Question

Average kinetic energy in one time period of a simple harmonic oscillator whose amplitude is $A$, angular velocity $\omega$ and mass $m$, is

• A. $\frac{1}{4}m{\omega }^2{A}^2$
• B. $\frac{1}{2}m{\omega }^2{A}^2$
• C. $m{\omega }^2{A}^2$
• D. $Zero$

Answer 1

The correct option is A $\frac{1}{4}m{\omega }^2{A}^2$

$x=Asin\omega t$
$V=A\omega cos\omega t$
$KE=\frac{1}{2}m{V}^2$
$=\frac{1}{2}m{A}^2{\omega }^{2}co{s}^2\omega t$
$AverageKE=\frac{{\int }_0^T\frac{1}{2}m{A}^2{\omega }^{2}co{s}^2\omega tdt}{{\int }_0^{T}dt}$
$=\frac{{\int }_0^T\frac{1}{4}(1+cos2\omega t)m{A}^2{\omega }^2}{{\int }_0^{T}dt}$
$=\frac{M{A}^2{\omega }^{2}T}{4T}$
$=\frac{M{A}^2{\omega }^2}{4}$
$4T=\frac{2\pi }{\omega }$