Physics Circle home Student Login

Home›Chapters›Oscillations PYQs›Velocity & Acceleration in SHM›Question 2

MHT CET 2026 (20 Apr, Shift 2) · Velocity & Acceleration in SHM

Velocity & Acceleration in SHM question from MHT CET 2026 (20 Apr, Shift 2)

Q2MHT CET 2026 · 20 APR · SHIFT 2Easy

Velocity & Acceleration in SHM

A particle is performing S.H.M. about x=0x = 0, with an amplitude aa and time period TT. The speed of the particle at x=a3x = \frac{a}{3} will be

Show answer and solution

Answer: (d)

  1. v=ωa2−x2v = \omega\sqrt{a^2 - x^2} with ω=2πT\omega = \frac{2\pi}{T}
  2. a2−a29=22a3\sqrt{a^2 - \frac{a^2}{9}} = \frac{2\sqrt{2}a}{3}
  3. v=2πT⋅22a3=42πa3Tv = \frac{2\pi}{T}\cdot\frac{2\sqrt{2}a}{3} = \frac{4\sqrt{2}\pi a}{3T}
  4. So the answer is (d) 42πa3T\frac{4\sqrt{2}\pi a}{3T}.
Report a mistake

More questions on Velocity & Acceleration in SHM