JEE Mains · Physics · STD 11 - 13. oscillations
A particle is executing simple harmonic motion \((SHM)\) of amplitude \(A,\) along the \(x-\) axis, about \(x = 0.\) When its potential energy \((PE)\) equals kinetic energy \((KE),\) the position of the particle will be
- A \(\frac {A}{2}\)
- B \(\frac {A}{2\sqrt 2}\)
- C \(\frac {A}{\sqrt 2}\)
- D \(A\)
Answer & Solution
Correct Answer
(C) \(\frac {A}{\sqrt 2}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{PE}=\mathrm{KE}\) \(\Rightarrow \quad \frac{1}{2} m \omega^{2}\left(A^{2}-x^{2}\right)=\frac{1}{2} m \omega^{2} x^{2}\) \(\Rightarrow \quad x=\frac{A}{\sqrt{2}}\)
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