A particle executes linear simple harmonic motion with amplitude 3 cm. When the particle is 2 cm from the mean position, the numerical magnitude of its velocity is equal to that of its acceleration. Then its time period in seconds is:
Options

Question

A particle executes linear simple harmonic motion with amplitude 3 cm. When the particle is 2 cm from the mean position, the numerical magnitude of its velocity is equal to that of its acceleration. Then its time period in seconds is:

Options

G1 Physics · Accepted Answer

Step 1: For SHM at displacement x: speed v = ω√(A² − x²), while acceleration magnitude |a| = ω²|x|. Step 2: With A = 3 cm and x = 2 cm, equating the numerical values gives ω√(9 − 4) = ω² × 2. Step 3: Therefore ω = √5/2 s⁻¹ in the numerical-unit convention used in the source. Step 4: Time period T = 2π/ω = 4π/√5 s. Final answer: (C) 4π/√5.