This MCQ module is based on: Shm Equations
TOPIC 25 OF 33
Shm Equations
🎓 Class 11
Physics
CBSE
Theory
Ch 13 – Oscillations
⏱ ~14 min
🌐 Language: [gtranslate]
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Shm Equations
13.5 SHM as Projection of Uniform Circular Motion
An astonishing geometric fact: simple harmonic motion is just the shadow of uniform circular motion on any diameter. Imagine a particle P moving uniformly on a circle of radius A with angular speed ω. Drop a perpendicular from P onto the x-axis to get the foot N. As P revolves, N oscillates back and forth — and N's motion is exactly SHM.
Let θ(t) = ωt + φ be the angle of P measured anti-clockwise from the x-axis. Then:
x_N(t) = A cos(ωt + φ) ; y_N(t) = A sin(ωt + φ)
The reference-circle picture lets us read off all SHM quantities geometrically.
13.6 Velocity in SHM
Differentiating the displacement once with respect to time:
\[v(t) = \dfrac{dx}{dt} = -A\,\omega\,\sin(\omega t + \varphi)\]
Using \(\sin^2 + \cos^2 = 1\):
v² = A²ω² − ω² x² ⇒ v(x) = ± ω √(A² − x²)
So velocity is maximum at the equilibrium position (x = 0): v_max = Aω. Velocity is zero at the extreme positions (x = ±A), where the particle momentarily reverses.
13.7 Acceleration in SHM
Differentiating again:
\[a(t) = \dfrac{d^2x}{dt^2} = -A\,\omega^2\,\cos(\omega t + \varphi) = -\omega^2\, x(t)\]
Acceleration is always directed opposite to displacement — towards the equilibrium position. This is the kinematic statement of the linear restoring force. Maximum acceleration |a|_max = Aω² occurs at the extremes (x = ±A); a = 0 at equilibrium.
| Quantity | Expression | Maximum | Zero at |
|---|---|---|---|
| Displacement x | A cos(ωt + φ) | A (at extremes) | x = 0 (equilibrium) |
| Velocity v | −A ω sin(ωt + φ) | A ω (at equilibrium) | extremes (x = ±A) |
| Acceleration a | −A ω² cos(ωt + φ) | A ω² (at extremes) | x = 0 (equilibrium) |
13.8 Force in SHM
By Newton's second law, F = m a = −m ω² x. Comparing with F = −k x (Hooke's law form):
\[k = m\omega^2 \quad\Longrightarrow\quad \omega = \sqrt{\dfrac{k}{m}},\quad T = 2\pi\sqrt{\dfrac{m}{k}},\quad \nu = \dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}\]
Heavier mass → slower oscillation; stiffer spring → faster oscillation. Note that amplitude does not affect the period — this is the famous isochronism of SHM.
Worked Examples
Example 13.4: Velocity and acceleration at an instant
An SHM has x(t) = 0.10 cos(4t + π/6) m. Find x, v and a at t = 0.
A = 0.10 m, ω = 4 rad/s, φ = π/6.
x(0) = 0.10 cos(π/6) = 0.10 × 0.866 = 0.0866 m.
v(0) = −0.10 × 4 × sin(π/6) = −0.4 × 0.5 = −0.20 m/s.
a(0) = −0.10 × 16 × cos(π/6) = −1.6 × 0.866 = −1.39 m/s² (= −ω² x ✓).
x(0) = 0.10 cos(π/6) = 0.10 × 0.866 = 0.0866 m.
v(0) = −0.10 × 4 × sin(π/6) = −0.4 × 0.5 = −0.20 m/s.
a(0) = −0.10 × 16 × cos(π/6) = −1.6 × 0.866 = −1.39 m/s² (= −ω² x ✓).
Example 13.5: Speed at given displacement
For an SHM with A = 0.05 m and T = 2 s, find the speed when x = 0.025 m.
ω = 2π/T = π rad/s. v = ω √(A² − x²) = π × √(0.05² − 0.025²) = π × √(0.001875) = π × 0.0433 = 0.136 m/s.
Example 13.6: Spring oscillator period
A 0.50 kg block is attached to a spring of force constant 200 N/m on a frictionless surface. Find ω, T, ν.
ω = √(k/m) = √(200/0.50) = √400 = 20 rad/s.
T = 2π/ω = 2π/20 = 0.314 s.
ν = 1/T = 3.18 Hz.
T = 2π/ω = 2π/20 = 0.314 s.
ν = 1/T = 3.18 Hz.
Interactive: Reference Circle Visualiser L3 Apply
Animate a particle on a circle and watch its x-projection trace SHM in real time. Adjust ω.
Activity 13.2 — Shadow Theatre of SHM
L4 Analyse
Predict: If you mount a torch on the wall and rotate a pencil tied to a thread (held vertical), where does its shadow on the floor move?
- Tie a small object to a string ~30 cm long.
- Spin it horizontally at a steady rate while shining a torch from the side onto a wall.
- Trace the back-and-forth motion of the shadow on the wall.
Observation: Despite the object moving uniformly on a circle, its shadow oscillates back and forth — exactly along a diameter. This is the geometric demonstration of SHM as the projection of UCM. Period of shadow = period of rotation; amplitude of shadow = circle radius.
Competency-Based Questions
A particle in SHM has amplitude A = 0.2 m and angular frequency ω = 5 rad/s, starting from x = +A at t = 0.
Q1. L1 Remember Write the equations for x(t), v(t), a(t).
x(t) = 0.2 cos(5t); v(t) = −1.0 sin(5t); a(t) = −5.0 cos(5t) = −25 x. (φ = 0 because x(0) = A.)
Q2. L2 Understand Why is the velocity zero at the extremes?
At the extremes the particle momentarily reverses direction; the velocity must change sign continuously, so it must pass through zero. Equivalently, all the energy is potential at the extremes and kinetic = 0.
Q3. L3 Apply Find v_max, a_max, and the speed at x = 0.1 m.
v_max = Aω = 0.2 × 5 = 1.0 m/s; a_max = Aω² = 0.2 × 25 = 5.0 m/s².
v(x=0.1) = ω√(A²−x²) = 5 × √(0.04 − 0.01) = 5 × 0.173 = 0.866 m/s.
v(x=0.1) = ω√(A²−x²) = 5 × √(0.04 − 0.01) = 5 × 0.173 = 0.866 m/s.
Q4. L4 Analyse At what time does the particle first reach x = +A/2?
A/2 = A cos(5t) ⇒ cos(5t) = 1/2 ⇒ 5t = π/3 ⇒ t = π/15 ≈ 0.209 s. The motion is decreasing from +A at t = 0 toward 0 at t = T/4 = π/10 = 0.314 s.
Q5. L5 Evaluate A student claims "doubling the amplitude doubles the period of an SHM." Critique using the relevant equations.
Wrong. T = 2π√(m/k) is independent of A — this is isochronism, the very property that makes pendulum clocks accurate. Doubling A doubles v_max and a_max but does not change ω or T.
Assertion-Reason Questions
Assertion (A): Acceleration in SHM is always opposite to displacement.
Reason (R): Acceleration = −ω² x.
Answer: A. The minus sign in a = −ω² x makes a antiparallel to x at every moment. R explains A directly.
Assertion (A): The maximum velocity in SHM is Aω.
Reason (R): v(t) = −A ω sin(ωt + φ), whose magnitude peaks at sin = ±1.
Answer: A. Direct mathematical fact.
Assertion (A): The displacement and acceleration in SHM differ by a phase of π/2.
Reason (R): The velocity differs from displacement by a phase of π/2.
Answer: D. A is false — displacement and acceleration differ by π (they are exactly opposite). R is true (v is the time-derivative of x, hence π/2 ahead).
Frequently Asked Questions - Shm Equations
What is the main concept covered in Shm Equations?
In NCERT Class 11 Physics Chapter 13 (Oscillations), "Shm Equations" 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 Shm Equations useful in real-life applications?
Real-life applications of Shm Equations 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 Shm Equations?
Key formulas in Shm Equations (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. Shm Equations 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 Shm Equations?
CBSE board questions from Shm Equations 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 Shm Equations 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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