A mass m attached to free end of a spring executes SHM with a period of 1s. If the mass is increased by 3 kg the period of the oscillation increases by one second, the value of mass m is ___________ kg.

Solution

Physics sir · 4 steps

Do not hunt for k: take the ratio of the two periods and the spring constant cancels out.

1. Period of a spring. For a mass on a spring, T = 2π√(m/k). The same spring is used both times, so k stays fixed and T ∝ √m.
2. Two cases. With mass m the period is 1 s. With mass m + 3 it rises by one second, so it becomes 2 s.

   T₂/T₁ = √((m+3)/m) = 2/1

3. Square and solve. Squaring gives (m+3)/m = 4, so m + 3 = 4m and 3m = 3.
4. So. m = 1 kg, so the answer is 1.

The common slip: 3. Taking T ∝ m instead of T ∝ √m gives (m+3)/m = 2 and m = 3, not 1. The period goes as the square root of the mass.

The idea to keep. When the same spring carries different masses, use T ∝ √m and take a ratio. Doubling the period needs four times the mass, not twice.

Answer from the NTA final answer key. Solution by PrepWiser.