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 executes a simple harmonic motion of time period $T$. Find the time taken by the particle to go directly from its mean position to half the amplitude
PhysicsOscillationsJEE Main
Options:
  • A $T / 2$
  • B $T / 4$
  • C $T / 8$
  • D $T / 12$
Solution:
2293 Upvotes Verified Answer
The correct answer is: $T / 12$
$y=A \sin \omega t=\frac{A \sin 2 \pi}{T} t \Rightarrow \frac{A}{2}=A \sin \frac{2 \pi t}{T} \Rightarrow t=\frac{T}{12}$

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.