Solution: A/√2 from the mean position. where A is amplitude.
We know
Potential energy = (1/2) kx2 = (1/2) mω 2x2
Kinetic energy = (1/2) mv2 = (1/2) m[ω √ (A2-x2)]2 = (1/2) mω 2(A2-x2)
Where x is the required position about the mean position,
A is amplitude of SHM,
m is mass of particle executing SHM,
ω is the angular frequency of the particle.
Given, P.E. = K.E. & A = 10√2
so, (1/2) mω 2x2 = (1/2) mω 2(A2-x2)
so, x2 = A2/2
so, x = A/√2
so, x = 10√2/√2 = 10cm
Hence, at distance x=10cm from the mean position, the K.E is equal to P.E
