Formula
Energy in SHM
KE(t) = ½ m ω² A² sin²(ωt+φ); PE(t) = ½ k x² = ½ m ω² A² cos²(ωt+φ); Total E = ½ k A² = ½ m ω² A². Energy oscillates between KE and PE; total constant.
-- NCERT Class 11 Physics, Ch. 13, p. 269Energy in SHM
8 MCQs3 revision cards9-step worked example
Source: NCERT Oscillations and WavesOfficial key: NTA-verifiedLast updated: 25 Sep 2026
Energy in SHM, explained for NEET
The trap: aspirants know that total energy in SHM is E = ½kA². They then see a question where amplitude doubles and confidently answer "energy doubles." It quadruples. Energy scales as the square of amplitude — and this is a high-frequency NEET distractor.
The concept. In simple harmonic motion, mechanical energy continuously converts between kinetic and potential forms. At the mean position (x = 0), all energy is kinetic: KE = ½mω²A². At the extreme positions (x = ±A), all energy is potential: PE = ½kA². At any intermediate displacement x:
- KE = ½k(A² − x²)
- PE = ½kx²
- Total E = KE + PE = ½kA² = ½mω²A²
The total is constant — independent of x, independent of time. This follows directly from the conservative nature of the restoring force (NCERT Class 11 Physics, Oscillations chapter, page 265).
Two critical facts from NCERT (Oscillations chapter, page 264): (1) KE and PE each oscillate at frequency 2ω (twice the oscillation frequency), and (2) their time-averaged values are equal, each ½E.
The amplitude-squared dependence is what catches students. Since E = ½kA², doubling A means E → 4E, not 2E. Tripling A means E → 9E. The period T = 2π√(m/k) does not depend on amplitude — but the energy emphatically does. This asymmetry is a common confusion (trap: period-amplitude independence does NOT mean energy-amplitude independence).
Watch-out for NEET: questions that change amplitude and ask for the new total energy, or ask at what displacement KE equals PE (answer: x = A/√2).
Can you answer these Energy in SHM MCQs?
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
MCQ 1Easy RecallPractice
In simple harmonic motion, at which position is the kinetic energy maximum?
Show answer and why every option is right or wrong
Answer: A. A is correct. At x = 0, the entire energy is kinetic: KE = ½kA². PE is zero at the mean position (NCERT Class 11 Physics, Oscillations chapter, page 265).
Why B is wrong: B is wrong because at the extreme position x = ±A, KE is zero and all energy is potential.
Why C is wrong: C is wrong because at x = A/2, KE = ½k(A² − A²/4) = ¾ × ½kA², which is less than the maximum value ½kA² at x = 0.
Why D is wrong: D is wrong because at x = A/√2, KE = PE = ½E. This is the equal-partition point, not the KE maximum.
MCQ 2Easy RecallPractice
The total mechanical energy of a particle in ideal SHM (no damping) is:
Show answer and why every option is right or wrong
Answer: C. C is correct. In ideal SHM, the restoring force is conservative. Total energy E = KE + PE = ½kA² remains constant at every point in the oscillation (NCERT Class 11 Physics, Oscillations chapter, page 265).
Why A is wrong: A is wrong because KE is maximum at the mean position, not total energy. Total energy is constant everywhere.
Why B is wrong: B is wrong because PE is maximum at the extremes, not total energy. Total energy does not change with position.
Why D is wrong: D is wrong because total energy is never zero at any point in ideal SHM. E = ½kA² is a fixed positive value independent of displacement.
MCQ 3Easy RecallPractice
In SHM, the kinetic energy and potential energy each oscillate with a frequency equal to:
Show answer and why every option is right or wrong
Answer: C. C is correct. KE = ½mω²A²cos²(ωt + φ) and PE = ½mω²A²sin²(ωt + φ). Using the identity cos²θ = (1 + cos 2θ)/2, both KE and PE oscillate at angular frequency 2ω, i.e., frequency 2ω/(2π) = twice the particle's oscillation frequency (NCERT Class 11 Physics, Oscillations chapter, page 264).
Why A is wrong: A is wrong because this is the frequency of displacement oscillation. KE and PE involve squared trig functions, which double the frequency.
Why B is wrong: B is wrong because there is no physical mechanism that halves the energy oscillation frequency relative to displacement.
Why D is wrong: D is wrong because the cos² identity gives frequency 2ω, not 3ω. There is no third-harmonic component in ideal SHM energy.
MCQ 4Direct ApplicationPractice
A block on a frictionless surface oscillates on a spring with amplitude A and total energy E. If the amplitude is doubled to 2A (same spring, same mass), the new total energy is:
Show answer and why every option is right or wrong
Answer: B. B is correct. E = ½kA². When A → 2A, E_new = ½k(2A)² = 4 × ½kA² = 4E. Energy scales as the square of amplitude (NCERT Class 11 Physics, Oscillations chapter, page 265).
Why A is wrong: A is wrong because it assumes energy scales linearly with amplitude (E ∝ A). The correct relation is E ∝ A², so doubling A quadruples E (trap: amplitude-squared dependence).
Why C is wrong: C is wrong because energy is not independent of amplitude. Period is independent of amplitude, but energy depends on A².
Why D is wrong: D is wrong because √2 scaling would apply if energy were proportional to √(A²) = A, which it is not. E ∝ A² gives a factor of 4.
MCQ 5Direct ApplicationPractice
A particle in SHM has amplitude A. At what displacement from the mean position is the kinetic energy equal to the potential energy?
Show answer and why every option is right or wrong
Answer: B. B is correct. KE = PE means ½k(A² − x²) = ½kx², giving A² − x² = x², so x² = A²/2 and x = A/√2 (NCERT Class 11 Physics, Oscillations chapter, page 265).
Why A is wrong: A is wrong because at x = A/2: PE = ½k(A/2)² = kA²/8, and KE = ½k(A² − A²/4) = 3kA²/8. Since KE ≠ PE, this is not the equal-partition point.
Why C is wrong: C is wrong because at x = A/4: PE = kA²/32, KE = 15kA²/32. KE is much larger than PE here.
Why D is wrong: D is wrong because at x = A√3/2: PE = 3kA²/8, KE = kA²/8. Here PE > KE, not equal.
MCQ 6Direct ApplicationPractice
A spring-mass system has spring constant k = 200 N/m and oscillates with amplitude 0.10 m. The total energy of the system is:
Show answer and why every option is right or wrong
Answer: A. A is correct. E = ½kA² = ½ × 200 × (0.10)² = ½ × 200 × 0.01 = 1.0 J (NCERT Class 11 Physics, Oscillations chapter, page 265).
Why B is wrong: B is wrong because it likely results from using E = kA² (missing the ½ factor). The correct formula is E = ½kA².
Why C is wrong: C is wrong because it results from using E = ½kA (linear in A instead of A²) or a computational error. ½ × 200 × 0.10 = 10, but the formula requires A², not A.
Why D is wrong: D is wrong because it likely results from an arithmetic error such as ½ × 200 × (0.10)² with a misplaced decimal, or using ½ × k × A instead of ½ × k × A².
MCQ 7CalculationPractice
A 0.50 kg block attached to a spring (k = 200 N/m) oscillates with amplitude 5.0 × 10⁻² m. What is the maximum speed of the block?
Show answer and why every option is right or wrong
Answer: D. D is correct. Step 1: Total energy E = ½kA² = ½ × 200 × (5.0 × 10⁻²)² = 0.25 J. Step 2: At x = 0, all energy is kinetic: ½mv²_max = E, so v_max = √(2E/m) = √(2 × 0.25/0.50) = √1.0 = 1.0 m/s (NCERT Class 11 Physics, Oscillations chapter, page 265).
Why A is wrong: A is wrong because 0.50 m/s is E/m = 0.25/0.50, which drops both the 2 and the square root of v_max = √(2E/m).
Why B is wrong: B is wrong because 0.25 is the energy E = ½kA² = 0.25 J itself, reported as a speed. Energy and speed have different dimensions.
Why C is wrong: C is wrong because 2.0 m/s comes from taking the full swing, 2A = 0.10 m, as the amplitude: ω × 2A = 20 × 0.10. The amplitude is measured from the centre, so v_max = ωA = 0.05 × 20 = 1.0 m/s.
MCQ 8CalculationPractice
A particle executes SHM with amplitude A and angular frequency ω. At displacement x = A/2 from the mean position, the ratio of kinetic energy to potential energy is:
Show answer and why every option is right or wrong
Answer: D. D is correct. KE = ½k(A² − x²) = ½k(A² − A²/4) = ½k × 3A²/4. PE = ½kx² = ½k × A²/4. Ratio KE:PE = (3A²/4):(A²/4) = 3:1 (NCERT Class 11 Physics, Oscillations chapter, page 265).
Why A is wrong: A is wrong because 1:3 reverses the ratio. At x = A/2, the particle is closer to the mean position than to the extreme, so KE > PE. The correct ratio is 3:1, not 1:3 (trap: confusing which energy dominates near the centre vs. near the extreme).
Why B is wrong: B is wrong because 2:1 would require A² − x² = 2x², giving x² = A²/3, i.e., x = A/√3 ≈ 0.577A. At x = A/2 = 0.5A, the ratio is 3:1.
Why C is wrong: C is wrong because KE = PE occurs at x = A/√2, not at x = A/2. At A/2, the particle is well within the high-KE zone.
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Energy in SHM: quick recall before you leave
How do you solve a Energy in SHM question? A worked example
Pattern: Given a spring-mass system with changed amplitude, find the ratio of total energies. (Based on the amplitude-squared energy dependence — the core trap for this topic.)
- 1
Given
A spring-mass oscillator has mass m = 0.40 kg and spring constant k = 160 N/m. It oscillates first with amplitude A₁ = 0.050 m, then the amplitude is increased to A₂ = 0.15 m.
- 2
Required
Find the ratio E₂/E₁.
- 3
Concept
Total energy in SHM is E = ½kA². Energy depends on the square of amplitude. Spring constant and mass remain unchanged.
- 4
Formula
E = ½kA², so E₂/E₁ = A₂²/A₁².
- 5
Substitution
E₂/E₁ = (0.15)²/(0.050)² = 0.0225/0.0025
- 6
Calculation
E₂/E₁ = 9
Note: k and m are the same in both cases, so they cancel. The ratio depends purely on the amplitude ratio squared: (A₂/A₁)² = (0.15/0.050)² = 3² = 9. - 7
Final answer
E₂/E₁ = 9. The new total energy is 9 times the original.
Note on exact values: the ratio 3 (= 0.15/0.050) is exact arithmetic, and the squaring gives an exact integer 9. No sig-fig ambiguity arises. - 8
Common trap
A student who thinks energy scales linearly with amplitude would answer E₂/E₁ = 3 (trap: period is amplitude-independent, but energy is NOT — it scales as A²). Another common error is confusing the energy ratio with the speed ratio (v_max ∝ Aω, so the speed ratio is 3, not 9).
- 9
Similar NEET-style question
A block oscillates on a spring with total energy 2.0 J and amplitude 4.0 × 10⁻² m. If the amplitude is halved to 2.0 × 10⁻² m (same spring), find the new total energy. [Answer: E_new = 2.0 × (1/2)² = 0.50 J]
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What to remember before solving Energy in SHM questions
Key Fact
Phase relation in SHM
Velocity leads displacement by π/2: v = -A ω sin(ωt+φ). Acceleration leads displacement by π: a = -A ω² cos(ωt+φ) = -ω² x. KE max at x=0; PE max at x=±A.
-- NCERT Class 11 Physics, Ch. 13, p. 267Which Energy in SHM formulas do you need for NEET?
1 formula — click to collapse
Total energy in SHM
Total mechanical energy is constant. Oscillates between KE (max at x=0) and PE (max at x=±A).
| Symbol | Quantity | SI Unit |
|---|---|---|
| E | total energy | J |
| k | spring constant | N/m |
| A | amplitude | m |
| m | mass | kg |
| omega | angular frequency | rad/s |
Valid when
- Conservative SHM (no damping)
- Elastic regime
More in Oscillations and Waves: 6 exam traps and mistakes · 9 formulas · 6 question patterns from its other lessons.
Energy in SHM questions from past NEET papers
No question in our NEET 2020–2025 set targets this topic directly.
Sources
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
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