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(a) At what displacement from equilibrium is the energy of a SHO one-sixth KE and five-sixths PE? (Answer in terms of the amplitude A.)

(b) What fraction of the total energy of a SHO is kinetic and what fraction potential when the displacement is four-fifths of the amplitude?

Added Mon, 22 Jun '15

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a)

We know

PE = 1/2 k x^2 ; where x is the distance form the equilibrium;

And KE = 1/2 k (A^2 -x^2)

Given,

PE = 5/6 * TE

=>1/2 k(x^2) = 5/6 *1/2k A^2

=>x^2 =5/6A^2

=>x= sqrt(5/6)A;

b)x= 4/5 A

So, PE = 1/2 K 16/25 A^2 = 16/25 TE

So,

fraction = 16/25

Hence

KE = 1 - 16/25 = 9/25

Added Mon, 22 Jun '15

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