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When a particle is performing linear SHM its KE is two times its PE at a position ‘A’ and its PE is two times its KE at another position ‘B’. Find ratio of ${KE}_{A}$ to ${KE}_{B}$

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Correct option is B

at position A, KE=2PE⇒PE=KE2, TE=KE+PE⇒TE=KE+KE2⇒TEA=3KEA2 at position B, PE=2KE⇒ TE=KE+PE⇒TE=KE+2KE⇒TEB=3KEB By principle of conservation of energy TEA=TEB On equating the total energy equations, KEAKEB=21

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When a particle is performing linear SHM its KE is two times its PE at a position ‘A’ and its PE is two times its KE at another position ‘B’. Find ratio of KEA to KEB

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When a particle is performing linear SHM its KE is two times its PE at a position ‘A’ and its PE is two times its KE at another position ‘B’. Find ratio of ${KE}_{A}$ to ${KE}_{B}$