STANDARD Physics Simple Harmonic Motion MCQ Question
A particle executing simple harmonic motion with an amplitude 5 cm and a time period 0.2 s. The velocity and acceleration of the particle when the displacement is 5 cm is
0.5π m s⁻¹, 0 m s⁻²
0.5 m s⁻¹, -5π² m s⁻²
0 m s⁻¹, -5π² m s⁻²
0.5π m s⁻¹, -0.5π² m s⁻²
Correct Answer
Detailed Explanation
For SHM, ω = 2π/T = 10π rad s⁻¹. Velocity v = ω√(A² - x²) and acceleration a = -ω²x. When x = 5 cm, v = 0 m s⁻¹ and a = -5π² m s⁻².
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