MHT CET20249 May 2024Evening ShiftPhysicsOscillationsActual
A particle is executing a linear simple harmonic motion. Let ' V ₁ ' and ' V ₂ ' are its speed at distance ' x₁ ' and ' x₂ ' from the equilibrium position. The amplitude of oscillation is
Options
- AV₁^2 x₂^2-V₂^2 x₂^2 V₁^2-V₂^2
- BV₁^2-V₂^2 V₁^2 x₂^2-V₂^2 x₁^2
- CV₁^2 x₂^2-V₂^2 x₁^2 V₁^2-V₂^2
- DV₁^2 x₁^2-V₂^2 x₂^2 V₁^2-V₂^2
Correct answer
A. V₁^2 x₂^2-V₂^2 x₂^2 V₁^2-V₂^2
Step-by-step solution
For S.H.M, velocity is given by, aligned & V= A^2-x^2 V^2= ^2 (A^2-x^2 )...(i) & V₁^2= ^2 (A^2-x₁^2 ) & and V₂^2= ^2 (A^2-x₂^2 ) ...[From(i)] & V₁^2 V₂^2 = ^2 (A^2-x₁^2 ) ^2 (A^2-x₂^2 ) & V₁^2 (A^2-x₂^2 )=V₂^2 (A^2-x₁^2 ) & A= V₁^2 x₂^2-V₂^2 x₁^2 V₁^2-V₂^2 aligned