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
If $x=5 \sin \left(\pi t+\frac{\pi}{3}\right) \mathrm{m}$ represents the motion of a particle executing simple harmonic motion, the amplitude and time period of motion, respectively, are
PhysicsOscillationsNEETNEET 2024
Options:
  • A $5 \mathrm{~m}, 2 \mathrm{~s}$
  • B $5 \mathrm{~cm}, 1 \mathrm{~s}$
  • C $5 \mathrm{~m}, 1 \mathrm{~s}$
  • D $5 \mathrm{~cm}, 2 \mathrm{~s}$
Solution:
1686 Upvotes Verified Answer
The correct answer is: $5 \mathrm{~m}, 2 \mathrm{~s}$
$\begin{aligned} & \therefore x=5 \sin \left(\pi t+\frac{\pi}{3}\right) \mathrm{m} \\ & \text { Amplitude }=5 \mathrm{~m} \\ & \omega=\pi=\frac{2 \pi}{T} \\ & T=\frac{2 \pi}{\pi}=2 \mathrm{~s}\end{aligned}$

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.