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Simple Harmonic Motion

Telangana Open School (TOSS) · Class 12 · Physics

Flashcards for Simple Harmonic Motion — Telangana Open School (TOSS) Class 12 Physics. Quick Q&A cards covering key concepts, definitions, and formulas.

42 questions30 flashcards5 concepts

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30 Flashcards
Card 1Periodic Motion

What is periodic motion?

Answer

Periodic motion is a type of motion that repeats itself after equal intervals of time. The smallest interval after which the motion repeats is called the time period (T). Example: motion of the hands

Card 2Oscillatory Motion

Define oscillatory motion. How is it different from periodic motion?

Answer

Oscillatory motion is a to-and-fro motion about a mean position. All oscillatory motions are periodic, but not all periodic motions are oscillatory. For example, the motion of Earth around the Sun is

Card 3Simple Harmonic Motion

What is Simple Harmonic Motion (SHM)?

Answer

Simple Harmonic Motion is a type of oscillatory motion where the acceleration of the particle is directly proportional to its displacement from the mean position and is always directed towards it. Mat

Card 4Conditions for SHM

State the condition for a motion to be simple harmonic.

Answer

For a motion to be simple harmonic, the restoring force must be directly proportional to the displacement and opposite in direction: F = -kx. This leads to acceleration a = -(k/m)x = -ω²x.

Card 5Displacement in SHM

A particle executes SHM with amplitude 5 cm and angular frequency 4 rad/s. Write the displacement equation assuming zero initial phase.

Answer

Given: A = 5 cm = 0.05 m, ω = 4 rad/s, φ = 0. Using: y(t) = A sin(ωt + φ) So, y(t) = 0.05 sin(4t) meters.

Card 6Velocity in SHM

The displacement of a particle in SHM is given by x = 4 sin(3t + π/6) cm. Find its maximum velocity.

Answer

Given: x = 4 sin(3t + π/6) cm → A = 4 cm = 0.04 m, ω = 3 rad/s Maximum velocity v_max = Aω = 0.04 × 3 = 0.12 m/s. Answer: 0.12 m/s.

Card 7Acceleration in SHM

A particle in SHM has a maximum acceleration of 18 m/s² and amplitude 0.06 m. Find its angular frequency.

Answer

Formula: a_max = Aω² Given: a_max = 18 m/s², A = 0.06 m So, 18 = 0.06 × ω² → ω² = 18 / 0.06 = 300 → ω = √300 ≈ 17.32 rad/s. Answer: 17.32 rad/s.

Card 8Simple Pendulum

What is the time period of a simple pendulum of length 1 m? (Take g = 9.8 m/s²)

Answer

Formula: T = 2π√(l/g) Given: l = 1 m, g = 9.8 m/s² T = 2π√(1/9.8) = 2π√(0.102) ≈ 2π × 0.319 ≈ 2.00 s. Answer: 2.00 seconds.

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