Shm Energy

Energy in simple harmonic motion, as developed in NCERT Class 11, Chapter 14 (Oscillations), splits into potential energy $U=\frac{1}{2}m{\omega }^2{x}^2$ and kinetic energy $K=\frac{1}{2}m{\omega }^2(A^2-x^2)$, whose sum is the constant total $E=\frac{1}{2}m{\omega }^2{A}^2$. At $x=A/2$, the kinetic energy is $K=\frac{1}{2}m{\omega }^2(A^2-A^2/4)=\frac{1}{2}m{\omega }^2\cdot \frac{3A^2}{4}$. Dividing by the total $E=\frac{1}{2}m{\omega }^2{A}^2$ gives $K/E=3/4$. So three-quarters of the energy is kinetic and the remaining one-quarter is potential. The option one-quarter confuses kinetic with potential, since $U/E=(A/2{)}^2/A^2=1/4$. The option one-half would only hold at $x=A/\sqrt{2}$, not at $A/2$. The option of the full energy applies only at the mean position where $x=0$. To see the pattern more generally, the kinetic fraction is $K/E=1-(x/A{)}^2$ and the potential fraction is $U/E=(x/A{)}^2$; substituting $x/A=1/2$ gives $1-1/4=3/4$ kinetic and $1/4$ potential, which is the algebraic route to the same answer. This also explains why the kinetic share falls off faster than linearly as the particle moves outward, since it depends on the square of the displacement ratio. Checking magnitude: at half amplitude the particle is still fairly fast, so kinetic energy dominating at 75% is reasonable, and the fractions correctly sum to unity as energy conservation demands.