Physics Circle home Student Login

Home›Chapters›Oscillations PYQs›Periodic Motion & SHM Basics›Question 75

JEE Main 2007 (AIEEE) · Periodic Motion & SHM Basics

Periodic Motion & SHM Basics question from JEE Main 2007 (AIEEE)

Q75JEE Main 2007 · AIEEEEasy

Periodic Motion & SHM Basics

The displacement of an object attached to a spring and executing simple harmonic motion is given by x=2×10−2cos⁡πtx = 2 \times 10^{-2}\cos \pi t metre. The time at which the maximum speed first occurs is

Show answer and solution

Answer: (b)

  1. ω=π\omega = \pi, so T=2ππ=2 sT = \frac{2\pi}{\pi} = 2\ \text{s}.
  2. x=Acos⁡πtx = A\cos\pi t starts at the extreme position; maximum speed is at the mean position.
  3. First reached after T4=0.5 s\frac{T}{4} = 0.5\ \text{s}.
  4. So the answer is (b) 0.5 s.
Report a mistake

More questions on Periodic Motion & SHM Basics