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Question: Answered & Verified by Expert
The period of oscillation of a mass $M$ suspended from a spring of negligible mass is $\mathrm{T}$. If along with it another mass $M$ is also suspended, the period of oscillation will now be
PhysicsOscillationsNEETNEET 2010 (Screening)
Options:
  • A $\mathrm{T}$
  • B $\mathrm{T} / \sqrt{2}$
  • C $2 \mathrm{~T}$
  • D $\sqrt{2} \mathrm{~T}$
Solution:
2365 Upvotes Verified Answer
The correct answer is: $\sqrt{2} \mathrm{~T}$
Time period of spring pendulum, $T=2 \pi \sqrt{\frac{M}{k}}$. If now mass in doubled $\mathrm{T}^{\prime}=2 \pi \sqrt{\frac{2 \mathrm{M}}{\mathrm{k}}}=\sqrt{2} \mathrm{~T}$

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