Question

Consider a particle executing in simple harmonic motion with the displacement of, = sin ( + ). Analyze the kinetic and potential energy variations at different points with the variation of time.​

Answer 1

Explanation:

Correct option is

D

II, IV

Given : x=Acosωt

Potential energy, U=21mω2x2=21mω2A2cos2ωt

On plotting a graph between potential energy U and time t, we shall get a curve II and graph between U and x we shall get curve IV.

Answer 2

Answer:

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

Explanation:

hope it's helpful dear mark as brainliest plss