This MCQ module is based on: Energy Shm
TOPIC 26 OF 33
Energy Shm
🎓 Class 11
Physics
CBSE
Theory
Ch 13 – Oscillations
⏱ ~14 min
🌐 Language: [gtranslate]
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Energy Shm
13.9 Energy in Simple Harmonic Motion
For an SHM with restoring force F = −k x, the potential energy stored in the restoring agent (e.g. the spring) is:
\[U(x) = -\int_0^x F\,dx' = \int_0^x k\,x'\,dx' = \tfrac{1}{2}\, k\, x^2\]
A simple parabolic well — quadratic in displacement. We choose U = 0 at the equilibrium x = 0.
The kinetic energy is K = ½ m v². Using v(x) = ω √(A² − x²) and k = m ω²:
\[K(x) = \tfrac{1}{2}\, m\, v^2 = \tfrac{1}{2}\, m\, \omega^2\, (A^2 - x^2) = \tfrac{1}{2}\, k\, (A^2 - x^2)\]
Adding K and U:
\[\boxed{\,E = K + U = \tfrac{1}{2}\, k\, A^2 = \tfrac{1}{2}\, m\, \omega^2\, A^2\,}\]
The total mechanical energy of an SHM is constant (no friction, no losses), depends on A² (and ω²), and is independent of x and t.
13.10 Energy as a Function of Time
With x(t) = A cos(ωt + φ) and v(t) = −Aω sin(ωt + φ):
U(t) = (1/2) k A² cos²(ωt + φ)
K(t) = (1/2) k A² sin²(ωt + φ)
E = U + K = (1/2) k A² [constant]
K(t) = (1/2) k A² sin²(ωt + φ)
E = U + K = (1/2) k A² [constant]
Note that K and U each oscillate with frequency 2ω (twice the displacement frequency) — because cos² and sin² complete a full cycle in time T/2. They take turns being maximum:
| Position | U(x) | K(x) | Total E |
|---|---|---|---|
| Equilibrium (x = 0) | 0 | (1/2) k A² (max) | (1/2) k A² |
| Half-amplitude (x = A/2) | (1/8) k A² | (3/8) k A² | (1/2) k A² |
| Extremes (x = ±A) | (1/2) k A² (max) | 0 | (1/2) k A² |
13.11 Average Energies
Time-averaging cos²(ωt) over one period gives ½. Hence:
⟨U⟩ = (1/2) × (1/2) k A² = (1/4) k A²
⟨K⟩ = (1/4) k A²
⟨U⟩ = ⟨K⟩ = E/2
⟨K⟩ = (1/4) k A²
⟨U⟩ = ⟨K⟩ = E/2
On average, half the total energy of an SHM is kinetic and half is potential — the virial theorem for a quadratic potential.
Worked Examples
Example 13.7: Maximum KE of a spring oscillator
A 0.20 kg block on a spring with k = 80 N/m oscillates with amplitude 0.10 m. Find the total energy, maximum speed, and KE at x = 0.05 m.
E = (1/2) k A² = 0.5 × 80 × 0.01 = 0.40 J.
v_max = √(2E/m) = √(2 × 0.40 / 0.20) = √4 = 2.0 m/s.
At x = 0.05: U = (1/2)(80)(0.05²) = 0.10 J, so K = 0.40 − 0.10 = 0.30 J.
v_max = √(2E/m) = √(2 × 0.40 / 0.20) = √4 = 2.0 m/s.
At x = 0.05: U = (1/2)(80)(0.05²) = 0.10 J, so K = 0.40 − 0.10 = 0.30 J.
Example 13.8: Position when K = U
For an SHM of amplitude A, find the displacement at which kinetic energy equals potential energy.
K = U ⇒ ½k(A² − x²) = ½kx² ⇒ A² − x² = x² ⇒ x² = A²/2 ⇒ x = ±A/√2 ≈ ±0.707 A.
At this displacement, U = K = E/2 — the energy is split exactly equally.
At this displacement, U = K = E/2 — the energy is split exactly equally.
Example 13.9: Energy when amplitude doubles
If the amplitude of an SHM is doubled keeping ω constant, what happens to (i) total energy, (ii) maximum velocity, (iii) period?
(i) E ∝ A² ⇒ E becomes 4 times. (ii) v_max = Aω ⇒ doubles. (iii) T = 2π/ω is independent of A — unchanged (isochronism).
Interactive: Energy-vs-Displacement Explorer L3 Apply
Vary x (between −A and +A); see how K and U partition E = ½kA². Try doubling A!
Activity 13.3 — Energy Trading Demo with a Pendulum
L4 Analyse
Predict: Lift a pendulum bob to amplitude A and release. At what point does the bob have all kinetic energy and zero potential energy? When is the reverse true?
- Set up a pendulum (~50 cm string, small bob) hanging at rest.
- Lift the bob a small angle and release; observe.
- Ask: at what positions does the speed peak? At what positions is the speed zero?
Observation: Maximum speed occurs at the bottom of the swing (equilibrium): K = K_max, U = 0. Zero speed at the highest points (extremes ±A): K = 0, U = K_max. Energy continually converts back and forth — same kA² total, just oscillating between forms. Total mechanical energy is conserved (small frictional losses cause amplitude to slowly decay — Part 4).
Competency-Based Questions
A 0.5 kg block oscillates on a spring (k = 50 N/m) with amplitude 0.06 m on a frictionless surface.
Q1. L1 Remember Write the formula for the total energy of an SHM in terms of A.
E = (1/2) k A² = (1/2) m ω² A². Independent of x and t.
Q2. L2 Understand Why do K and U oscillate at twice the displacement frequency?
K ∝ sin²(ωt) and U ∝ cos²(ωt). Squared sinusoids complete a full cycle in time T/2, so their angular frequency is 2ω — twice the displacement.
Q3. L3 Apply Find E and v_max for the spring-block system.
E = (1/2)(50)(0.06²) = (1/2)(50)(0.0036) = 0.090 J. v_max = √(2E/m) = √(0.18/0.5) = √0.36 = 0.60 m/s.
Q4. L4 Analyse What fraction of the total energy is kinetic when the block is at x = A/2?
U = (1/2)k(A/2)² = (1/8)kA² = E/4. So K = E − U = 3E/4; the kinetic fraction is 0.75.
Q5. L5 Evaluate A student claims the average kinetic energy and average potential energy of an SHM are unequal because K and U have different shapes. Is this correct?
No. Although K(t) and U(t) have different time profiles (sin² vs cos²), both have the same time-average ½ over a period. Hence ⟨K⟩ = ⟨U⟩ = E/2 (the virial theorem for a quadratic potential). The shapes look different but their averages are identical.
Assertion-Reason Questions
Assertion (A): The total energy of an SHM is independent of position.
Reason (R): K and U continuously trade off in such a way that K + U remains constant.
Answer: A. In a frictionless SHM, mechanical energy E = ½kA² is conserved; K and U trade off perfectly.
Assertion (A): Doubling the amplitude of an SHM quadruples the energy.
Reason (R): E ∝ A².
Answer: A. E = ½kA² ⇒ E ∝ A². Doubling A multiplies E by 4.
Assertion (A): At x = A/√2, kinetic energy equals potential energy.
Reason (R): ⟨K⟩ = ⟨U⟩ in an SHM.
Answer: B. A is the instantaneous statement K(x=A/√2)=U(x=A/√2)=E/2. R is the time-averaged statement. Both true; R is not the cause — the instantaneous equality follows from the explicit ½kx² = ½k(A²−x²) algebra, not from the average.
Frequently Asked Questions - Energy Shm
What is the main concept covered in Energy Shm?
In NCERT Class 11 Physics Chapter 13 (Oscillations), "Energy Shm" covers core principles and equations needed for board exam success. The MyAiSchool lesson explains the topic with definitions, derivations, worked examples, and interactive simulations. Key formulas and dimensional analysis are included to build conceptual depth and problem-solving skills aligned with the CBSE 2025-26 syllabus.
How is Energy Shm useful in real-life applications?
Real-life applications of Energy Shm from NCERT Class 11 Physics Chapter 13 include engineering design, satellite mechanics, sports biomechanics, transportation safety, and electrical/electronic devices. The MyAiSchool lesson links every concept to a tangible example so students see physics as a problem-solving framework for the physical world, not as abstract formulas.
What are the key formulas in Energy Shm?
Key formulas in Energy Shm (NCERT Class 11 Physics Chapter 13 Oscillations) are derived step-by-step in the MyAiSchool lesson. Students should memorize the final formula AND understand its derivation for full board marks. Each formula is listed with its dimensional formula, SI unit, applicability range, and common pitfalls. The Summary section at the end of each part includes a quick-reference formula card.
How does this part connect to other parts of Chapter 13?
NCERT Class 11 Physics Chapter 13 (Oscillations) is structured so each part builds on the previous one. Energy Shm connects directly to neighbouring parts via shared definitions, units, and methodology. The MyAiSchool lesson cross-references related concepts with internal links so students can navigate the whole chapter as one connected story rather than disconnected fragments.
What types of CBSE board questions come from Energy Shm?
CBSE board questions from Energy Shm typically include: (1) 1-mark MCQs on definitions and formulas, (2) 2-mark short-answer derivations or applications, (3) 3-mark numerical problems with units, (4) 5-mark long-answer derivations followed by application. The MyAiSchool lesson tags each Competency-Based Question (CBQ) with Bloom level (L1-L6) so students know how to study for each weight.
How can students use the interactive simulation effectively?
The interactive simulation in the Energy Shm lesson allows students to adjust input parameters (sliders or selectors) and see physical quantities update in real time. To use it effectively: (1) try extreme values to understand limiting cases, (2) compare with the analytical formula, (3) check unit consistency, (4) test special configurations from worked examples. The simulation reinforces conceptual intuition that pure formula manipulation cannot.
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