Physics Circle home Student Login

Home›Chapters›Oscillations PYQs›Velocity & Acceleration in SHM›Question 72

NEET 2015 (AIPMT, Cancelled Paper) · Velocity & Acceleration in SHM

Velocity & Acceleration in SHM question from NEET 2015 (AIPMT, Cancelled Paper)

Q72NEET 2015 · AIPMT · CANCELLED PAPERMedium

Velocity & Acceleration in SHM

A particle is executing SHM along a straight line. Its velocities at distances x1 and x2 from the mean position are V1 and V2 respectively. Its time period is

Show answer and solution

Answer: (d)

  1. V12=ω2(A2−x12)V_1^2 = \omega^2(A^2 - x_1^2), V22=ω2(A2−x22)V_2^2 = \omega^2(A^2 - x_2^2).
  2. Subtract: V12−V22=ω2(x22−x12)V_1^2 - V_2^2 = \omega^2(x_2^2 - x_1^2).
  3. T=2πω=2πx22−x12V12−V22T = \frac{2\pi}{\omega} = 2\pi\sqrt{\frac{x_2^2 - x_1^2}{V_1^2 - V_2^2}}.
  4. So the answer is (d).
Report a mistake

More questions on Velocity & Acceleration in SHM