MHT CET202526 Apr 2025Evening ShiftPhysicsOscillationsActual
A mass 'M' attached to a horizontal spring executes S.H.M. of amplitude A₁ . When the mass M passes through its mean position, then a smaller mass ' m ' is placed over it and both of them move together with amplitude A₂ . The ratio ( A₁ A₂ ) is
Options
- AM+m M
- BM M+m
- C( M+m M )^ 1 2
- D( M M+m )^ 1 2
Correct answer
C. ( M+m M )^ 1 2
Step-by-step solution
Solution: Let the spring constant be k , with initial angular frequency ₁ = k M for mass M oscillating with amplitude A₁ . The maximum velocity at mean position is v₁ = A₁ ₁ = A₁ k M . When mass m is added at mean position, momentum conservation yields (M + m)v₂ = M v₁ , so the new maximum velocity becomes v₂ = M M + m v₁ . The new angular frequency is ₂ = k M + m , and the corresponding maximum velocity is v₂ = A₂ ₂ = A₂ k M + m . Equating expressions for v₂ : A₂ k M + m = M M + m A₁ k M . Simplifying and solving