(a) At what displacement from equilibrium is the energy of a SHO one-sixth KE an

Question

(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.)

5/6????A(which is the correct answer)

(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?

kinetic fraction: (I got 16/25)

potential fraction: (I got 9/25)

I figured out how to do (a), and I thought I figured out the correct way to do (b), but unfortunately, it says I'm wrong. How would you do part (b).

Am I correct? What am I doing wrong?

Explanation / Answer

1) 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

So, you have your answers interchanged

Related Questions

Navigate

• Browse (All)
• Browse #
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.