The period of oscillation of a mass M suspended from a spring of negligible mass is T. If along with it another mass M is also suspended, the period of oscillation will now be

Options

(a) T
(b) T / √2
(c) 2T
(d) √2T

Correct Answer:

√2T

Explanation:

T = 2π √(m / K)
T₁ / T₂ = √(M₁ /M₂)
T₂ = T₁√(M₂ / M₁) = T₁ √(2M / M)
T₂ = T₁ √2 = √2 T (where T₁ = T)