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Question

A particle’s displacement is \( x = 3 \cos (2\pi t - \frac{\pi}{4}) \) (in m). What is its velocity at
\( t = 0.25 \, \text{s} \)? (Take \( \sin 45^\circ = \frac{\sqrt{2}}{2} \))

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Explanation

**Effect of damping** is gradual amplitude reduction while period remains nearly constant for light damping. Mechanical energy decreases as work done against damping force, E(t) = ½ k A(t)² decaying exponentially, and motion ceases without external energy input, distinguishing from ideal undamped SHM. Velocity: v = -ω A sin (ω t + Φ) . A = 3 m, ω = 2π s⁻¹, Φ = -(π/4) . At t = 0.25 : 2π × 0.25 - (π/4) = (π/2) - (π/4) = (π/4) . v = -2π × 3 sin (π/4) = -6π × (√(2)/2) ≈ -13.32 m/s . Applying x = A cos(ωt + φ),

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