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Physics · Simple Harmonic Motion

JEE Advanced 2024 — Paper 1 — Question 21

A block of mass 5 kg moves along the xx-direction subject to the force F=(−20x+10)NF=(-20 x+10) \mathrm{N}, with the value of xx in metre. At time t=0 st=0 \mathrm{~s}, it is at rest at position x=1 mx=1 \mathrm{~m}. The position and momentum of the block at t=(π/4)st=(\pi / 4) \mathrm{s} are

  1. Option A:

    −0.5 m,5 kg m/s-0.5 \mathrm{~m}, 5 \mathrm{~kg} \mathrm{~m} / \mathrm{s}

  2. Option B:

    0.5 m,0 kg m/s0.5 \mathrm{~m}, 0 \mathrm{~kg} \mathrm{~m} / \mathrm{s}

  3. Option C:

    0.5 m,−5 kg m/s0.5 \mathrm{~m},-5 \mathrm{~kg} \mathrm{~m} / \mathrm{s}

    Correct
  4. Option D:

    −1 m,5 kg m/s-1 \mathrm{~m}, 5 \mathrm{~kg} \mathrm{~m} / \mathrm{s}

Answer: C

Step-by-step solution

a=−20x−105=−4(x−12)a=-\frac{20 x-10}{5}=-4\left(x-\frac{1}{2}\right) So, x=12+12cos⁡(2t)mx=\frac{1}{2}+\frac{1}{2} \cos (2 t) m At t=π4sect=\frac{\pi}{4} \mathrm{sec} x=0.5 m\mathrm{x}=0.5 \mathrm{~m} v=−12×2=−1 m/sv=-\frac{1}{2} \times 2=-1 \mathrm{~m} / \mathrm{s} So, momentum P=−5 kg−m/sP=-5 \mathrm{~kg}-\mathrm{m} / \mathrm{s}

Solution figure

Answer key and solution verified before publishing.

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Exam
JEE Advanced 2024
Paper
Paper 1
Subject
Physics
Chapter
Simple Harmonic Motion
Topic
Kinematics of SHM, Phase and Energy in SHM

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