JEE Mains · Physics · STD 11 - Oscillations
- JEE Mains 2023
- Medium
- MCQ
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
Pick an option to check your answer.
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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