Question

A linear harmonic oscillator has a total mechanical energy of 200 J. Potential energy of it at mean position is 50 J. Find,

(i) the maximum kinetic energy

(ii) the minimum potential energy

(iii) the potential energy at extreme positions.

Answer 1

At mean position, potential energy is minimum and kinetic energy is maximum. Hence,

U_min = 50 J (at mean position)

and K_max = E – U_min = 200 – 50

K_max = 150 J (at mean position)

At extreme positions, kinetic energy is zero and potential energy is maximum

∴ U_max = E

U_max = 200 J (at extreme position)