Question

At mean position during simple harmonic motion

P.E is min and K.E is max?

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Answer 1

Answer:

Yes.

Explanation:

You know, KE = $\frac{1}{2} k(A^{2} -x^{2} )$

at mean x=0, so KE is maximum.

Since Total Mechanical Energy is always conserved.

TME = KE + PE

Therefore PE has to be minimum.

Answer 2

Answer:

At mean position kinetic energy is maximum and potential energy is minimum