Practice question
Question
A particle in SHM has \( x = 5 \sin (4t - \frac{\pi}{6}) \) (in m). What is its acceleration at \( t =
0.25 \, \text{s} \)? (Take \( \cos 30^\circ = \frac{\sqrt{3}}{2} \))
Explanation
**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 = -ω² x . ω = 4 s⁻¹, x = 5 sin (4 × 0.25 - (π/6)) = 5 sin (1 - (π/6)) ≈ 5 sin 0.476 ≈ 2.3 m . a = -4² × 2.3 = -16 × 2.3 ≈ -36.8 m/s² . Applying x = A
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