JEE Main 2024 Physics question
A particle of mass 0.50 kg executes simple harmonic motion under force F = −50(Nm⁻¹)x. The time period of oscillation is s. The value of x is ................ (Given π = 22/7)
Answer: 22
Solution
ma = −50x gives a = −100x, so ω² = 100 and ω = 10 rad/s. The time period is
See the full step-by-step solution
Create a free account to read the complete working, and practise questions like this in a timed test.
Create a free account