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#trigonometry

5 public questions tagged with this topic.

Equations of SHM, Phase and Angular Frequency

A particle’s displacement is \( x = 5 \sin (\pi t + \frac{\pi}{3}) \) (in m). What is its velocity at \( t = 0 \, \text{

**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 cos (ω t + Φ) . A = 5 m, ω = π s⁻¹, Φ = (π/3) . At t = 0 : v = π × 5 cos ((π/3)) = 5π × 0.5 = 2.5π ≈ 7.85 m/s . Applying x = A cos(ωt + φ), v = -ωA sin(ωt + φ), a = -ω²x and

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

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Periodic Motion and SHM Basic Concepts

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

**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 A sin (ω t + Φ) . A = 3 m, ω = 4π s⁻¹, Φ = (π/3) . At t = 0 : v = -4π × 3 sin (π/3) = -12π × (√(3)/2) ≈ -32.58 m/s . 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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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 \( x = 6 \sin (3t - \frac{\pi}{4}) \) (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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