JEE Mains · Physics · STD 11 - 13. oscillations
A particle executes SHM of amplitude A. The distance from the mean position when its's kinetic energy becomes equal to its potential energy is on
- A \(\sqrt{2\,A }\)
- B \(2\,A\)
- C \(\frac{1}{\sqrt{2}}\,A\)
- D \(\frac{1}{2}\,A\)
Answer & Solution
Correct Answer
(C) \(\frac{1}{\sqrt{2}}\,A\)
Step-by-step Solution
Detailed explanation
\(KE = PE\) \(\frac{1}{2} M \omega^2\left( A ^2- x ^2\right)=\frac{1}{2} M \omega^2 x ^2\) \(A ^2- x ^2= x ^2 \Rightarrow A ^2=2 \times 2\) \(\Rightarrow x = \pm \frac{ A }{\sqrt{2}}\)
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