Join the Most Relevant JEE Main 2025 Test Series & get 99+ percentile! Join Now
Search any question & find its solution
Question: Answered & Verified by Expert
A particle of mass $200 \mathrm{gm}$ executes S.H.M. The restoring force is provided by a spring of force constant $80 \mathrm{~N} / \mathrm{m}$. The time period of oscillations is
PhysicsOscillationsJEE Main
Options:
  • A $0.31 \mathrm{sec}$
  • B $0.15 \mathrm{sec}$
  • C $0.05 \mathrm{sec}$
  • D $0.02 \mathrm{sec}$
Solution:
2360 Upvotes Verified Answer
The correct answer is: $0.31 \mathrm{sec}$
$T=2 \pi \sqrt{\frac{m}{k}}=2 \pi \sqrt{\frac{0.2}{80}}=0.31 \mathrm{sec}$

Looking for more such questions to practice?

Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.