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
- A$\sqrt{2\,A }$
- B$2\,A$
- ✓$\frac{1}{\sqrt{2}}\,A$
- D$\frac{1}{2}\,A$
✓
Answer
Correct option: C.
$\frac{1}{\sqrt{2}}\,A$
c
$KE = PE$
$KE = PE$
$\frac{1}{2} M \omega^2\left( A ^2- x ^2\right)=\frac{1}{2} M \omega^2 x ^2$
$A ^2- x ^2= x ^2 \Rightarrow A ^2=2 \times 2$
$\Rightarrow x = \pm \frac{ A }{\sqrt{2}}$
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