A mass is oscillating with amplitude A at the end of a spring. How far (in terms

Q: A mass is oscillating with amplitude A at the end of a spring. How far (in terms

Apply, elastic potential energy = (1/3)kinetic energy 0.5*k*d^2 = (1/3)*0.5*m*v^2 but maximum potential energy is equal to some of potential and kinetic energy. 0.5*k*A^2 = 0.5*k*d^2 + (1/3)*0.5*m*v^2 0.5*k*A^2 = 0.5*k*d^2 + 0.5*k*d^2 0.5*k*A^2 = k*d^2 0.5*A^2 = d^2 ==> d = A/sqrt(2)

ID: 1401653 • Letter: A

Question

A mass is oscillating with amplitude A at the end of a spring.

How far (in terms of A) is this mass from the equilibrium position of the spring when the elastic potential energy equals a third the kinetic energy?

d= (answer) A

I have attempted this problem many times incorrectly. The answers I have inputted have been A/sqrt2, sqrt2, A sqrt(1/2), A/2, 1/sqrt2, .707. Apprently the correct answer does not depend on A in the solution because A is already listed in d= (answer) A. Please help and don't answer with any of the above previously incorrectly used solutions.

Upvote and positive rating for correct answer. Thank you!

Explanation / Answer

Apply,

elastic potential energy = (1/3)kinetic energy

0.5*k*d^2 = (1/3)*0.5*m*v^2

but maximum potential energy is equal to some of potential and kinetic energy.

0.5*k*A^2 = 0.5*k*d^2 + (1/3)*0.5*m*v^2

0.5*k*A^2 = 0.5*k*d^2 + 0.5*k*d^2

0.5*k*A^2 = k*d^2

0.5*A^2 = d^2

==> d = A/sqrt(2)

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A mass is oscillating with amplitude A at the end of a spring. How far (in terms

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