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Question:

A body of mass m is attached to the lower end of a spring whose upper end is fixed. The spring has negligible mass. When the mass m is slightly pulled down and released, it oscillates with a time period of 3s. When the mass m is increased by 1kg, the time period of oscillations becomes 5s. Then the value of m in kg is?

43

169

916

34

Solution:

For SHM of a hanging mass by a spring, F = −ksx where ks = mω² ⇒ ω = √(ks/m) Hence time period of SHM = T = 2πω = 2π√(m/ks) ∝ √m Hence 5/3 = √((m+1)/m) ⇒ m = 9/16 kg