A particle is vibrating in a simple harmonic motion with an amplitude of 4 cm. At what displacement from the equilibrium position, is its energy half potential and half kinetic
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a
1 cm
b
2cm
c
3 cm
d
22cm
answer is D.
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Detailed Solution
Let x be the point where K.E. = P.E. Hence 12mω2(a2−x2)=12mω2x2⇒2x2=a2 ⇒x=a2=42=22cm