JEE Main 2024 Physics question
A particle executes simple harmonic motion with an amplitude of 4 cm. At the mean position, velocity of the particle is 10 cm/s. The distance of the particle from the mean position when its speed becomes 5 cm/s is cm, where α = .
Answer: 12
Solution
= ωA = 10 cm/s. From v = ω√(A² − x²): 5 = 10√(1 − x²/16), so x² = 12 cm² and x = √12 cm, giving α = 12.