Oscillations for NEET UG
Anything that wobbles back and forth around a balance point oscillates. Simple harmonic motion is the cleanest case, and it describes springs, pendulums and even atoms in a solid.
PhysicsClass 1162 MCQs3 lessons10 flashcards3 memory tricks
What you'll learn
- Simple harmonic motion
Recognise SHM from its force law and write displacement, velocity and acceleration. - Springs and pendulums
Find the period of a mass on a spring and of a simple pendulum. - Damping and resonance
Explain damped oscillations, forced oscillations and resonance.
Key formulas
| Displacement | x = A sin(ωt + φ) |
|---|---|
| Velocity | v = ω√(A² − x²) |
| Acceleration | a = −ω²x |
| Angular frequency | ω = 2π/T = 2πν |
| Mass on a spring | T = 2π√(m/k) |
| Simple pendulum (small angles) | T = 2π√(L/g) |
| Springs in series | 1/k = 1/k₁ + 1/k₂ |
| Springs in parallel | k = k₁ + k₂ |
Memory tricks
In SHM, speed is maximum at the mean position (where acceleration is zero); acceleration is maximum at the extremes (where speed is zero).
T = 2π√(l/g) does not depend on the mass of the bob (for small swings). Longer string, slower swing.
Series: 1/k = 1/k₁ + 1/k₂ (softer). Parallel: k = k₁ + k₂ (stiffer). For a spring–mass system T = 2π√(m/k).
Try 3 NEET-style questions
Q1. The length of a simple pendulum with a period of 2 s at a place where g = 9.8 m/s² is about
- 9.8 m
- 0.25 m
- 0.99 m
- 2 m
Show answer
Answer: (3) 0.99 m
T = 2π√(l/g) gives l = gT²/(4π²) = 9.8 × 4 / (4 × 9.87) ≈ 0.99 m.
Q2. A block hung from a vertical spring stretches it by 9.8 cm at equilibrium. Taking g = 9.8 m/s², the period of small vertical oscillations is
- 0.2π s
- 2π s
- 0.02π s
- 0.1π s
Show answer
Answer: (1) 0.2π s
At equilibrium mg = kΔl, so m/k = Δl/g. T = 2π√(Δl/g) = 2π√(0.098/9.8) = 2π × 0.1 = 0.2π s.
Q3. In SHM, the total mechanical energy
- remains constant, while kinetic and potential energies continuously interchange
- is maximum at the mean position only
- is zero at the extreme positions
- keeps increasing with time
Show answer
Answer: (1) remains constant, while kinetic and potential energies continuously interchange
In undamped SHM, total energy ½kA² is constant. KE is maximum at the mean position, PE is maximum at the extremes.
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