#harmonic motion

Forced Oscillations and Resonance

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

**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 = 3 m, ω = 2π s⁻¹, Φ = (π/3) . At t = 0.5 : 2π × 0.5 + (π/3) = π + (π/3) = (4π/3) . v = -2π × 3 sin (4π/3) = -6π sin (180° - 60°) = -6π (-(√(3)/2)) ≈ 16.31 m/s . Applying x =

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

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Forced Oscillations and Resonance

A particle in SHM has \( a = -64 x \) (in SI units). What is its period?

**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. For SHM, a = -ω² x . Given a = -64 x , ω² = 64 ⇒ ω = 8 rad/s . Period: T = (2π/ω) = (2π/8) = (π/4) ≈ 0.785 s . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 0.785 s follows, reflecting SHM dependence

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

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Forced Oscillations and Resonance

A simple pendulum has a length of \( 2.25 \, \text{m} \) and oscillates with \( g = 9.8 \, \text{m/s}^2 \). What is its

**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. Period: T = 2π √((L/g)) = 2π √((2.25/9.8)) ≈ 2 × 3.14 √(0.2296) ≈ 3.01 s . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 3.01 s follows, reflecting SHM dependence on amplitude A, ω and system parameters.

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

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Energy in SHM - Kinetic, Potential and Total

A particle’s x-projection from circular motion is \( x = 9 \cos (\pi t) \) (in m). What is its maximum speed?

**Energy in SHM** interconverts between kinetic K = ½ m v² = ½ m ω² (A² - x²) and potential U = ½ k x² = ½ m ω² x², total E = K + U = ½ k A² = ½ m ω² A² constant, independent of time. At mean position x=0, E = K_max = ½ m ω² A², at extremes x=±A, E = U_max = ½ k A². Maximum speed: vₘₐₓ = ω A . A = 9 m, ω = π s⁻¹ . vₘₐₓ = π × 9 ≈ 28.26 m/s . Applying x = A cos(ωt + φ), v =

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

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Periodic Motion and SHM Basic Concepts

A particle’s x-projection from circular motion is \( x = 7 \cos (3t) \) (in m). What is its maximum 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). Maximum acceleration: aₘₐₓ = ω² A . A = 7 m, ω = 3 s⁻¹ . aₘₐₓ = 3² × 7 = 9 × 7 = 63 m/s² . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 63 m/s² follows, reflecting SHM dependence on amplitude A, ω and system parameters.

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

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Spring-Mass System and Combination of Springs

Two identical springs (\( k = 90 \, \text{N/m} \)) are attached to a \( 1.8 \, \text{kg} \) mass as in Fig. 13.14. What

**Mass-spring dynamics** show T depends on mass and stiffness, independent of amplitude for ideal spring. Given T and m, k = 4π² m/T² extracted, and energy E = ½ k A² connects amplitude to total mechanical energy, illustrating isochronism. Effective kₑff = 2k = 2 × 90 = 180 N/m . T = 2π √((m/kₑff)) = 2π √((1.8/180)) = 2π √(0.01) = 2π × 0.1 ≈ 0.628 s . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 0.628 s follows, reflecting SHM dependence on amplitude A, ω and system parameters.

Ref: NCERT > Physics Book > Oscillations > Spring-Mass System and Combination of Springs

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Energy in SHM - Kinetic, Potential and Total

A mass of \( 1 \, \text{kg} \) on a spring with \( k = 100 \, \text{N/m} \) has \( A = 15 \, \text{cm} \). What is the k

**Conservation of mechanical energy** in undamped SHM implies total energy proportional to amplitude squared A² and spring constant k. E = ½ k A² allows amplitude determination from known E and k via A = √(2E/k), with k = m ω² linking dynamical and energetic descriptions for spring-mass system. Total energy: E = (1/2) k A² = 0.5 × 100 × (0.15)² = 1.125 J . Potential energy: U = (1/2) k x² = 0.5 × 100 × (0.075)² = 0.28125 J . Kinetic energy: K = E - U = 1.125 - 0.28125 = 0.84375 J . Applying x = A cos(ωt + φ),

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

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Simple Pendulum and Angular SHM

A simple pendulum has a length of \( 1.44 \, \text{m} \) and oscillates with \( g = 9.8 \, \text{m/s}^2 \). What is its

**Angular SHM** of pendulum results from restoring torque τ = -m g L sinθ ≈ -m g L θ for small θ, analogous to linear SHM with ω = √(g/L). This relation allows period determination from length and local gravity, illustrating gravitational influence on oscillation. Period: T = 2π √((L/g)) = 2π √((1.44/9.8)) ≈ 2 × 3.14 √(0.1469) ≈ 2.406 s . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 2.406 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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Periodic Motion and SHM Basic Concepts

A particle’s x-projection from circular motion is \( x = 5 \cos (2t) \) (in m). What is its maximum speed?

**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. Maximum speed: vₘₐₓ = ω A . A = 5 m, ω = 2 s⁻¹ . vₘₐₓ = 2 × 5 = 10 m/s . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 10 m/s

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

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Displacement, Velocity and Acceleration in SHM

Which of the following functions represents SHM? (Assume \( \omega \) is a positive constant)

**Velocity and acceleration in SHM** follow from differentiation, showing 90° phase lead of v over x and 180° for a over x. At mean position x=0, a=0, v=±ωA maximum; at extremes x=±A, v=0, a=∓ω²A maximum magnitude, illustrating energy conversion. For SHM, acceleration a = -ω² x . Check by differentiating twice: (a) x = sin ω t + cos 2ω t : Not SHM (different frequencies). (b) x = 2 sin (ω t + (π/4)) : v = 2ω cos (ω t + (π/4)), a = -2ω² sin (ω t + (π/4)) = -ω² x . SHM. (c) x = e⁻ω t : Not periodic,

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

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Simple Pendulum and Angular SHM

Why does the period of a spring-mass system differ fundamentally from that of a simple pendulum in terms of gravitationa

**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. The spring-mass period ( T = 2π √((m/k)) ) lacks gravitational dependence, relying on elasticity, while the pendulum’s period ( T = 2π √((L/g)) ) varies with gravity. Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result The restoring force excludes gravity 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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Simple Pendulum and Angular SHM

What distinguishes the restoring force in a spring-mass system from that in a simple pendulum at small amplitudes?

**Angular SHM** of pendulum results from restoring torque τ = -m g L sinθ ≈ -m g L θ for small θ, analogous to linear SHM with ω = √(g/L). This relation allows period determination from length and local gravity, illustrating gravitational influence on oscillation. The spring-mass system uses elastic force ( F = -kx ), constant with displacement, while the pendulum’s force ( F = -mg sin θ ≈ -mg θ ) derives from gravity and varies with angle. Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result It is

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

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