Question

Starting from mean position body oscillates simple harmonically with a period of 2s. After what time will its kinetic energy will be 75% of the total energy

Answer 1

Kinetic energy K.E= (1/2)mv²

For an object in SHM,

K.E= (1/2)K(A²-x²)

     But, k= mω²

So K.E= (1/2)m(A.ωcosωt)²

Total energy, E= (1/2)mA²ω²

According to given condition

K.E= (75/100)E

(1/2)mA²ω²cos²ωt= (75/100)×(1/2)mA²ω²

cos²ωt= 3/4

cosωt= √3/2

ωt= π/6

t= π/ω6

We know, ω= 2πf= 2π/T

T= 2s (given)

 So, t=2π/(6×2π)= 1/6 s.