13.8
The Simple Pendulum
A simple pendulum consists of a small bob (mass m) tied to a massless inextensible string of length L. For small angular displacements (θ ≤ 20°, where sinθ ≈ θ in radians), the restoring torque gives SHM:
T = 2π√(L/g)
T = Period | L = Length of pendulum (m) | g = acceleration due to gravity (m/s²)
The restoring force is the tangential component: F = −mg sinθ ≈ −mgθ (for small θ). The period does NOT depend on mass m or amplitude (for small oscillations).
📌 Key Facts — Simple Pendulum
Valid only for small angular displacements (θ ≤ 20°).
Period increases if g decreases (e.g., on Moon: g_moon = 1.7 m/s², T increases).
Period is independent of mass of the bob and amplitude (for small θ).
A seconds pendulum (T = 2 s) has length L = gT²/4π² = 9.8×4/(4π²) ≈ 1 m.
For a pendulum in a car/elevator: effective g changes if the vehicle accelerates.
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Insert Diagram: Simple pendulum showing restoring torque mg sinθ and centripetal force T − mg cosθ
Recommended: Fig 13.17 from NCERT — Bob oscillating about mean position with force diagram
🎯 Practice MCQs — Simple Pendulum
AAI ATC Style
1 The time period of a simple pendulum on the surface of Moon is: (T_earth = 3.5 s, g_earth = 9.8 m/s², g_moon = 1.7 m/s²)
A 3.5 s
B 5.0 s
C 8.4 s
D 14.2 s
✅ T_moon/T_earth = √(g_earth/g_moon) = √(9.8/1.7) = √5.76 = 2.4. T_moon = 3.5 × 2.4 ≈ 8.4 s. Since T ∝ 1/√g, lower gravity on Moon means longer period. Pendulum clocks run slower on the Moon.
2 A simple pendulum ticks seconds (T = 2 s). Its length is: (g = 9.8 m/s²)
A 0.5 m
B 0.25 m
C 1.0 m
D 2.0 m
✅ L = gT²/4π² = 9.8×4/(4π²) = 9.8/π² ≈ 9.8/9.87 ≈ 1.0 m. This is the standard seconds pendulum — exactly 1 m long for T = 2 s at standard gravity. Historical clocks used this as their reference.
3 The motion of a simple pendulum is simple harmonic only when:
A The length of the string is very large
B The mass of the bob is small
C The angular displacement is small (sinθ ≈ θ)
D There is no damping
✅ SHM condition requires sinθ ≈ θ, i.e., small angular displacement (θ ≤ 20°). For large angles, the restoring torque is −mgL sinθ, which is NOT proportional to θ (not linear), so the motion is not SHM. Period also depends on amplitude for large oscillations.
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