The total mechanical energy of a particle of mass $m$ in SHM with amplitude $A$ and angular frequency $\omega$ is:
A$\dfrac{1}{2} m \omega A$
B$\dfrac{1}{2} m \omega^2 A^2$
C$m \omega^2 A$
D$\dfrac{1}{2} m A^2$
Answer & Solution
Correct answer: B. $\dfrac{1}{2} m \omega^2 A^2$
Total energy is conserved during SHM and equals the maximum kinetic energy (at the mean position) or the maximum potential energy (at the extreme):
$E = \dfrac{1}{2} m v_{\max}^2 = \dfrac{1}{2} m (A\omega)^2 = \dfrac{1}{2} m \omega^2 A^2$.
Equivalently, $E = \dfrac{1}{2} k A^2$ for a spring-mass system, using $\omega^2 = k/m$.
Quadratic in amplitude: doubling the amplitude *quadruples* the total energy. Same scaling as kinetic energy with velocity.
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