#sinusoidal motion

Forced Oscillations and Resonance

A particle’s motion is \( x = 4 \sin (3t - \frac{\pi}{3}) \) (in m). What is its velocity at \( t = \frac{\pi}{6} \, \te

**Driven system** exhibits amplitude-frequency response peaking at resonance, phase shift between drive and displacement varying 0° to 180° across resonance. Forced oscillations sustain motion against damping, amplitude controlled by detuning |ω_d - ω₀| and damping strength b. Velocity: v = ω A cos (ω t + Φ) . A = 4 m, ω = 3 s⁻¹, Φ = -(π/3) . At t = (π/6) : 3 × (π/6) - (π/3) = (π/2) - (π/3) = (π/6) . v = 3 × 4 cos (π/6) = 12 × (√(3)/2) = 6√(3) ≈ 10.39 m/s . Applying x = A cos(ωt + φ), v = -ωA sin(ωt

Ref: NCERT > Physics Book > Oscillations > Forced Oscillations and Resonance

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Forced Oscillations and Resonance

A particle’s displacement is \( x = 5 \sin (2\pi t + \frac{\pi}{6}) \) (in m). What is its velocity at \( t = 0.5 \, \te

**Driven system** exhibits amplitude-frequency response peaking at resonance, phase shift between drive and displacement varying 0° to 180° across resonance. Forced oscillations sustain motion against damping, amplitude controlled by detuning |ω_d - ω₀| and damping strength b. Velocity: v = ω A cos (ω t + Φ) . A = 5 m, ω = 2π s⁻¹, Φ = (π/6) . At t = 0.5 : 2π × 0.5 + (π/6) = π + (π/6) = (7π/6) . v = 2π × 5 cos (7π/6) = 10π cos (180° - 30°) = 10π (-(√(3)/2)) ≈ -27.14 m/s . Applying x = A cos(ωt + φ), v

Ref: NCERT > Physics Book > Oscillations > Forced Oscillations and Resonance

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Damped Oscillations

A particle in SHM follows \( x = 2 \sin (4t + \frac{\pi}{2}) \) (in m). What is its acceleration at \( t = 0 \, \text{s}

**Real oscillators** experience damping, amplitude decreasing with time. Critical damping returns to equilibrium fastest without oscillation, overdamping slows return, underdamping shows decaying oscillations, classification based on b relative to 2mω₀, important for practical systems. Acceleration: a = -ω² x . ω = 4 s⁻¹, x(0) = 2 sin ((π/2)) = 2 × 1 = 2 m . a = -4² × 2 = -16 × 2 = -32 m/s² . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result -32 m/s² follows, reflecting SHM dependence on amplitude A, ω and system parameters.

Ref: NCERT > Physics Book > Oscillations > Damped Oscillations

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Equations of SHM, Phase and Angular Frequency

A mass oscillates with \( v = -12 \sin (6t) \) (in m/s). What is its amplitude?

**Angular frequency** ω = 2πf = 2π/T characterizes rapidity, independent of amplitude for SHM. Phase constant φ shifts sine/cosine, allowing any initial condition, e.g., x(0)=0 requires φ=0 for sine form. Displacement, velocity, acceleration share ω but differ in phase by 90° and 180°. Velocity: v = -ω A sin (ω t) . ω = 6 s⁻¹, vₘₐₓ = ω A = 12 ⇒ A = (12/6) = 2 m . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 2.0 m follows, reflecting SHM dependence on amplitude A, ω and system parameters.

Ref: NCERT > Physics Book > Oscillations > Equations of SHM, Phase and Angular Frequency

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Periodic Motion and SHM Basic Concepts

A particle in SHM has \( x = 5 \sin (4t - \frac{\pi}{6}) \) (in m). What is its acceleration at \( t = 0.25 \, \text{s}

**Distinction between periodic and oscillatory** clarifies all SHM is periodic but not all periodic is SHM. SHM requires linear restoring force and inertia, a ∝ -x, with ω = √(k/m). Functions like sin²ωt have period π/ω but lack a = -ω² x, thus periodic not SHM, while uniform circular motion is periodic without linear oscillation. Acceleration: a = -ω² x . ω = 4 s⁻¹, x = 5 sin (4 × 0.25 - (π/6)) = 5 sin (1 - (π/6)) ≈ 5 sin 0.476 ≈ 2.3 m . a = -4² × 2.3 = -16 × 2.3 ≈ -36.8 m/s² . Applying x = A

Ref: NCERT > Physics Book > Oscillations > Periodic Motion and SHM Basic Concepts

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Periodic Motion and SHM Basic Concepts

A particle in SHM has \( x = 3 \sin (2t) \) (in m). What is its speed at \( x = 1.5 \, \text{m} \)?

**Periodic motion** repeats after fixed period T, x(t+T)=x(t), while oscillatory motion involves to-and-fro about equilibrium. SHM is special periodic motion where restoring force proportional to displacement, F = -k x, acceleration a = -ω² x, leading to sinusoidal displacement x = A cos(ωt + φ). Velocity: v = ± ω √(A² - x²) . A = 3 m, ω = 2 s⁻¹, x = 1.5 m . v = 2 √(3² - 1.5²) = 2 √(9 - 2.25) = 2 √(6.75) ≈ 5.2 m/s . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA²

Ref: NCERT > Physics Book > Oscillations > Periodic Motion and SHM Basic Concepts

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Displacement, Velocity and Acceleration in SHM

A particle’s displacement is \( x = 8 \sin (2\pi t - \frac{\pi}{3}) \) (in m). What is its velocity at \( t = 0.25 \, \t

**Sinusoidal description** links amplitude A and angular frequency ω to instantaneous values. Given x(t) = A cos(ωt), max acceleration ω²A quantifies force requirement F_max = m ω²A, and velocity at arbitrary x is v = ±ω√(A² - x²) from energy conservation. Velocity: v = ω A cos (ω t + Φ) . A = 8 m, ω = 2π s⁻¹, Φ = -(π/3) . At t = 0.25 : 2π × 0.25 - (π/3) = (π/2) - (π/3) = (π/6) . v = 2π × 8 cos (π/6) = 16π × (√(3)/2) ≈ 43.54 m/s . Applying x = A cos(ωt + φ), v =

Ref: NCERT > Physics Book > Oscillations > Displacement, Velocity and Acceleration in SHM

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Displacement, Velocity and Acceleration in SHM

A mass oscillates with \( v = -10 \sin (5t) \) (in m/s). What is its amplitude?

**SHM kinematics** given by x = A cos(ωt + φ), velocity v = dx/dt = -ω A sin(ωt + φ), acceleration a = dv/dt = -ω² A cos(ωt + φ) = -ω² x, maxima v_max = ωA at mean position x=0, a_max = ω²A at extremes x=±A. Phase φ determines initial position, ω = 2π/T = √(k/m). Velocity: v = -ω A sin (ω t) . ω = 5 s⁻¹, vₘₐₓ = ω A = 10 ⇒ A = (10/5) = 2 m . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² =

Ref: NCERT > Physics Book > Oscillations > Displacement, Velocity and Acceleration in SHM

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Displacement, Velocity and Acceleration in SHM

A particle in SHM has \( x = 3 \sin (4t + \frac{\pi}{3}) \) (in m). What is its acceleration at \( t = 0.25 \, \text{s}

**SHM kinematics** given by x = A cos(ωt + φ), velocity v = dx/dt = -ω A sin(ωt + φ), acceleration a = dv/dt = -ω² A cos(ωt + φ) = -ω² x, maxima v_max = ωA at mean position x=0, a_max = ω²A at extremes x=±A. Phase φ determines initial position, ω = 2π/T = √(k/m). Acceleration: a = -ω² x . ω = 4 s⁻¹, x = 3 sin (4 × 0.25 + (π/3)) = 3 sin (1 + (π/3)) ≈ 3 sin 1.571 ≈ 3 m . a = -4² × 3 = -16 × 3 = -48 m/s² . Applying x

Ref: NCERT > Physics Book > Oscillations > Displacement, Velocity and Acceleration in SHM

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Displacement, Velocity and Acceleration in SHM

A particle in SHM has \( x = 5 \sin (2t) \) (in m). What is its speed at \( x = 2.5 \, \text{m} \)?

**SHM kinematics** given by x = A cos(ωt + φ), velocity v = dx/dt = -ω A sin(ωt + φ), acceleration a = dv/dt = -ω² A cos(ωt + φ) = -ω² x, maxima v_max = ωA at mean position x=0, a_max = ω²A at extremes x=±A. Phase φ determines initial position, ω = 2π/T = √(k/m). Velocity: v = ± ω √(A² - x²) . A = 5 m, ω = 2 s⁻¹, x = 2.5 m . v = 2 √(5² - 2.5²) = 2 √(25 - 6.25) = 2 √(18.75) ≈ 8.66 m/s . Applying x = A cos(ωt + φ), v

Ref: NCERT > Physics Book > Oscillations > Displacement, Velocity and Acceleration in SHM

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Displacement, Velocity and Acceleration in SHM

A mass oscillates with \( v = -15 \sin (6t) \) (in m/s). What is its amplitude?

**Sinusoidal description** links amplitude A and angular frequency ω to instantaneous values. Given x(t) = A cos(ωt), max acceleration ω²A quantifies force requirement F_max = m ω²A, and velocity at arbitrary x is v = ±ω√(A² - x²) from energy conservation. Velocity: v = -ω A sin (ω t) . ω = 6 s⁻¹, vₘₐₓ = ω A = 15 ⇒ A = (15/6) = 2.5 m . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 2.5 m follows, reflecting SHM dependence on amplitude A, ω and system parameters.

Ref: NCERT > Physics Book > Oscillations > Displacement, Velocity and Acceleration in SHM

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