An SHM has amplitude 10√2cm.At what dist.from mean position its K.E

1.	An SHM has amplitude 10√2cm.At what dist.from mean position its K.E is equal to P.E?
Answer» 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

An SHM has amplitude 10√2cm.At what dist.from mean position its K.E is equal to P.E?

Answer»

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

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