When a particle is performing linear SHM its K.E. is two times its P.E. at a position A and its P.E. is two times its K.E. at another position B. Find the ratio of K.E, at A to K.E. at B.

When a particle is performing linear SHM its K.E. is two times its P.E

1.	When a particle is performing linear SHM its K.E. is two times its P.E. at a position A and its P.E. is two times its K.E. at another position B. Find the ratio of K.E, at A to K.E. at B.
Answer» SOLUTION :TE at A = K.E. at A+P.E at A. But K.E. at A= 2(P.E. at A) `P.E. "at" A= (1)/(2)(K.E. at A)= (3)/(2)K.E. at A` Similarly, T.E at B = K.E. at B + P.E. at B. But P.E, at B = 2(K.E. at B) T.E. at B = K.E. at B+ 2 (K.E. at B) = 3 K.E. at B. By the principle of CONSERVATION of energy T.E. at A=T.E. at `B= (3)/(2)K.E`. at `A= 3K.E at B= ("K.E. at A")/("K.E. at B")`