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A)Derivetheexpressionsforthekineticenergyandpotentialenergyofasimple harmonic oscillator.​

Answer»

Answer:

Explanation:

For SHM,

acceleration , a = -ω²y

F = ma = -mω²y

now, work , W = F.dy cos180° { because displacement and acceleration are in OPPOSITE direction so, cos180° taken }

W = ∫mω²y.dy = mω²y²/2

use standard form of SHM , y = Asin(ωt ± Ф)

W = mω²A²/2 sin²(ωt ± Ф)

We know, POTENTIAL energy is work DONE stored in system .

so, P.E = W = mω²A²/2 sin²(ωt ± Ф)

again, Kinetic energy , K.E = 1/2mv² , here v is velocity

we know, v = ωAcos(ωt ± Ф)

so, K.E = mω²A²/2cos²(ωt ± Ф)

TOTAL energy = K.E + P.E

= mω²A²/2 cos²(ωt ± Ф) + mω²A²/2 sin²(ωt ± Ф)

= mω²A²/2 [ cos²(ωt ± Ф) + sin²(ωt ± Ф) ] = mω²A²/2 [ ∵sin²α + cos²α = 1]

= mω²A²/2 = constant


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