Question

Derive an expression for K.E and P.E and total energy and represent variation of these with displacement for a particle executing SHM.

Answer 1

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

Hence, total energy is always constant .