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BITSAT 2023Paper 1
Exam
BITSAT
Year
2023
Paper
Paper 1
Subject
Physics
Chapter
Oscillations

BITSAT 2023 · Paper 1 · Physics · Oscillations

A particle executes SHM of amplitude A. The distance from the mean...

A particle executes SHM of amplitude A. The distance from the mean position when it's kinetic energy becomes equal to its potential energy is

Options

  1. A2A\sqrt{2 \mathrm{ A}}
  2. B2A2 \mathrm{ A}
  3. C12A\frac{1}{\sqrt{2}} \mathrm{ A}
  4. D12A\frac{1}{2} \mathrm{ A}

Correct answer

C. 12A\frac{1}{\sqrt{2}} \mathrm{ A}

Step-by-step solution

Let the distance from the mean position is X\mathrm{X}.
Given KE=PE\mathrm{KE}=\mathrm{PE}
So, 12Mω2(A2−x2)=12Mω2x2\frac{1}{2} M \omega^2\left(A^2-x^2\right)=\frac{1}{2} M \omega^2 x^2
A2−x2=x2⇒A2=2×2A^2-x^2=x^2 \Rightarrow A^2=2 \times 2
∴x=±A2\therefore \mathrm{x}= \pm \frac{\mathrm{A}}{\sqrt{2}}
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