Question

A particle starts oscillating simple harmonically from its equilibrium position then the ratio of kinetic energy and potential energy of the particle at the same time T/12 is : (T = time period)

• A. 2:1
• B. 3:1
• C. 4:1
• D. 1:4

Answer 1

The correct option is B 3:1

Since $x=a\sin \omega t$, we have
$x(t=\frac{T}{12})=a\sin (\frac{2\pi }{T}\times \frac{T}{12})=a\sin \left(\frac{\pi }{6}\right)=\frac{a}{2}$
then the ratio of the kinetic energy to the potential energy at $x=\frac{a}{2}$will be given by
$\frac{K.E.}{P.E.}=\frac{\frac{1}{2}k(a^2-x^2)}{\frac{1}{2}k{x}^2}=\frac{a^2-x^2}{x^2}\Rightarrow \frac{K.E.}{P.E.}=\frac{a^2-{\left(\frac{a}{2}\right)}^2}{{\left(\frac{a}{2}\right)}^2}=\frac{3}{1}$