JEE Main 2024 Physics question
The position, velocity and acceleration of a particle executing simple harmonic motion are found to have magnitudes of 4 m, 2 ms⁻¹ and 16 ms⁻² at a certain instant. The amplitude of the motion is √x m where x is .
Answer: 17
Solution
|a| = ω²x gives 16 = ω²(4), so ω = 2 rad/s. From v = ω√(A² − x²): A² = v²/ω² + x² = 1 + 16 = 17, so A = √17 m and x = 17.