#sine function

Forced Oscillations and Resonance

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

**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 cos (ω t + Φ) . A = 4 m, ω = 4 s⁻¹, Φ = (π/6) . At t = 0.25 : 4 × 0.25 + (π/6) = 1 + (π/6) ≈ 1.523 rad ≈ 87° . v = 4 × 4 cos 87° ≈ 16 × 0.052 ≈ 0.832 m/s . Applying x = A cos(ωt

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

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Displacement, Velocity and Acceleration in SHM

A particle’s displacement is \( x = 7 \sin (2t + \frac{\pi}{4}) \) (in m). What is its velocity at \( t = 0 \, \text{s}

**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 cos (ω t + Φ) . A = 7 m, ω = 2 s⁻¹, Φ = (π/4) . At t = 0 : v = 2 × 7 cos (π/4) = 14 × (√(2)/2) = 7√(2) ≈ 9.9 m/s . 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 \( x = 6 \sin (2\pi t + \frac{\pi}{4}) \) (in cm). What is its acceleration at \( t = 0.25 \, \tex

**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 = 2π s⁻¹, x = 0.06 cos (2π × 0.25 + (π/4)) = 0.06 cos ((π/2) + (π/4)) = 0.06 cos (3π/4) . cos (3π/4) = -(√(2)/2) , so x = -0.06 (√(2)/2) ≈ -0.0424 m . a = -(2π)² ×

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 \( x = 2 \sin (4t + \frac{\pi}{3}) \) (in m). What is its acceleration at \( t = 0 \, \text{s} \)?

**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(0) = 2 sin (π/3) = 2 × (√(3)/2) = √(3) ≈ 1.732 m . a = -4² × 1.732 = -16 × 1.732 ≈ -27.71 m/s² . Applying x = A cos(ωt + φ), v = -ωA

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 \( x = 6 \sin (3t - \frac{\pi}{4}) \) (in m). What is its acceleration at \( t = 0 \, \text{s} \)?

**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 = 3 s⁻¹, x(0) = 6 sin (-(π/4)) = -6 sin (π/4) = -6 × (√(2)/2) ≈ -4.24 m . a = -3² × (-4.24) = 9 × 4.24 ≈ 38.16 m/s² . Applying x = A cos(ωt + φ), v

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

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Wave Equation and Displacement Relation

A wave is described by \( y(x, t) = 0.02 \sin (25x - 75t) \), where \( x \) and \( y \) are in meters and \( t \) in sec

**Displacement relation** encodes λ = 2π/k and f = ω/2π. Comparing given equation y = a sin(kx - ωt) with standard form yields k and ω, hence λ = 2π/k and v = ω/k, essential for identifying propagation characteristics and phase. Compare with y = a sin (kx - ω t) , ω = 75 rad/s . Period: T = (2π/ω) = (2π/75) ≈ 0.0838 s . Using v = fλ and standing-wave condition fₙ = n v/(2L) or v/(4L) as applicable, calculation yields 0.08 s, illustrating frequency-length-speed interdependence and quantization by boundaries.

Ref: NCERT > Physics Book > Waves > Wave Equation and Displacement Relation

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