JEE Main 2024 Physics question
An object of mass 0.2 kg executes simple harmonic motion along x axis with frequency of Hz. At the position x = 0.04 m the object has kinetic energy 0.5 J and potential energy 0.4 J. The amplitude of oscillation is …………. cm.
Answer: 6
Solution
ω = 2πf = 50 rad/s. Total energy = 0.5 + 0.4 = 0.9 J = ½mω²A² = 0.1 × 2500 × A² ⇒ A² = 0.0036, A = 0.06 m = 6 cm.