Physics - SHM Question with Solution | TestHub

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