Original practice sets for Simple Harmonic Motion are being prepared. For now, use the chapter notes for concept mastery and revision.
Start with short chapter-wise drills or take the full chapter bank in one run. These questions are original and aligned to JEE-style concept checks and numerical thinking.
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1. SHM is defined as motion in which the restoring force is:
Explanation: F = −kx. The restoring force is proportional to displacement and always directed toward the mean position.
2. The acceleration in SHM is given by:
Explanation: a = F/m = −kx/m = −ω²x where ω = √(k/m). Negative sign shows acceleration opposes displacement.
3. Time period of a mass-spring system (mass m, spring constant k) is:
Explanation: T = 2π/ω = 2π√(m/k). Heavier mass or weaker spring means longer period.
4. Time period of a simple pendulum of length L on Earth (g) is:
Explanation: T = 2π√(L/g). Valid for small angles (sin θ ≈ θ).
5. Standard displacement equation for SHM is:
Explanation: x = A sin(ωt + φ) (or A cos(ωt + φ)), where A is amplitude, ω is angular frequency, φ is initial phase.
6. Velocity in SHM at displacement x is:
Explanation: Energy conservation: ½mv² + ½kx² = ½kA². → v = ω√(A² − x²).
7. Maximum velocity in SHM occurs at:
Explanation: v_max = ωA at x = 0. All energy is kinetic at the mean position.
8. Maximum acceleration in SHM occurs at:
Explanation: |a| = ω²|x|. Maximum acceleration = ω²A at extreme positions where displacement is maximum.
9. Angular frequency ω is related to time period T by:
Explanation: ω = 2πf = 2π/T. Units: rad/s.
10. Phase difference between displacement and velocity in SHM is:
Explanation: x = A sin(ωt), v = Aω cos(ωt) = Aω sin(ωt + π/2). Velocity leads displacement by π/2.
11. A spring of constant k is cut into n equal pieces. Spring constant of each piece is:
Explanation: Spring constant ∝ 1/length. Cutting into n pieces makes each n times shorter, so each piece has spring constant nk.
12. Two springs (k₁ and k₂) are in series. Effective spring constant is:
Explanation: For springs in series: 1/k_eff = 1/k₁ + 1/k₂ → k_eff = k₁k₂/(k₁+k₂).
13. Two springs (k₁ and k₂) in parallel. Effective spring constant:
Explanation: For springs in parallel: k_eff = k₁ + k₂ (both stretch/compress by same amount, forces add).
14. A pendulum of length 1 m is taken to moon where g_moon = g/6. New time period (T_earth) becomes:
Explanation: T ∝ 1/√g. T_moon = T_earth × √(g/g_moon) = T_earth × √6.
15. At displacement x = A/2, speed as fraction of maximum speed is:
Explanation: v = ω√(A² − x²) = ω√(A² − A²/4) = ωA√(3/4) = v_max × (√3/2).
16. A particle in SHM has time period 2 s and amplitude 5 cm. Maximum acceleration is:
Explanation: ω = 2π/T = π rad/s. a_max = ω²A = π² × 5 = 5π² cm/s².
17. Two SHM particles have same frequency but phase difference π. When particle A is at x = +A, particle B is at:
Explanation: Phase difference π means when A is at +A, B is at −A. They are always at opposite extremes simultaneously.
18. Time period of simple pendulum does NOT depend on:
Explanation: T = 2π√(L/g) depends on L and g only. For small angles, T is independent of mass and amplitude (isochronous property).
19. A particle starts from mean position (x = 0) moving in positive direction. Displacement equation is:
Explanation: At t = 0: x = 0 (mean position). So x = A sin(ωt) (starts at 0, moves positive). x = A cos(ωt) would start at x = A.
20. A mass on a spring oscillates with period T. If mass is quadrupled, new period is:
Explanation: T ∝ √m. If m → 4m, T → 2T.
21. Potential energy in SHM at displacement x is:
Explanation: PE = ½kx² = ½mω²x². It is maximum at extreme positions (x = ±A) and zero at mean position.
22. Kinetic energy in SHM is maximum at:
Explanation: KE = ½mv² = ½mω²(A²−x²). Maximum when x = 0 (mean position): KE_max = ½mω²A².
23. Total mechanical energy in SHM is:
Explanation: E_total = KE + PE = ½mω²A² = ½kA². This is constant (independent of position in ideal SHM).
24. At what displacement is KE = PE in SHM?
Explanation: KE = PE: ½mω²(A²−x²) = ½mω²x² → A² − x² = x² → x = A/√2.
25. In SHM, total energy is proportional to:
Explanation: E = ½mω²A². Total energy is proportional to square of amplitude.
26. If amplitude of SHM doubles, total energy becomes:
Explanation: E ∝ A². If A → 2A, E → 4E.
27. In SHM, when particle is at x = A (extreme), kinetic energy is:
Explanation: At x = A (extreme), velocity = 0 → KE = 0. All energy is stored as potential energy.
28. Potential energy in SHM varies with time as:
Explanation: x = A sin(ωt), PE = ½kx² = ½kA²sin²(ωt). PE varies as sin² (or cos²) at frequency 2ω.
29. Average kinetic energy over a complete cycle of SHM is:
Explanation: KE = ½mω²(A²−x²). Average KE = ½ × total energy = ½ × ½mω²A² = ¼mω²A². Same for average PE.
30. In SHM, which quantity remains constant throughout?
Explanation: In ideal SHM (no damping), total energy E = ½kA² = constant. KE and PE exchange but their sum stays fixed.
31. A particle of mass 0.1 kg oscillates with angular frequency 10 rad/s and amplitude 5 cm. Total energy is:
Explanation: E = ½mω²A² = ½ × 0.1 × 100 × (0.05)² = ½ × 0.1 × 100 × 0.0025 = 0.0125 J.
32. Frequency of variation of PE with time in SHM of frequency f is:
Explanation: PE ∝ sin²(ωt) = (1 − cos2ωt)/2. Frequency of PE variation is 2f (twice the SHM frequency).
33. Spring of constant k is stretched by x₀ and released. When stretched by x₀/2 during oscillation, KE is:
Explanation: Total E = ½kx₀² (initial PE). At x = x₀/2: PE = ½k(x₀/2)² = ⅛kx₀² = E/4. KE = E − E/4 = 3E/4.
34. Two springs k₁ and k₂ in series support mass m. Time period of oscillation is:
Explanation: k_eff = k₁k₂/(k₁+k₂). T = 2π√(m/k_eff) = 2π√(m(k₁+k₂)/(k₁k₂)).
35. If total energy of SHM is E, then energy when displacement is half the amplitude is:
Explanation: At x = A/2: PE = ½k(A/2)² = ½kA²/4 = E/4. KE = E − E/4 = 3E/4.
36. In SHM, particle has speed v₁ at x₁ and v₂ at x₂. Amplitude is:
Explanation: v² = ω²(A²−x²). v₁² = ω²(A²−x₁²) and v₂² = ω²(A²−x₂²). Eliminate ω²: A² = (v₁²x₂²−v₂²x₁²)/(v₁²−v₂²).
37. A mass m between two springs (k₁ left, k₂ right) is displaced and released. Effective spring constant is:
Explanation: When mass is displaced, one spring is compressed and one extended — both exert restoring forces on same side. Effective k = k₁ + k₂ (parallel).
38. In SHM, PE = KE at t = 0 (particle starts from x = A/√2). Time to reach mean position first is:
Explanation: x = A/√2 corresponds to ωt = π/4. Starting from x = A/√2 (positive), mean position is at ωt = π/2... actually from x = A/√2: x = A sin(ωt+φ) at t=0, x=A/√2 → sin(φ)=1/√2 → φ = π/4. x = 0 when ωt + π/4 = π → t = 3π/(4ω) = 3T/8. First: ωt + π/4 = π gives t= 3T/8. The answer depends on direction.
39. If frequency of SHM doubles (same amplitude), total energy:
Explanation: E = ½mω²A² = 2π²mf²A². If f doubles, E → 4E.
40. At what phase angle (ωt + φ) does KE = 75% of total energy?
Explanation: KE/E = cos²(ωt+φ) = 0.75. cos(ωt+φ) = √0.75 = √3/2 → ωt+φ = 30°.
41. Simple pendulum clock runs slow. To correct it:
Explanation: T = 2π√(L/g). Running slow means T is too large → L is too large. Shorten the pendulum to reduce T and make the clock run at correct speed.
42. Time period of a pendulum depends on:
Explanation: T = 2π√(L/g). Independent of mass, material, or small amplitude (isochronous).
43. A compound (physical) pendulum has time period T = 2π√(I/mgd) where d is:
Explanation: d = distance from the pivot (suspension point) to the centre of mass. I is moment of inertia about the pivot.
44. Liquid column of length L oscillates in a U-tube with time period:
Explanation: For a U-tube oscillation, effective length = L/2. T = 2π√(L/2g) = π√(2L/2g)... T = 2π√(L/(2g)). With total liquid length L: T = π√(L/g).
45. A mass on a spring oscillates on a frictionless inclined plane (angle θ). Time period:
Explanation: On an inclined plane, equilibrium position shifts but spring constant remains k. Period T = 2π√(m/k), unchanged (gravity changes equilibrium, not the restoring constant).
46. Two identical pendulums coupled by a spring: normal modes of oscillation are:
Explanation: Coupled pendulums have two normal modes: symmetric (in-phase, spring unstretched) and antisymmetric (anti-phase, spring stretched). Complex motions are superposition of these.
47. A torsional pendulum has time period T = 2π√(I/C) where C is:
Explanation: Torsional pendulum: τ = −Cθ. C = torsional constant (N·m/rad). T = 2π√(I/C) analogous to mass-spring T = 2π√(m/k).
48. LC circuit oscillates analogously to mass-spring SHM. Angular frequency is:
Explanation: LC circuit: ω = 1/√(LC). Analogy: L↔m (inertia), C↔1/k (compliance), charge↔displacement, current↔velocity.
49. A pendulum oscillates with period T on Earth. Same pendulum in a lift accelerating upward at g/4 has period:
Explanation: Effective g in upward accelerating lift = g + a = g + g/4 = 5g/4. T' = 2π√(L/(5g/4)) = 2π√(4L/5g) = T√(4/5).
50. In damped SHM, amplitude decreases with time as:
Explanation: Amplitude A(t) = A₀ e^(−γt) where γ = b/(2m) and b is the damping coefficient. Exponential decay.
51. Resonance occurs when the driving frequency equals:
Explanation: At resonance, f_drive = f_natural. Amplitude becomes maximum (limited only by damping). Energy transfer is most efficient.
52. At resonance, amplitude in a lightly damped system is:
Explanation: At resonance, amplitude = F₀/(mω₀) × Q where Q = quality factor = ω₀m/b. High Q (low damping) → very large amplitude.
53. A clock pendulum (T = 2 s on Earth's surface). Pendulum is taken to altitude h = R/2 (R = Earth's radius). New period approximately:
Explanation: g ∝ 1/r². At height R/2: g' = g×R²/(3R/2)² = g×4/9. T' = T/√(4/9) = T × 3/2 = 3 s.
54. Quality factor Q of a damped oscillator is:
Explanation: Q = 2π × (energy stored)/(energy dissipated per cycle) = ω₀/Δω (ratio of resonant frequency to bandwidth). Both definitions are equivalent.
55. Phase of forced oscillations relative to driving force:
Explanation: At exact resonance, response lags force by 90°. Below resonance: lag 90° → approaches 180°.
56. A block of mass m attached to spring (k) is placed on a surface with friction μ. Block oscillates. Amplitude decreases by Δa each half cycle where Δa is:
Explanation: Each half cycle, friction removes energy. Amplitude reduction per half cycle: ΔA = 2μmg/k. (Friction force = μmg over half cycle length.)
57. SHM can be regarded as the projection of:
Explanation: A reference particle moves in a circle of radius A with uniform angular speed ω. Its projection on any diameter is SHM with amplitude A and frequency ω.
58. A particle dropped into a tunnel through the centre of Earth oscillates with period:
Explanation: Inside Earth, g(r) = gr/R. This is SHM with ω² = g/R. T = 2π/ω = 2π√(R/g) ≈ 84 min — equal to orbital period at Earth's surface!
59. A cylinder of mass m, area A, density ρ_c floats upright in liquid (density ρ_l). When pushed down by x and released, time period is:
Explanation: Restoring force = ρ_l × A × x × g. F = −(ρ_l Ag)x → k_eff = ρ_l Ag. T = 2π√(m/k_eff) = 2π√(m/(ρ_l Ag)).
60. Barton's pendulum experiment shows:
Explanation: In Barton's experiment, various pendulums of different lengths are driven by a heavy driver. The pendulum with the same natural frequency (same length) as the driver achieves maximum amplitude at resonance.
61. Time period of simple pendulum on surface of moon compared to Earth (g_moon = g_earth/6):
Explanation: T = 2π√(L/g). Lower g → larger T. Pendulum swings slower on moon.
62. If spring constant is 400 N/m and mass 1 kg, angular frequency of oscillation is:
Explanation: ω = √(k/m) = √(400/1) = 20 rad/s.
63. x = 5 sin(3πt + π/6) cm. Amplitude, frequency, and initial phase are:
Explanation: A = 5 cm. ω = 3π rad/s → f = ω/(2π) = 3π/(2π) = 1.5 Hz. Initial phase φ = π/6.
64. A particle in SHM has amplitude 10 cm and frequency 2 Hz. Speed at x = 6 cm is:
Explanation: ω = 2π×2 = 4π rad/s. v = ω√(A²−x²) = 4π×√(100−36) = 4π×8 = 32π cm/s. Hmm — check: √64 = 8. v = 4π×8 = 32π cm/s. Closest: none exactly. v = 4π√64 = 32π.
65. In SHM, if the amplitude is halved, maximum speed becomes:
Explanation: v_max = ωA ∝ A. If A → A/2, v_max → v_max/2.
66. Maximum acceleration in SHM if amplitude = 3 cm and ω = 10 rad/s is:
Explanation: a_max = ω²A = 100 × 0.03 = 3 m/s².
67. A spring stretches 5 cm when mass of 0.5 kg is attached statically. Time period of oscillation is:
Explanation: k = mg/x = 0.5×10/0.05 = 100 N/m. T = 2π√(0.5/100) = 2π×0.0707 = 0.444 s ≈ 0.449 s.
68. A uniform rod (length L, mass m) pivoted at one end oscillates as a physical pendulum. Time period:
Explanation: I = mL²/3, d = L/2. T = 2π√(I/mgd) = 2π√(mL²/3 / (mg×L/2)) = 2π√(2L/3g).
69. x₁ = A sinωt and x₂ = A sin(ωt + π/3). The resultant x = x₁ + x₂ has amplitude:
Explanation: Two SHM of same amplitude A with phase difference π/3. Resultant amplitude = 2A cos(π/6) = 2A × (√3/2) = A√3.
70. In underdamped oscillation, the oscillation frequency ω_d compared to natural frequency ω₀ is:
Explanation: ω_d = √(ω₀² − γ²)
71. Total energy of SHM is 0.1 J. When displacement is half the amplitude, PE is:
Explanation: PE = ½kx² = ½k(A/2)² = ¼ × ½kA² = E/4 = 0.1/4 = 0.025 J.
72. A pendulum of length 1 m swings with amplitude 4 cm. Maximum speed of bob is approximately (g = 10 m/s²):
Explanation: ω = √(g/L) = √10 ≈ 3.16 rad/s. v_max = ωA = 3.16 × 0.04 ≈ 0.126 m/s ≈ 0.12 m/s.
73. A spring-mass system (m = 0.5 kg, k = 200 N/m) is given initial amplitude 4 cm. Maximum acceleration is:
Explanation: ω = √(200/0.5) = 20 rad/s. a_max = ω²A = 400 × 0.04 = 16 m/s².
74. Two perpendicular SHMs of same frequency and 90° phase difference produce:
Explanation: x = A sinωt, y = B sin(ωt+π/2) = B cosωt. If A = B: circle (x²+y²=A²). If A ≠ B: ellipse (x²/A²+y²/B²=1).
75. Velocity-displacement graph (v vs x) for SHM is:
Explanation: v² = ω²(A²−x²) → v²/ω²A² + x²/A² = 1. This is the equation of an ellipse in v-x space.
76. A mass hanging on a spring is in equilibrium. If spring elongation at equilibrium is d, time period of vertical oscillation is:
Explanation: At equilibrium: kd = mg → k/m = g/d. T = 2π/ω = 2π/√(k/m) = 2π√(d/g).
77. Tacoma Narrows Bridge (1940) collapsed due to:
Explanation: Wind vortex shedding at the natural frequency of the bridge caused resonant oscillations that grew until the bridge collapsed — a dramatic example of mechanical resonance.
78. Reversible pendulum is used to measure g accurately because it has two pivots where:
Explanation: Kater's reversible pendulum: adjusted so T is equal from both pivot points, which corresponds to the equivalent simple pendulum length L = distance between pivots. This gives accurate g measurement.
79. A test tube of mass m, area A, floats in liquid. When pushed down and released, effective spring constant is:
Explanation: When pushed down by x, extra buoyant force = ρ_l × A × x × g. This is the restoring force: k_eff = ρ_l Ag.
80. A particle moves in circle of radius R at angular speed ω. Projection on a diameter starts at x = R (t=0) and moves to x = 0 in time:
Explanation: x = R cosωt. Reaches x = 0 when cosωt = 0 → ωt = π/2 → t = π/(2ω) = T/4.
81. A particle under SHM has displacement x = 4sin(2t + π/6) m. At t = 0, KE and PE are (m = 1 kg):
Explanation: At t=0: x = 4sin(π/6) = 2 m. v = 4×2×cos(π/6) = 8×(√3/2) = 4√3 m/s. KE = ½×1×(4√3)² = 24 J. PE = ½×mω²x² = ½×4×4 = 8 J. Hmm: E = ½mω²A² = ½×1×4×16 = 32 J. PE = ½mω²x² = ½×1×4×4 = 8 J. KE = 32−8 = 24 J. Total 32 J = 24+8.
82. A particle executes SHM along x and y axes with frequencies in ratio 1:2 and same initial phase. The Lissajous figure traced is:
Explanation: For f_x:f_y = 1:2 with zero phase difference, the Lissajous figure is a figure-eight. The specific shape depends on the phase difference and frequency ratio.
83. Three masses connected by springs (linear chain) have how many normal modes of oscillation?
Explanation: N coupled masses have N normal modes. Each mode has a specific frequency (ω_n) and pattern. General motion is a superposition of all normal modes.
84. Speed of sound in an ideal gas is √(γP/ρ) because compression and rarefaction in sound are adiabatic. If γ doubles (at same P, ρ), speed:
Explanation: v_sound = √(γP/ρ) ∝ √γ. If γ doubles, v → √2 × v.
85. For a pendulum with large amplitude (θ not small), time period compared to small-angle period is:
Explanation: For large angles, restoring force (mg sinθ
86. Energy of a damped oscillator decays as:
Explanation: Amplitude A(t) = A₀e^(−γt). Energy ∝ A² ∝ e^(−2γt). Energy decays at twice the rate of amplitude.
87. Superposition of x = A sinωt and y = A sin(ωt + φ) gives a circle when φ equals:
Explanation: x = A sinωt, y = A sin(ωt + π/2) = A cosωt → x² + y² = A². Perfect circle. Any other phase: ellipse or line.
88. A block on a frictionless surface is connected to two walls by springs k₁ and k₂. Normal mode frequency is:
Explanation: Both springs restore simultaneously in parallel: k_eff = k₁ + k₂. f = (1/2π)√((k₁+k₂)/m).
89. Critical damping occurs when:
Explanation: Critical damping: γ = ω₀ (or b = 2mω₀ = 2√(km)). System returns to equilibrium as fast as possible without oscillating. Used in door dampers, galvanometers.
90. A child pumping a swing increases amplitude by:
Explanation: Standing/squatting on a swing changes its effective length at 2f (twice per cycle). This is parametric resonance — energy is pumped into the oscillator by modulating a parameter.
91. A particle in SHM has v₁ = 4 cm/s at x = 3 cm and v₂ = 3 cm/s at x = 4 cm. Amplitude A and ω are:
Explanation: v² = ω²(A²−x²): 16 = ω²(A²−9) and 9 = ω²(A²−16). Dividing: 16/9 = (A²−9)/(A²−16). 16A²−256 = 9A²−81 → 7A² = 175 → A² = 25 → A = 5 cm. ω² = 16/(25−9) = 1 → ω = 1 rad/s.
92. Two identical pendulums weakly coupled show 'beats' phenomenon because:
Explanation: Coupling splits the degenerate frequency into two close frequencies ω₁ and ω₂. Beat frequency = |ω₁−ω₂|. Energy transfers from one pendulum to the other at the beat frequency.
93. For SHM x = A sin(ωt), time spent near the extreme positions is more than near the mean because:
Explanation: dt = dx/v. Near extremes, v → 0, so dx/v becomes large — particle spends more time near ±A. This is why SHM probability density peaks at extremes.
94. An anharmonic oscillator has potential U = ½kx² + bx³. For small oscillations about equilibrium:
Explanation: Wait — for small x, bx³
95. A particle in SHM crosses mean position every 0.5 s. Time period is:
Explanation: Particle crosses mean position twice per complete cycle (once going right, once going left). So T/2 = 0.5 s → T = 1 s.
96. Vertical oscillation of a mass on spring: equilibrium is displaced downward by d = mg/k. If mass is pulled extra distance x₀ and released, amplitude of oscillation is:
Explanation: In vertical spring-mass: the equilibrium position is the new origin. Additional displacement from new equilibrium = x₀. Amplitude of oscillation about new equilibrium = x₀.
97. In a Melde's experiment (string vibrations), resonance occurs when:
Explanation: Melde's experiment: frequency of the vibrating fork drives transverse waves in the string. Resonance (standing waves) occurs when the driving frequency matches one of the string's normal mode frequencies.