Q.
The displacement of a particle of mass 3g executing simple harmonic motion is given by Y=3sin(0.2t) in SI units. The kinetic energy of the particle at a point which is at a distance equal to 31 of its amplitude from its mean position is
Equation of SHM,Y=3sin(0.2t)
Comparing with Y=asinωt, we have a=3m,ω=0.2s−1
Mass of the particle =3g=3×10−3kg
Therefore, kinetic energy of the particle is K=21mω−3(a2−x2) =21×3×10−3×(0.2)2(32−12) (∵x=31) =0.48×10−3J