Chapter 13 – Oscillations | AAI ATC Exam Prep

Topics Covered

Periodic & Oscillatory Motion Period & Frequency (T = 1/ν) Displacement in SHM SHM Equation: x = A cos(ωt + φ) Amplitude, Angular Frequency, Phase SHM & Uniform Circular Motion Velocity in SHM: v = −ωA sin(ωt + φ) Acceleration: a = −ω²x Force Law: F = −kx Spring-Mass System: T = 2π√(m/k) KE & PE in SHM Total Energy E = ½kA² Simple Pendulum: T = 2π√(L/g) Phase Difference Damped & Forced Oscillations

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Section A — Multiple Choice Questions

Q.1–25 · 1 mark each

The displacement of a particle in SHM is given by x(t) = A cos(ωt + φ). Here A represents: Easy

(A)Angular frequency

(B)Phase constant

(C)Amplitude (maximum displacement)

(D)Time period

Solution

✓ Correct: (C)

x(t) = A cos(ωt + φ) A = amplitude = maximum displacement from mean position ω = angular frequency (rad/s) φ = phase constant (rad) (ωt + φ) = phase at time t

The amplitude A is always positive and equals the maximum value of |x|. Since cosine varies between −1 and +1, displacement varies between −A and +A.

Relation between angular frequency ω, time period T and frequency ν is: Easy

(A)ω = T/2π

(B)ω = 2π/T = 2πν

(C)ω = 2π × T

(D)ω = ν/2π

Solution

✓ Correct: (B)

ω = 2π/T = 2πν T = 2π/ω, ν = 1/T = ω/2π SI unit of ω: rad/s

Angular frequency ω tells how many radians are swept per second. One complete oscillation = 2π radians. So in time T, ωT = 2π → ω = 2π/T.

The acceleration of a particle in SHM is related to its displacement x by: Easy

(A)a = +ω²x

(B)a = −ω²x

(C)a = ω²/x

(D)a = −ωx

Solution

✓ Correct: (B)

x(t) = A cos(ωt + φ) v(t) = dx/dt = −ωA sin(ωt + φ) a(t) = dv/dt = −ω²A cos(ωt + φ) = −ω²x(t)

The negative sign is crucial — acceleration is always directed opposite to displacement (towards the mean position). This is the defining property of SHM. Maximum acceleration = ω²A (at extremes).

The restoring force in SHM is given by F = −kx. The angular frequency ω of oscillation is: Moderate

(A)ω = √(k/m)

(B)ω = √(m/k)

(C)ω = km

(D)ω = k/m

Solution

✓ Correct: (A)

F = ma = −kx → a = −(k/m)x Comparing with a = −ω²x: ω² = k/m ω = √(k/m) T = 2π/ω = 2π√(m/k)

Stiffer spring (larger k) → higher ω → shorter period. Heavier mass (larger m) → lower ω → longer period. T is independent of amplitude A!

The time period of a simple pendulum of length L is: Easy

(A)T = 2π√(g/L)

(B)T = 2π√(L/g)

(C)T = π√(L/g)

(D)T = 2π(L/g)

Solution

✓ Correct: (B)

T = 2π√(L/g) Valid for small oscillations (θ < 15°) only Effective g = restoring force per unit mass per unit displacement

Period depends on L and g, not on the mass of the bob or amplitude (for small angles). On the Moon (g smaller), pendulum swings slower (T larger). A "seconds pendulum" has T = 2 s → L ≈ 1 m on Earth.

In SHM, velocity is maximum at: Easy

(A)Extreme positions (x = ±A)

(B)Mean position (x = 0)

(C)x = A/2

(D)Velocity is constant throughout

Solution

✓ Correct: (B)

v(t) = −ωA sin(ωt + φ) |v|_max = ωA (when sin = ±1, i.e., x = 0) |v|_min = 0 (when sin = 0, i.e., x = ±A) v as function of x: v = ω√(A² − x²)

At mean position: all energy is KE, PE = 0. At extremes: KE = 0, all energy is PE. This is consistent with energy conservation — KE and PE constantly interchange.

Total mechanical energy of a particle in SHM (amplitude A, spring constant k) is: Moderate

(A)E = kA

(B)E = ½kA²

(C)E = kA²

(D)E = ½mωA

Solution

✓ Correct: (B)

KE = ½mω²A² sin²(ωt + φ) = ½kA² sin²(ωt + φ) PE = ½kx² = ½kA² cos²(ωt + φ) E = KE + PE = ½kA²[sin² + cos²] = ½kA²

Total energy is constant (independent of time) and proportional to A². Doubling amplitude → 4× energy. This is true for any conservative force.

In SHM, the phase difference between displacement and velocity is: Moderate

(A)Zero

(B)π/4

(C)π/2

(D)π

Solution

✓ Correct: (C)

x(t) = A cos(ωt + φ) [displacement] v(t) = −ωA sin(ωt + φ) = ωA cos(ωt + φ + π/2) Phase difference (velocity leads displacement by π/2) Acceleration lags displacement by π (they are anti-phase)

Velocity leads displacement by π/2 (90°). Acceleration and displacement are in antiphase (π or 180°). This is why when displacement is maximum, velocity is zero and acceleration is maximum.

Which of the following correctly identifies SHM from the force-displacement relationship? Moderate

(A)F = +kx (force in direction of displacement)

(B)F = −kx (restoring force)

(C)F = −kx² (nonlinear)

(D)F = constant

Solution

✓ Correct: (B)

F = −kx is the SHM force law. Two key features: (1) F is proportional to displacement (linear), (2) F is directed towards mean position (negative sign). F = +kx would give unstable equilibrium (not oscillation). F = −kx² is a non-linear oscillator.

Q10

The projection of uniform circular motion (radius A, angular speed ω) on a diameter gives: Moderate

(A)Uniform motion

(B)Simple harmonic motion

(C)Projectile motion

(D)Non-periodic motion

Solution

✓ Correct: (B)

If particle P moves on a circle of radius A with angular speed ω, its x-projection P′ has position x(t) = A cos(ωt + φ) — exactly the SHM equation. This is the reference circle concept. The reference particle has speed ωA = maximum speed of P′ in SHM.

Q11

A body executes SHM with x = 5 cos[2πt + π/4] (SI units). Its angular frequency and time period are: Moderate

(A)ω = 2 rad/s, T = π s

(B)ω = 2π rad/s, T = 1 s

(C)ω = π rad/s, T = 2 s

(D)ω = 4π rad/s, T = 0.5 s

Solution

✓ Correct: (B)

x = 5 cos[2πt + π/4] Comparing with x = A cos(ωt + φ): A = 5 m, ω = 2π rad/s, φ = π/4 T = 2π/ω = 2π/2π = 1 s ν = 1/T = 1 Hz

The coefficient of t inside the cosine argument is ω. Here it's 2π, so T = 1 s. At t = 1.5 s, displacement = 5 cos(3π + π/4) = −5 × cos(π/4) = −3.535 m.

Q12

In SHM, the potential energy (PE) is maximum when: Easy

(A)Displacement is zero (at mean position)

(B)Velocity is maximum

(C)Displacement is maximum (at extreme)

(D)PE is always constant

Solution

✓ Correct: (C)

PE = ½kx² = ½kA²cos²(ωt + φ) PE_max = ½kA² (when x = ±A, displacement maximum) PE_min = 0 (when x = 0, mean position) KE = ½mv² = ½kA²sin²(ωt + φ) KE + PE = ½kA² = constant (Total Energy)

Q13

If the amplitude of SHM is doubled, the total energy becomes: Moderate

(A)Doubled

(B)Halved

(D)Unchanged

Solution

✓ Correct: (C)

E = ½kA² ∝ A² If A → 2A: E_new = ½k(2A)² = ½k(4A²) = 4 × ½kA² = 4E

Energy is proportional to A². Doubling amplitude quadruples energy. Note: period T = 2π√(m/k) is independent of amplitude — a key feature of SHM not shared by most other oscillations.

Q14

Two springs of constants k₁ and k₂ are connected in series. The equivalent spring constant is: Moderate

(A)k = k₁ + k₂

(B)k = k₁k₂/(k₁ + k₂)

(C)k = (k₁ + k₂)/2

(D)k = √(k₁k₂)

Solution

✓ Correct: (B)

Series: 1/k = 1/k₁ + 1/k₂ → k = k₁k₂/(k₁+k₂) [k < k₁, k₂] Parallel: k = k₁ + k₂ [k > k₁, k₂] Two identical springs k in series: k_eq = k/2 → T = 2π√(2m/k) Two identical springs k in parallel: k_eq = 2k → T = 2π√(m/2k)

Q15

The period of a simple pendulum is independent of: Easy

(A)Length of pendulum

(B)Value of g

(C)Mass of the bob

(D)Latitude of location

Solution

✓ Correct: (C)

T = 2π√(L/g) Independent of: mass of bob (m), amplitude (for small θ) Depends on: L (length), g (gravity at that location)

Mass does not appear in T = 2π√(L/g). This is analogous to all objects falling with same acceleration in gravity regardless of mass. T depends on latitude because g varies slightly (larger at poles).

Q16

The velocity of a particle in SHM at displacement x from mean position is: Moderate

(A)v = ω√(A² + x²)

(B)v = ω√(A² − x²)

(C)v = ω(A − x)

(D)v = ωA

Solution

✓ Correct: (B)

From energy conservation: ½mv² + ½kx² = ½kA² ½mv² = ½k(A² − x²) v² = (k/m)(A² − x²) = ω²(A² − x²) v = ω√(A² − x²) At x = 0: v_max = ωA ✓ At x = ±A: v = 0 ✓

Q17

Which function of time does NOT represent SHM? Moderate

(A)A sin(ωt + φ)

(B)A cos(ωt) + B sin(ωt)

(C)sin(ωt) + sin(2ωt)

(D)A sin(ωt − π/4)

Solution

✓ Correct: (C)

sin(ωt) + sin(2ωt) — Two different frequencies! This is PERIODIC (with period 2π/ω) but NOT simple harmonic. For SHM, there must be ONE single sinusoidal component. A cos + B sin = D sin(ωt + φ) ← this IS SHM (same ω)

Option (B) can be written as D sin(ωt + φ) — still SHM. Option (C) has TWO different angular frequencies ω and 2ω — it's periodic but not SHM. SHM requires a single frequency sinusoidal function.

Q18

Two identical springs (each k) are attached to a block of mass m on both sides (one to wall on each side). When displaced and released, the time period is: Moderate

(A)T = 2π√(m/k)

(B)T = 2π√(m/2k)

(C)T = 2π√(2m/k)

(D)T = π√(m/k)

Solution

✓ Correct: (B)

When mass displaced by x: Left spring: F₁ = −kx (pulls back) Right spring: F₂ = −kx (pushes back) Net F = −2kx → effective k_eff = 2k T = 2π√(m/k_eff) = 2π√(m/2k)

Both springs act as parallel springs, so k_eff = 2k. The period is shorter than a single spring — the system is stiffer.

Q19

If the length of a simple pendulum is increased 4 times, the time period: Moderate

(A)Doubles

(B)Becomes 4 times

(C)Becomes half

(D)Remains same

Solution

✓ Correct: (A)

T = 2π√(L/g) ∝ √L If L → 4L: T_new = 2π√(4L/g) = 2 × 2π√(L/g) = 2T

T ∝ √L, so quadrupling L only doubles T. To halve T, you need to reduce L to L/4. This square-root relationship is important for AAI exam.

Q20

The kinetic energy of a particle in SHM at displacement x from mean position is: Moderate

(A)KE = ½k(A² + x²)

(B)KE = ½kA²

(C)KE = ½k(A² − x²)

(D)KE = ½kx²

Solution

✓ Correct: (C)

KE = ½mv² = ½m × ω²(A² − x²) = ½k(A² − x²) PE = ½kx² KE + PE = ½kA² = E (constant) ✓

KE is maximum (= ½kA²) at x = 0 and zero at x = ±A. PE is zero at x = 0 and maximum (= ½kA²) at x = ±A. Together they always add to E = ½kA².

Q21

The pendulum clock loses or gains time when taken to a mountain because: Moderate

(A)Temperature changes at altitude

(B)g decreases at higher altitudes → T increases → clock loses time

(C)g increases at higher altitudes → T decreases → clock gains time

(D)Mass of the bob changes

Solution

✓ Correct: (B)

T = 2π√(L/g) At mountain: g decreases (farther from Earth's centre) → T increases (each oscillation takes longer) → Clock loses time (fewer oscillations per day)

On Moon (g ≈ 1.63 m/s²), T would be √(9.8/1.63) ≈ 2.45 times longer than on Earth. At Earth's poles, g is slightly larger → pendulum oscillates slightly faster → clock gains time compared to equator.

Q22

The maximum acceleration in SHM of amplitude A and angular frequency ω is: Easy

(A)a_max = ωA

(B)a_max = ω²A

(C)a_max = ω/A

(D)a_max = A/ω²

Solution

✓ Correct: (B)

a(t) = −ω²A cos(ωt + φ) |a|_max = ω²A (at x = ±A, extreme positions) |a|_min = 0 (at x = 0, mean position) v_max = ωA, a_max = ω²A = ω × v_max

Q23

A bob of a pendulum oscillating with small amplitude in a car. If the car accelerates forward, the effective time period: Hard

(A)Increases

(B)Decreases

(C)Remains same

(D)Becomes zero

Solution

✓ Correct: (B)

In accelerating car (acceleration a forward): Effective g = √(g² + a²) > g T = 2π√(L/g_eff) Since g_eff > g → T_new < T (period decreases)

The resultant of gravity g (down) and pseudo force a (backward) gives a larger effective gravity at an angle. A larger effective g means shorter period. Similarly, car braking → effective g is also larger → T decreases.

Q24

Energy of both KE and PE in SHM varies with period: Moderate

(A)T (same as displacement)

(B)T/2 (half the period)

(D)Energy doesn't vary

Solution

✓ Correct: (B)

KE = ½kA²sin²(ωt) varies with period T/2 PE = ½kA²cos²(ωt) varies with period T/2 Both sin² and cos² complete one cycle in π, i.e., in T/2

KE and PE both pass through their maximum and minimum values twice per complete oscillation. So they repeat with period T/2. Only total energy E = ½kA² is truly constant.

Q25

Simple harmonic motion of a simple pendulum is valid only for: Easy

(A)All angular displacements

(B)Large angular displacements (θ > 45°)

(C)Small angular displacements (θ < ~15°)

(D)Only when θ = 90°

Solution

✓ Correct: (C)

For SHM: we need sinθ ≈ θ (in radians) At θ = 5°: sinθ = 0.087, θ = 0.087 → error ~0% At θ = 15°: sinθ = 0.259, θ = 0.262 → error ~1.2% At θ = 20°: sinθ = 0.342, θ = 0.349 → error ~2% At θ = 45°: sinθ = 0.707, θ = 0.785 → error ~11% (too large!)

The approximation sinθ ≈ θ breaks down for large angles. For θ < 15°, error is less than 1% — acceptable for SHM. Beyond 15–20°, motion is still periodic but not simple harmonic.

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Section B — Numerical Problems

N1–N15 · 3–5 marks each

A human heart beats 75 times per minute. Find its (a) frequency and (b) time period. Easy

Solution

Beats = 75 per minute = 75/60 per second (a) Frequency ν = 75/60 = 1.25 Hz = 1.25 s⁻¹ (b) Time period T = 1/ν = 1/1.25 = 0.8 s

The heart beats 1.25 times per second on average. A healthy resting heart rate is 60–100 bpm. This corresponds to periods of 0.6–1.0 s.

A body oscillates with SHM: x = 5 cos(2πt + π/4) m. At t = 1.5 s, find (a) displacement, (b) velocity, (c) acceleration. Moderate

Solution

ω = 2π rad/s, A = 5 m, T = 1 s At t = 1.5 s: argument = 2π(1.5) + π/4 = 3π + π/4 (a) x = 5 cos(3π + π/4) = 5 × cos(π + 2π + π/4) = 5 × (−cos π/4) = −5 × (1/√2) = −5 × 0.707 = −3.535 m (b) v = −ωA sin(3π + π/4) = −(2π)(5) sin(3π + π/4) = −10π × (−sin π/4) = 10π × 0.707 = 22.2 m/s (c) a = −ω²x = −(2π)²(−3.535) = 4π² × 3.535 = 39.48 × 3.535 ≈ 140 m/s²

A spring of constant k = 1200 N/m has a mass of 3 kg attached. Mass is pulled 2.0 cm from equilibrium and released. Find (a) frequency, (b) maximum acceleration, (c) maximum speed. Moderate

Solution

k = 1200 N/m, m = 3 kg, A = 2.0 cm = 0.02 m ω = √(k/m) = √(1200/3) = √400 = 20 rad/s (a) ν = ω/2π = 20/2π = 20/6.283 ≈ 3.18 Hz T = 1/ν ≈ 0.314 s (b) a_max = ω²A = (20)²(0.02) = 400 × 0.02 = 8 m/s² (c) v_max = ωA = 20 × 0.02 = 0.4 m/s

A block (m = 1 kg) on spring (k = 50 N/m) is pulled to x = 10 cm and released. Find KE, PE and total energy when x = 5 cm. Moderate

Solution

k = 50 N/m, m = 1 kg, A = 10 cm = 0.10 m, x = 5 cm = 0.05 m ω = √(k/m) = √50 = 7.07 rad/s Total energy: E = ½kA² = ½ × 50 × (0.10)² = 0.25 J PE at x = 5 cm: U = ½kx² = ½ × 50 × (0.05)² = 0.0625 J KE at x = 5 cm: K = E − U = 0.25 − 0.0625 = 0.1875 J ≈ 0.19 J Verification: v = ω√(A²−x²) = 7.07√(0.01−0.0025) = 7.07 × √0.0075 = 7.07 × 0.0866 = 0.612 m/s KE = ½mv² = ½ × 1 × (0.612)² = 0.187 J ✓

Find the length of a simple pendulum that ticks seconds (T = 2 s). Take g = 9.8 m/s². Easy

Solution

T = 2π√(L/g) → T² = 4π²L/g → L = gT²/4π² L = (9.8 × 2²) / (4π²) = (9.8 × 4) / (4 × 9.87) = 39.2 / 39.48 = 0.993 m ≈ 1 m

A "seconds pendulum" (T = 2 s) has length approximately 1 m on Earth. Galileo is said to have measured pendulum periods using his heartbeat!

A pendulum with T = 3.5 s on Earth. Find its period on the Moon where g_moon = 1.7 m/s² (g_earth = 9.8 m/s²). Moderate

Solution

T = 2π√(L/g) → T ∝ 1/√g (L same on Moon) T_moon/T_earth = √(g_earth/g_moon) = √(9.8/1.7) = √5.765 = 2.401 T_moon = T_earth × 2.401 = 3.5 × 2.401 ≈ 8.4 s

The pendulum swings 2.4× slower on the Moon. A clock using this pendulum would run much slower — losing significant time compared to an Earth-calibrated clock.

A spring balance reads 0–50 kg over 20 cm. A body suspended from it oscillates with T = 0.6 s. Find the weight of the body. (g = 9.8 m/s²) Moderate

Solution

Spring constant: F = kx → k = F/x At max reading: F = 50 × 9.8 = 490 N, x = 0.20 m k = 490/0.20 = 2450 N/m From T = 2π√(m/k): T² = 4π²m/k → m = kT²/4π² m = 2450 × (0.6)² / (4π²) = 2450 × 0.36 / 39.48 = 882 / 39.48 = 22.34 kg Weight = mg = 22.34 × 9.8 ≈ 219 N

A particle executes SHM with amplitude 8 cm and period 6 s. Find its (a) maximum speed, (b) speed when displacement is 4 cm, (c) maximum acceleration. Moderate

Solution

A = 8 cm = 0.08 m, T = 6 s ω = 2π/T = 2π/6 = π/3 rad/s ≈ 1.047 rad/s (a) v_max = ωA = (π/3)(0.08) = 0.0838 m/s ≈ 8.38 cm/s (b) At x = 4 cm = 0.04 m: v = ω√(A² − x²) = (π/3)√(0.08² − 0.04²) = (π/3)√(0.0064 − 0.0016) = (π/3)√0.0048 = (π/3)(0.0693) = 1.047 × 0.0693 = 0.0726 m/s ≈ 7.26 cm/s (c) a_max = ω²A = (π/3)²(0.08) = (1.097)(0.08) = 0.0877 m/s²

A particle starts SHM at t = 0 from x = 0 moving towards positive x direction. Write x(t) if amplitude = 10 cm and ω = π rad/s. Find x at t = 1/3 s. Moderate

Solution

At t = 0: x = 0, v > 0 (moving in +x direction) Using x = A sin(ωt + φ): At t = 0: x = 0 → A sin(φ) = 0 → φ = 0 (since v > 0) v = ωA cos(ωt + φ) at t = 0: v = ωA cos(0) = ωA > 0 ✓ So: x(t) = 10 sin(πt) cm At t = 1/3 s: x = 10 sin(π/3) = 10 × (√3/2) = 10 × 0.866 = 8.66 cm

When starting from mean position with positive velocity, use sine (not cosine). Starting from maximum positive displacement uses cosine with φ = 0.

N10

A locomotive piston has stroke of 1.0 m (stroke = 2A) and angular frequency 200 rad/min. Find its maximum speed. Easy

Solution

Stroke = 2A = 1.0 m → A = 0.5 m ω = 200 rad/min = 200/60 rad/s = 10/3 rad/s v_max = ωA = (10/3)(0.5) = 5/3 m/s ≈ 1.67 m/s

Always convert angular frequency to rad/s before calculating. The maximum speed of the piston is about 1.67 m/s ≈ 6 km/h.

N11

A particle in SHM has velocity v₁ = 4 cm/s when displacement x₁ = 3 cm, and velocity v₂ = 3 cm/s when x₂ = 4 cm. Find amplitude and time period. Hard

Solution

Using v² = ω²(A² − x²): v₁² = ω²(A² − x₁²): 16 = ω²(A² − 9) ... (i) v₂² = ω²(A² − x₂²): 9 = ω²(A² − 16) ... (ii) Dividing (i) by (ii): 16/9 = (A² − 9)/(A² − 16) 16(A² − 16) = 9(A² − 9) 16A² − 256 = 9A² − 81 7A² = 175 A² = 25 → A = 5 cm From (i): ω² = 16/(A² − 9) = 16/(25 − 9) = 16/16 = 1 ω = 1 rad/s T = 2π/ω = 2π s ≈ 6.28 s

N12

A spring-mass system has m = 0.5 kg and k = 200 N/m. Find (a) time period, (b) total energy if amplitude = 5 cm, (c) velocity at x = 3 cm. Moderate

Solution

m = 0.5 kg, k = 200 N/m, A = 5 cm = 0.05 m, x = 3 cm = 0.03 m (a) ω = √(k/m) = √(200/0.5) = √400 = 20 rad/s T = 2π/ω = 2π/20 = π/10 s ≈ 0.314 s (b) E = ½kA² = ½ × 200 × (0.05)² = ½ × 200 × 0.0025 = 0.25 J (c) v = ω√(A² − x²) = 20 × √(0.05² − 0.03²) = 20 × √(0.0025 − 0.0009) = 20 × √0.0016 = 20 × 0.04 = 0.8 m/s

N13

A particle of mass 0.2 kg executes SHM with amplitude 0.1 m and frequency 5 Hz. Find (a) maximum restoring force, (b) total energy. Moderate

Solution

m = 0.2 kg, A = 0.1 m, ν = 5 Hz ω = 2πν = 2π × 5 = 10π rad/s k = mω² = 0.2 × (10π)² = 0.2 × 100π² = 20π² N/m (a) F_max = kA = 20π² × 0.1 = 2π² N ≈ 19.74 N (Also: F_max = mω²A = 0.2 × (10π)² × 0.1 = 2π² N) (b) E = ½kA² = ½ × 20π² × (0.1)² = ½ × 20π² × 0.01 = π²/10 J ≈ 0.987 J ≈ 1 J

N14

Two springs (k₁ = 3 N/m, k₂ = 6 N/m) are connected in series with a 2 kg block. Find (a) equivalent spring constant, (b) time period. Moderate

Solution

k₁ = 3 N/m, k₂ = 6 N/m, m = 2 kg (a) Series: 1/k_eq = 1/k₁ + 1/k₂ = 1/3 + 1/6 = 2/6 + 1/6 = 3/6 = 1/2 k_eq = 2 N/m (b) T = 2π√(m/k_eq) = 2π√(2/2) = 2π√1 = 2π s ≈ 6.28 s

Springs in series are weaker than either individual spring (k_eq = 2 < 3). This results in a longer period than either spring alone would give.

N15

A pendulum of length 1 m is taken inside a lift that accelerates upward at 5 m/s². Find the new time period. (g = 10 m/s²) Moderate

Solution

L = 1 m, g = 10 m/s², a_lift = 5 m/s² (upward) In upward accelerating lift: g_eff = g + a = 10 + 5 = 15 m/s² T = 2π√(L/g_eff) = 2π√(1/15) = 2π/√15 = 2π/3.873 = 6.283/3.873 ≈ 1.62 s Compare with Earth: T_earth = 2π√(1/10) = 2π/3.162 ≈ 1.99 s Period decreased because effective g increased.

In an upward-accelerating lift, the pendulum oscillates faster. In free fall (a = g downward), g_eff = 0 → T = ∞ (pendulum doesn't oscillate at all — weightlessness!).