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JEE Advanced · Physics · 15. Oscillations

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

  1. A \(-0.5 \mathrm{~m}, 5 \mathrm{~kg} \mathrm{~m} / \mathrm{s}\)
  2. B \(0.5 \mathrm{~m}, 0 \mathrm{~kg} \mathrm{~m} / \mathrm{s}\)
  3. C \(0.5 \mathrm{~m},-5 \mathrm{~kg} \mathrm{~m} / \mathrm{s}\)
  4. D \(-1 \mathrm{~m}, 5 \mathrm{~kg} \mathrm{~m} / \mathrm{s}\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(0.5 \mathrm{~m},-5 \mathrm{~kg} \mathrm{~m} / \mathrm{s}\)

Step-by-step Solution

Detailed explanation

\(\mathrm{F}=-20\left(\mathrm{x}-\frac{1}{2}\right)=-20 \mathrm{X} \quad\left(\mathrm{X}=\mathrm{x}-\frac{1}{2}\right)\)
\(\therefore\) Particle will perform SHM about \(\mathrm{x}=\frac{1}{2}\) with \(\omega=2 \mathrm{rad} / \mathrm{sec} \Rightarrow \mathrm{T}=\pi \mathrm{sec}\).
\(\therefore\) Phase covered in \(t=\frac{\pi}{4}\) second \(=90^{\circ}\).
Given particle is at rest at \(\mathrm{x}=1 \mathrm{~m} \Rightarrow \mathrm{x}=1\) is extreme position.
\(\therefore\) In \(\frac{\pi}{4}\) sec, it will be at equilibrium
\(\therefore \mathrm{x}=0.5 \mathrm{~m}\) and momentum \(=\mathrm{m} \omega \mathrm{A}=5 \times 2 \times 0.5=5 \mathrm{~kg} \mathrm{~m} / \mathrm{s}\) Direction will be towards -ve \(\mathrm{x}\).
Hence option (3)
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