#oscillations

Forced Oscillations and Resonance

What is the significance of the phase constant in the displacement equation of SHM?

**Forced oscillations** result when external periodic driving force F = F₀ cos(ω_d t) acts on oscillator, steady-state frequency equals driving frequency ω_d, amplitude A = F₀/√((k - m ω_d²)² + (b ω_d)²) depends on proximity to natural frequency ω₀ = √(k/m). Resonance when ω_d ≈ ω₀, amplitude maximum. The phase constant ( Φ in x = A cos (ω t + Φ) ) determines the initial position and velocity, setting the starting point of the oscillation cycle. Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result It fixes the initial position

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

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Forced Oscillations and Resonance

A particle in SHM has \( x = 4 \cos (2t + \frac{\pi}{3}) \) (in m). What is its speed at \( t = 0.5 \, \text{s} \)? (Tak

**Resonance phenomenon** amplifies response when driving frequency matches natural frequency ω_d ≈ ω₀, large amplitude even with small F₀, as damping limits growth. Natural frequency determined by system parameters, resonance condition crucial for understanding vibrations and energy absorption, e.g., bridge collapse, tuning. Velocity: v = -ω A sin (ω t + Φ) . A = 4 m, ω = 2 s⁻¹, Φ = (π/3) . At t = 0.5 : 2 × 0.5 + (π/3) = 1 + (π/3) ≈ 2.047 rad ≈ 117° . v = -2 × 4 sin 117° ≈ -8 sin (180° - 63°) ≈ -8 × 0.838 ≈ -6.7 m/s

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

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Forced Oscillations and Resonance

A body oscillates with SHM according to \( x = 4 \cos (2\pi t + \frac{\pi}{6}) \) (in SI units). What is its velocity at

**Resonance phenomenon** amplifies response when driving frequency matches natural frequency ω_d ≈ ω₀, large amplitude even with small F₀, as damping limits growth. Natural frequency determined by system parameters, resonance condition crucial for understanding vibrations and energy absorption, e.g., bridge collapse, tuning. Velocity: v(t) = -ω A sin (ω t + Φ) . Here, A = 4 m, ω = 2π s⁻¹, Φ = (π/6) . At t = 0.5 s : ω t + Φ = 2π × 0.5 + (π/6) = π + (π/6) = (7π/6) . sin (7π/6) = sin (180° + 30°) = -sin 30° = -(1/2) . v = -2π

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

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Forced Oscillations and Resonance

A particle in SHM has \( a = -25 x \) (in SI units). What is its period?

**Forced oscillations** result when external periodic driving force F = F₀ cos(ω_d t) acts on oscillator, steady-state frequency equals driving frequency ω_d, amplitude A = F₀/√((k - m ω_d²)² + (b ω_d)²) depends on proximity to natural frequency ω₀ = √(k/m). Resonance when ω_d ≈ ω₀, amplitude maximum. For SHM, a = -ω² x . Given a = -25 x , ω² = 25 ⇒ ω = 5 rad/s . Period: T = (2π/ω) = (2π/5) ≈ 1.256 s . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 1.256 s

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

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Equations of SHM, Phase and Angular Frequency

A spring-mass system has \( m = 1.6 \, \text{kg}, k = 640 \, \text{N/m} \). What is its angular frequency?

**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°. ω = √((k/m)) = √((640/1.6)) = √(400) = 20 rad/s . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 20 rad/s 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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Equations of SHM, Phase and Angular Frequency

What critical condition must the acceleration satisfy for a motion to be classified as simple harmonic?

**SHM representation** using sine or cosine equivalent with phase offset, ω relates to system parameters like mass and stiffness. Understanding ω and φ permits prediction of position at any time and comparison of two SHM via phase difference Δφ = φ₂ - φ₁. In SHM, acceleration must be proportional to displacement and directed opposite to it ( a = -ω² x ), ensuring harmonic oscillation about the mean position. Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result It is proportional to displacement and opposite in direction follows, reflecting SHM dependence

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

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Equations of SHM, Phase and Angular Frequency

Which function represents periodic motion but not SHM? (\( \omega \) is a positive constant)

**General equation of SHM** x = A sin(ωt + φ) or A cos(ωt + φ) includes amplitude A (m), angular frequency ω = √(k/m) (rad/s) for spring system, and initial phase φ (rad) setting t=0 position. Phase (ωt + φ) determines instantaneous state, phase difference Δφ governs interference of two SHM motions. (a) 3 sin (ω t + (π/4)) : SHM. (b) cos² ω t : Periodic (period (π/ω) ), not SHM (not sinusoidal). (c) 2 cos (2ω t) : SHM. (d) e⁻ω t : Not periodic. Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E

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

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Energy in SHM - Kinetic, Potential and Total

What distinguishes the energy transformation in SHM from that in uniform circular motion?

**Energy distribution** shows maximum kinetic at equilibrium and maximum potential at extremes, sum constant. This relation enables calculation of amplitude, velocity at any displacement via v = ±√(2(E-U)/m), and understanding of energy storage in oscillating system for NEET problems. In SHM, energy oscillates between kinetic and potential forms, while in uniform circular motion, kinetic energy remains constant due to constant speed. Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result SHM cycles between kinetic and potential energy follows, reflecting SHM dependence on amplitude A, ω and system parameters.

Ref: NCERT > Physics Book > Oscillations > Energy in SHM - Kinetic, Potential and Total

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Periodic Motion and SHM Basic Concepts

In SHM, if the displacement is at its maximum positive value, what can be said about the direction of the acceleration?

**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 in SHM is a = -ω² x . At maximum positive displacement ( x = A ), a = -ω² A , which is negative, meaning it acts towards the mean position. Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and

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

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Periodic Motion and SHM Basic Concepts

In SHM, what is the phase relationship between displacement and acceleration?

**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). Displacement ( x = A cos (ω t + Φ) ) and acceleration ( a = -ω² A cos (ω t + Φ) ) are 180° ( π radians) out of phase, as acceleration is the negative of displacement. 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 in SHM has \( x = 3 \cos (2t + \frac{\pi}{6}) \) (in m). What is its speed at \( t = 0.5 \, \text{s} \)? (Tak

**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 + Φ) . A = 3 m, ω = 2 s⁻¹, Φ = (π/6) . At t = 0.5 : 2 × 0.5 + (π/6) = 1 + (π/6) = (π/3) + (π/6) = (π/2) . v = -2 × 3 sin ((π/2)) = -6 × 1 = -6 m/s . Applying x = A cos(ωt

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

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Simple Pendulum and Angular SHM

A simple pendulum has a length of \( 0.4 \, \text{m} \) and oscillates with \( g = 9.8 \, \text{m/s}^2 \). What is its p

**Pendulum motion** exhibits isochronism for small amplitudes, period depends only on L and g. Using g = 9.8 m/s², T calculation requires √(L/g), frequency reciprocal of period. Angular frequency directly √(g/L), e.g., L = 0.25 m gives T ≈ 1.0 s. Period: T = 2π √((L/g)) = 2π √((0.4/9.8)) ≈ 2 × 3.14 √(0.0408) ≈ 1.27 s . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 1.27 s follows, reflecting SHM dependence on amplitude A, ω and system parameters.

Ref: NCERT > Physics Book > Oscillations > Simple Pendulum and Angular SHM

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