Simple Harmonic Motion
NIOS · Class 12 · Physics
Flashcards for Simple Harmonic Motion — NIOS Class 12 Physics. Quick Q&A cards covering key concepts, definitions, and formulas.
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Explore the full setState the definition and mathematical condition for Simple Harmonic Motion (SHM).
Answer
Definition: A particle executes SHM if it moves to and fro about a fixed mean position under a restoring force that is directly proportional to its displacement from the mean position and is always di…
What is the difference between periodic motion and oscillatory motion? Give one example of each.
Answer
Periodic Motion: Any motion that repeats itself after a fixed interval of time (time period T). → Example: Revolution of Earth around the Sun (repeats every 365 days) — Non-oscillatory periodic motion…
Write the general equations representing displacement in SHM and define each term.
Answer
General Equations of SHM: y = a sin(ωt + φ₀) OR y = a cos(ωt + φ₀) Where: • y = displacement from mean position at time t (m) • a = amplitude (maximum displacement) (m) • ω = angular frequency…
Derive the expression for the time period T of a simple harmonic oscillator using the circle of reference method.
Answer
Derivation Using Circle of Reference: Step 1: Consider a point M moving with constant speed v in a circle of radius a (amplitude). At t=0, M is at X. The phasor OM rotates with angular velocity ω = v…
A tray of mass 9 kg oscillates with time period 1.0 s on a spring. When a block of mass M is placed on it, the period becomes 2.0 s. Find the mass M of the block.
Answer
Given: • Mass of tray, m₁ = 9 kg, T₁ = 1.0 s • Mass of tray + block, m₂ = 9 + M, T₂ = 2.0 s Formula: T = 2π√(m/k) → T² = 4π²m/k → m = kT²/4π² Step 1: For empty tray: 9 = k(1)²/4π² → k = …
A spring of force constant 1600 N/m has a mass of 4.0 kg attached to it. It is pulled 4.0 cm and released. Calculate: (i) Frequency (ii) Maximum acceleration (iii) Maximum speed
Answer
Given: • k = 1600 N/m • m = 4.0 kg • Amplitude a = 4.0 cm = 0.04 m Step 1 – Angular Frequency: ω = √(k/m) = √(1600/4) = √400 = 20 rad/s Step 2 – Frequency: ν = ω/2π = 20/2π = 20/6.28 ≈ 3.18 Hz Step…
Derive the expression for the time period of a simple pendulum.
Answer
Derivation: Setup: A bob of mass m hangs from a string of length l. It is displaced by a small angle θ from vertical. Step 1: Forces on bob: • Weight mg (downward) • Tension T (along string, upward)…
Why is the time period of a simple pendulum independent of the mass of the bob? What factors does it depend on?
Answer
Why independent of mass: Restoring force F = −mg sinθ ≈ −(mg/l)x Acceleration a = F/m = −(g/l)x The mass m cancels out. So acceleration (and hence ω and T) does not depend on mass. T = 2π√(l/g) Fac…
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