This MCQ module is based on: Spring Pendulum Damped
TOPIC 27 OF 33
Spring Pendulum Damped
🎓 Class 11
Physics
CBSE
Theory
Ch 13 – Oscillations
⏱ ~14 min
🌐 Language: [gtranslate]
🧠 AI-Powered MCQ Assessment
▲
📝 Worksheet / Assessment
▲
This assessment will be based on: Spring Pendulum Damped
Upload images, PDFs, or Word documents to include their content in assessment generation.
Spring Pendulum Damped
13.12 Spring–Mass Oscillator
A block of mass m attached to a spring of force constant k on a frictionless surface obeys Hooke's law F = −kx, the prototype of SHM:
\[\boxed{\,T_{\text{spring}} = 2\pi\sqrt{\dfrac{m}{k}}\,}\]
Period grows as √m, decreases as √k. Amplitude does not enter — isochronism.
13.12.1 Springs in Series and Parallel
| Configuration | Effective k | Period |
|---|---|---|
| Two springs (k₁, k₂) in parallel (both hold the same mass) | k_eff = k₁ + k₂ | T = 2π√(m/(k₁+k₂)) |
| Two springs in series | 1/k_eff = 1/k₁ + 1/k₂ | T = 2π√(m(k₁+k₂)/(k₁k₂)) |
13.13 Simple Pendulum
A point mass m on a massless inextensible string of length L. For small angular displacement θ, restoring torque about pivot = −mgL sin θ ≈ −mgL θ. Newton's rotational law gives:
m L² (d²θ/dt²) = −m g L θ ⇒ d²θ/dt² + (g/L) θ = 0
This is SHM with ω² = g/L. Hence:
\[\boxed{\,T_{\text{pendulum}} = 2\pi\sqrt{\dfrac{L}{g}}\,}\]
Period depends only on length and local gravity, not on mass or amplitude — provided amplitude is small (≲ 10°). This is exact only in the small-angle limit (sin θ ≈ θ).
13.13.1 Quick Estimates
- L = 1.0 m, g = 9.8 m/s²: T = 2π × √(1/9.8) = 2π × 0.319 = 2.01 s — the famous "seconds pendulum".
- L = 0.25 m: T = 2π × 0.16 = 1.0 s.
- On the Moon (g = 1.62 m/s²) the same 1 m pendulum gives T = 2π × √(1/1.62) = 4.94 s — about 2.45× slower.
13.14 Damped Oscillations
Real oscillators lose energy to friction or air drag. With damping force proportional to velocity (F_damp = −b v), the equation of motion becomes:
m d²x/dt² + b dx/dt + k x = 0
For weak damping (b² < 4mk), the solution is:
\[x(t) = A\, e^{-bt/(2m)}\, \cos(\omega'\, t + \varphi),\quad \omega' = \sqrt{\dfrac{k}{m} - \left(\dfrac{b}{2m}\right)^2}\]
The amplitude decays exponentially with time constant τ = 2m/b. The angular frequency ω' is slightly smaller than the undamped ω₀ = √(k/m).
| Damping regime | Condition | Behaviour |
|---|---|---|
| Underdamped | b² < 4mk | Decaying oscillation (most musical instruments, pendulum in air) |
| Critically damped | b² = 4mk | Returns to equilibrium fastest, no oscillation (door closer) |
| Overdamped | b² > 4mk | Slow exponential return, no oscillation (heavy oil-filled meter) |
13.15 Forced Oscillations & Resonance
Driving a damped oscillator with an external sinusoidal force F₀ cos(ω_d t) gives:
m d²x/dt² + b dx/dt + k x = F₀ cos(ω_d t)
After transient die-down, the steady-state response is at the driving frequency:
x(t) = A(ω_d) cos(ω_d t − φ)
The amplitude depends sensitively on ω_d:
\[A(\omega_d) = \dfrac{F_0/m}{\sqrt{(\omega_0^2 - \omega_d^2)^2 + (b\omega_d/m)^2}}\]
The amplitude is maximised when ω_d ≈ ω₀ = √(k/m). This phenomenon is resonance.
Worked Examples
Example 13.10: Period of a long pendulum
Find the length of a pendulum whose period is exactly 2.0 s at g = 9.81 m/s².
T = 2π√(L/g) ⇒ L = g T²/(4π²) = 9.81 × 4 / 39.48 = 0.994 m. Almost exactly 1 m — Huygens' "seconds pendulum".
Example 13.11: Damped amplitude after 1 minute
An undriven oscillator has m = 200 g and damping coefficient b = 0.040 kg/s. By what factor does its amplitude shrink in 60 s?
A(t) = A₀ exp(−bt/2m). Exponent = b·t/(2m) = 0.04 × 60 / (2 × 0.20) = 2.4/0.40 = 6.0.
Ratio = e^(−6.0) = 2.48 × 10⁻³ ⇒ amplitude falls to 0.25 % of its initial value.
Ratio = e^(−6.0) = 2.48 × 10⁻³ ⇒ amplitude falls to 0.25 % of its initial value.
Example 13.12: Series springs
Two springs k₁ = 100 N/m and k₂ = 300 N/m are connected end-to-end with a 0.50 kg mass attached to the lower one. Find the period of vertical oscillation.
Series: 1/k = 1/100 + 1/300 = 4/300 ⇒ k_eff = 75 N/m.
T = 2π√(m/k) = 2π√(0.5/75) = 2π × 0.0816 = 0.513 s.
T = 2π√(m/k) = 2π√(0.5/75) = 2π × 0.0816 = 0.513 s.
Interactive: Pendulum Period & Damping Calculator L3 Apply
Set L and local g; compute T. Add damping b to see amplitude after time t.
Activity 13.4 — Resonance with Coupled Pendulums
L4 Analyse
Predict: Two pendulums of different lengths hang from a horizontal string. If you start one swinging, does the other start to swing? When is the energy transfer maximum?
- Tie a horizontal string between two supports (~50 cm apart).
- Hang two pendulums (length L₁, L₂) from this string with bobs of similar mass.
- Start pendulum 1 swinging; observe pendulum 2 over the next minute.
- Adjust L₂ until L₂ = L₁ — what changes?
Observation: Pendulum 2 begins to swing if and only if its natural period (= 2π√(L₂/g)) matches that of pendulum 1. When L₂ = L₁, energy transfers fully through the string within seconds — full resonance. When L₂ ≠ L₁, almost no energy is exchanged. This is exactly how tuning forks and mechanical filters select frequencies.
Competency-Based Questions
A simple pendulum of length 0.5 m and bob mass 0.02 kg swings with maximum angular displacement θ_0 = 0.10 rad. g = 9.8 m/s².
Q1. L1 Remember Write the formula for the period of a simple pendulum and a spring oscillator.
Pendulum: T = 2π√(L/g). Spring: T = 2π√(m/k). The pendulum period depends on g; the spring period does not.
Q2. L2 Understand Why does the pendulum formula fail at large angles?
The restoring torque is −mgL sin θ. The SHM approximation uses sin θ ≈ θ, valid only when θ is small (≲ 10°). At larger angles, the period grows; e.g. at θ_0 = 90°, T is about 18% greater than 2π√(L/g).
Q3. L3 Apply Compute T, ν and v_max of the pendulum.
T = 2π√(0.5/9.8) = 2π × 0.226 = 1.42 s; ν = 0.704 Hz. Linear amplitude A = L θ_0 = 0.05 m. v_max = Aω = 0.05 × (2π/1.42) = 0.05 × 4.43 = 0.22 m/s.
Q4. L4 Analyse The pendulum is now placed in a freely falling lift. What is its period?
In free fall, effective g = 0. T = 2π√(L/0) → ∞ — the pendulum does not oscillate at all (no restoring force in the freely falling frame). The bob remains where it is set.
Q5. L5 Evaluate Soldiers crossing a suspension bridge are ordered to break step. Why?
If their stepping frequency matches the bridge's natural frequency, resonance amplifies the bridge's oscillation amplitude — possibly catastrophic (e.g. London Millennium Bridge 2000). Breaking step destroys the periodic forcing, preventing resonance.
Assertion-Reason Questions
Assertion (A): The period of a simple pendulum on the Moon is greater than on Earth.
Reason (R): g is smaller on the Moon and T ∝ 1/√g.
Answer: A. g_Moon ≈ 1.62 m/s² < g_Earth, so T = 2π√(L/g) is larger by factor √(g_E/g_M) ≈ 2.45.
Assertion (A): Resonance amplitude grows without bound when ω_d = ω₀.
Reason (R): Damping always limits the resonant amplitude.
Answer: D. A is false in real systems — damping caps the peak amplitude at F₀/(b ω₀). Only in the idealised b = 0 case would amplitude grow indefinitely. R is true.
Assertion (A): A critically damped oscillator returns to equilibrium faster than an underdamped one.
Reason (R): Critical damping has the optimal balance — no oscillation overshoot, fastest return.
Answer: A. Critical damping (b² = 4mk) returns to x = 0 fastest without oscillation — used in door-closers, galvanometers, car suspensions.
Frequently Asked Questions - Spring Pendulum Damped
What is the main concept covered in Spring Pendulum Damped?
In NCERT Class 11 Physics Chapter 13 (Oscillations), "Spring Pendulum Damped" 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 Spring Pendulum Damped useful in real-life applications?
Real-life applications of Spring Pendulum Damped 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 Spring Pendulum Damped?
Key formulas in Spring Pendulum Damped (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. Spring Pendulum Damped 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 Spring Pendulum Damped?
CBSE board questions from Spring Pendulum Damped 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 Spring Pendulum Damped 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.
AI Tutor
Physics Class 11 Part II – NCERT (2025-26)
Ready