Quantrex Academy · Free JEE Advanced PYQ solutions
JEE Advanced Physics Oscillations 2022 JEE Advanced 2022 (Paper 2)

JEE Advanced Physics Question (2022) — Solution

Question

On a frictionless horizontal plane, a bob of mass m = 0 . 1   kg  is attached to a spring with natural length l 0 = 0 . 1   m . The spring constant is k 1 = 0 . 009   N   m - 1 when the length of the spring l > l 0  and is k 2 = 0 . 016   N   m - 1  when  l < l 0 . Initially the bob is released from l = 0 . 15   m . Assume that Hooke's law remains valid throughout the motion. If the time period of the full oscillation is T = n π s , then the integer closest to n is _______.

Options

  1. A. A
  2. B. B
  3. C. C
  4. D. D

Answer

A. A

Step-by-step solution

Angular frequency for different values of spring constant will be given by,  ω 1 = k 1 m  and  ω 2 = k 2 m Time period for a spring of fix spring constant,  k  is given by,  = 2 π m k . As per question, spring constant is having different values, hence we will have to find half of the time period for the given parts, first for the right half part of the oscillations & next for the left half part of the oscillation from the mean position. ∴   Time period  = π m k 1 + π m k 2 = π 0 . 1 0 . 009 + π 0 . 1 0 . 016 = π 0 . 3 + π 0 . 4 = π × 4 + 3 12 × 10 = 70 12 π = 5 . 83 π Therefore,  n = 6 .

Practice more on Quantrex App →

Related: Physics — Oscillations · All PYQ Banks