#cosine function

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 \( x = 4 \cos (2t - \frac{\pi}{6}) \) (in m). What is its speed at \( t = 0.5 \, \text{s} \)? (Tak

**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. Velocity: v = -ω A sin (ω t + Φ) . A = 4 m, ω = 2 s⁻¹, Φ = -(π/6) . At t = 0.5 : 2 × 0.5 - (π/6) = 1 - (π/6) ≈ 0.476 rad ≈ 27.3° . v = -2 × 4 sin (27.3°) ≈ -8 × 0.46 ≈ -3.68 m/s

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

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Forced Oscillations and Resonance

A particle’s x-projection from circular motion is \( x = 8 \cos (\pi t) \) (in m). What is its maximum speed?

**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. Maximum speed: vₘₐₓ = ω A . A = 8 m, ω = π s⁻¹ . vₘₐₓ = π × 8 ≈ 3.14 × 8 ≈ 25.12 m/s . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 25.12 m/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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Spring-Mass System and Combination of Springs

A particle’s motion is given by \( x = 2 \cos (5t - \frac{\pi}{4}) \) (in m). What is its kinetic energy at \( x = 1 \,

**Oscillation of mass attached to spring** follows Hooke's law F = -k x, SHM with ω = √(k/m). Effective stiffness for two identical springs in parallel doubles, k_eff = 2k, increasing frequency by √2, while series halves stiffness to k/2, lowering frequency. Energy E = ½ k_eff A². Total energy: E = (1/2) k A² = (1/2) m ω² A² = 0.5 × 1 × 5² × 2² = 50 J . Potential energy: U = (1/2) m ω² x² = 0.5 × 1 × 25 × 1² = 12.5 J . Kinetic energy: K = E - U = 50 - 12.5 = 37.5

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

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Energy in SHM - Kinetic, Potential and Total

A particle’s displacement is \( x = 4 \cos (3\pi t + \frac{\pi}{6}) \) (in m). What is its velocity at \( t = 0 \, \text

**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. Velocity: v = -ω A sin (ω t + Φ) . A = 4 m, ω = 3π s⁻¹, Φ = (π/6) . At t = 0 : v = -3π × 4 sin (π/6) = -12π × 0.5 ≈ -18.84 m/s . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ),

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

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Energy in SHM - Kinetic, Potential and Total

A particle in SHM has \( x = 4 \cos (3t) \) (in m). What is its maximum acceleration?

**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 acceleration: aₘₐₓ = ω² A . A = 4 m, ω = 3 s⁻¹ . aₘₐₓ = 3² × 4 = 9 × 4 = 36 m/s² . Applying x = A cos(ωt

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

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Periodic Motion and SHM Basic Concepts

A particle in SHM has \( x = 6 \cos (4t) \) (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 = 6 m, ω = 4 s⁻¹ . aₘₐₓ = 4² × 6 = 16 × 6 = 96 m/s² . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 96 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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Periodic Motion and SHM Basic Concepts

A particle in SHM has a displacement \( x = 3 \cos (4t + \frac{\pi}{3}) \) (in meters). What is its acceleration at \( t

**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(t) = -ω² x(t) . Here, ω = 4 s⁻¹, x(0) = 3 cos ((π/3)) = 3 × 0.5 = 1.5 m . a(0) = -4² × 1.5 = -16 × 1.5 = -24 m/s² . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ),

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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Displacement, Velocity and Acceleration in SHM

A particle in SHM has \( x = 6 \cos (2\pi t + \frac{\pi}{4}) \) (in m). What is its speed at \( t = 0.25 \, \text{s} \)?

**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 = 6 m, ω = 2π s⁻¹, Φ = (π/4) . At t = 0.25 : 2π × 0.25 + (π/4) = (π/2) + (π/4) = (3π/4) . v = -2π × 6 sin (3π/4) = -12π × (√(2)/2) ≈ -26.64 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 \( v = -15 \cos (3t) \) (in m/s). What is its amplitude?

**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. Velocity: v = -ω A sin (ω t) , but given v = -15 cos (3t) . ω = 3 s⁻¹, vₘₐₓ = ω A = 15 ⇒ A = (15/3) = 5 m . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and E = ½kA² = ½mω²A², result 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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