A) \[2\sqrt{2}\]cm
B) 2 cm
C) \[\sqrt{2}\]cm
D) \[\frac{3}{\sqrt{2}}\]cm
Correct Answer: A
Solution :
If amplitude of S.H.M. is a, then at a distance \[\frac{a}{\sqrt{2}}\] from mean position half energy is as P.E. and half is as K.E. If \[a=4\]then \[\frac{4}{\sqrt{2}}=2\sqrt{2}cm\] At \[2\sqrt{2}\]half energy is as RE. and half is as K.E.
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