JEE Mains · Physics · STD 11 - 13. oscillations
The kinetic energy and potential energy of a particle executing simple harmonic motion will be equal, when displacement (amplitude = \(a\)) is
- A \(\frac{a}{2}\)
- B \(a\sqrt 2 \)
- C \(\frac{a}{{\sqrt 2 }}\)
- D \(\frac{{a\sqrt 2 }}{3}\)
Answer & Solution
Correct Answer
(C) \(\frac{a}{{\sqrt 2 }}\)
Step-by-step Solution
Detailed explanation
(c) Suppose at displacement \(y\) from mean position potential energy = kinetic energy ==> \(\frac{1}{2}m({a^2} - {y^2}){\omega ^2} = \frac{1}{2}m{\omega ^{\rm{2}}}{y^2}\) ==> \({a^2} = 2{y^2}\) ==> \(y = \frac{a}{{\sqrt 2 }}\)
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