6 Problem Types — Fully Solved
Every type of SHM question that has ever appeared in an exam. For each: what's given, what's tested, which concept to select, complete solution, and shortcut insight.
CBSE: Direct formula, Derivation, and 2–3 step numericals. Focus on Types 1, 2, and 3 below. No assertion-reasoning in CBSE SHM.
NEET: Direct formula (30%), Conceptual (40%), Graph-based (30%). All types appear. Types 1, 3, 4 are highest frequency. Time per question: 90 seconds max.
JEE Main: Multi-step (35%), Numerical value (25%), Graph-based (20%). Types 2, 3, 4 are key. Numerical questions have no negative marking — always attempt.
JEE Advanced: Types 5 & 6 dominate. Multiple correct, paragraph-based. One wrong selection = negative marks. Read ALL options before answering.
Type 1 — Direct Formula Application
Solved Example 1.1
Problem: A particle executes SHM with amplitude 0.1 m and time period 0.4 s. Calculate (i) maximum velocity, (ii) maximum acceleration, (iii) velocity when displacement is 0.06 m.
A = 0.1 m, T = 0.4 s, x = 0.06 m (for part iii)
Whether you know ω = 2π/T, and then apply v_max = Aω, a_max = Aω², and v = ω√(A²−x²). The multi-part structure tests systematic approach.
- Find ω: ω = 2π/T = 2π/0.4 = 5π rad/s ≈ 15.71 rad/s
- v_max = Aω = 0.1 × 5π = 0.5π ≈ 1.57 m/s
- a_max = Aω² = 0.1 × (5π)² = 0.1 × 25π² = 2.5π² ≈ 24.67 m/s²
- v at x = 0.06 m: v = ω√(A²−x²) = 5π × √(0.01 − 0.0036) = 5π × √0.0064 = 5π × 0.08 = 0.4π ≈ 1.26 m/s
Always compute ω first. Everything flows from ω. In an exam, write ω = 5π without computing the decimal — it saves time and avoids rounding errors in subsequent steps.
Type 2 — Conceptual Reasoning
Solved Example 2.1 — "Is this SHM?" Type
Problem: Which of the following equations represent SHM?
(A) x = A sin²ωt (B) x = A sin(ωt) + B cos(ωt) (C) x = A sin(ωt) + B sin(2ωt) (D) x = A/sinωt
SHM condition: x must satisfy d²x/dt² = −ω²x. Test each option by differentiating twice or using trig identity.
(A) x = A sin²ωt = A(1 − cos2ωt)/2 = A/2 − (A/2)cos2ωt
This oscillates but about x = A/2, not x = 0. It IS SHM with angular frequency 2ω but with shifted equilibrium. SHM ✓ (about x = A/2)
(B) x = A sinωt + B cosωt can be written as R sin(ωt + φ) where R = √(A²+B²). Single frequency, sinusoidal — SHM ✓
(C) x = A sinωt + B sin2ωt — two different frequencies. d²x/dt² ≠ −constant×x. Not SHM ✗ (oscillatory, but not SHM)
(D) x = A/sinωt = A cosecωt — not bounded, goes to infinity. Not oscillatory ✗
Option (A) trips many students — they see sin² and reject it. Use the identity sin²θ = (1−cos2θ)/2 first. Any function that reduces to R cos(ωt + φ) or R sin(ωt + φ) is SHM.
Rule: If after simplification, x = R sin(ωt + φ) + constant → SHM about a shifted equilibrium. If superposition of two DIFFERENT frequencies → Not SHM.
Type 3 — Multi-Step Problems
Solved Example 3.1 — Energy & System Change
Problem: A spring (k = 200 N/m) is attached to a 2 kg mass performing SHM with amplitude 0.05 m. At the moment when the mass is at x = 0.03 m and moving away from equilibrium, a 1 kg mass is gently placed on it (no momentum change to the spring system, just mass added). Find the new amplitude.
k = 200 N/m, m₁ = 2 kg, A₁ = 0.05 m, x₀ = 0.03 m at moment of collision. New total mass: m₂ = 3 kg
Concept: when mass is added, velocity of mass at that instant doesn't change (gently placed = no impulse). But new ω changes. Find new amplitude from energy conservation at the new equilibrium.
- Find v at x = 0.03 m (before collision):
ω₁ = √(k/m₁) = √(200/2) = 10 rad/s
v = ω₁√(A₁²−x₀²) = 10√(0.0025−0.0009) = 10√0.0016 = 10×0.04 = 0.4 m/s - After mass addition: velocity remains 0.4 m/s (gently placed). New ω₂ = √(200/3) = √(66.67) ≈ 8.165 rad/s
- New amplitude A₂ from energy conservation:
½m₂v² + ½kx₀² = ½kA₂²
½(3)(0.16) + ½(200)(0.0009) = ½(200)A₂²
0.24 + 0.09 = 100A₂²
A₂² = 0.33/100 = 0.0033 → A₂ = √0.0033 ≈ 0.0574 m ≈ 5.74 cm
If mass added at equilibrium (x=0): A₂ = v/ω₂ = (ω₁A₁)/ω₂ = A₁√(m₁/m₂) — much simpler! Know this formula. Examiners often specify "at equilibrium" to simplify the problem.
Type 4 — Graph-Based Problems
Solved Example 4.1 — From Graph to Formula
Problem: From the a-x graph of an SHM particle, the graph passes through (x=0, a=0) with slope = −16. Find: (i) ω, (ii) Time period T, (iii) Maximum acceleration if amplitude = 0.5 m.
Key equation: a = −ω²x. The a-x graph is a straight line with slope = −ω². Given slope = −16 → ω² = 16 → ω = 4 rad/s.
- From a = −ω²x: slope of a-x graph = −ω² = −16 → ω = 4 rad/s
- T = 2π/ω = 2π/4 = π/2 s ≈ 1.57 s
- a_max = Aω² = 0.5 × 16 = 8 m/s²
In any a-x graph: ω² = |slope|. This is the fastest extraction. For v-x graphs: the shape is an ellipse, and ω = (semi-axis in v direction)/(semi-axis in x direction) = v_max/A.
Type 5 — Assertion & Reason (JEE/NEET)
Options: (A) Both A and R are true; R is correct explanation of A | (B) Both A and R are true; R is NOT correct explanation | (C) A is true but R is false | (D) A is false but R is true
Example 5.1
Assertion (A): The time period of a simple pendulum increases when it is taken to a higher altitude.
Reason (R): At higher altitude, the value of g decreases.
Step 1: Is Assertion true? T = 2π√(L/g) → if g decreases, T increases. ✓ True.
Step 2: Is Reason true? g decreases with altitude (g = GM/(R+h)²). ✓ True.
Step 3: Does R explain A? Yes — T increases BECAUSE g decreases at altitude. → Answer: (A)
This is where most students lose marks in assertion-reason. They identify both as true but select (B). Always verify causal connection: Does the Reason directly explain the Assertion's mechanism? If yes → (A). If R and A are unrelated truths → (B).
Example 5.2 — Tricky One
Assertion (A): The total energy of SHM is independent of amplitude.
Reason (R): E = ½kA² and k is constant for a given spring.
Step 1: Is A true? E = ½kA² → E depends ON amplitude A. A is FALSE.
Step 2: Is R true? Yes, k is constant. But A is false, so answer is → (D): A is false, R is true.
Before reading the Reason, always verify the Assertion independently. If A is clearly false, jump to option (D). This saves 20 seconds per question.
Type 6 — Case-Based / Paragraph Type
Solved Paragraph 6.1
PASSAGE:
A 1 kg mass is connected to a spring (k = 100 N/m) and oscillates on a frictionless horizontal surface. At t = 0, the mass is at x = 0.1 m with velocity 0 (held from rest and released). A charged metal plate is then switched on, creating a uniform force F₀ = 2 N in the +x direction throughout the motion.
Q1: What is the new equilibrium position?
At new equilibrium: kx_eq = F₀ → x_eq = F₀/k = 2/100 = 0.02 m from original equilibrium
Q2: What is the new amplitude of oscillation?
Initial position x₀ = 0.1 m, v₀ = 0. New equilibrium at x = 0.02 m. Distance from new equilibrium at release = 0.1 − 0.02 = 0.08 m. Since released from rest: New Amplitude A' = 0.08 m
Q3: What is the time period?
T = 2π√(m/k) = 2π√(1/100) = 2π/10 = π/5 s. Note: T does NOT change when a constant force is added — it only shifts the equilibrium.
Constant force in SHM → shifts equilibrium, keeps T same, changes amplitude based on initial displacement from NEW equilibrium. This is a JEE Advanced favourite because students forget to measure amplitude from the new equilibrium.