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Question

A particle in SHM has \( x = 6 \cos (2\pi t + \frac{\pi}{4}) \) (in m). What is its speed at \( t =
0.25 \, \text{s} \)? (Take \( \sin 45^\circ = \frac{\sqrt{2}}{2} \))

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Explanation

**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 =

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