Question

two similar simple pendulum are vibrating with time period 2s and 3swhat is the ratio of their total energies?

Answer 1

$$\begin{gathered} Totalenergyofasimplependulumisgivenby=E=\frac{1}{2}m{\omega }^2{r}^2 \\ wherer=Lengthofthestring=L \\ and\sin ceboththependulumsaresimilartherefore, \\ {m}_1=m_2=m \\ and, \\ {r}_1=r_2=L \\ andT=2\mathrm{\pi }\sqrt{\frac{\mathrm{L}}{\mathrm{g}}} \\ \mathrm{or}\mathrm{f}=\frac{1}{\mathrm{T}} \\ \mathrm{and},\mathrm{\omega }=2\mathrm{\pi f}=\frac{2\mathrm{\pi }}{\mathrm{T}} \\ \mathrm{therefore}, \\ {\mathrm{\omega }}_1=\frac{2\mathrm{\pi }}{{\mathrm{T}}_1}=\frac{2\mathrm{\pi }}{2}=\mathrm{\pi }\mathrm{rad}/\mathrm{s} \\ {\mathrm{\omega }}_2=\frac{2\mathrm{\pi }}{{\mathrm{T}}_2}=\frac{2\mathrm{\pi }}{3}\mathrm{rad}/\mathrm{s} \\ \mathrm{therefore}, \\ {\mathrm{E}}_1=\frac{1}{2}{\mathrm{m}}_1{{\mathrm{\omega }}_1}^2{{\mathrm{r}}_1}^2=\frac{1}{2}{\mathrm{m\pi }}^2{\mathrm{L}}^2 \\ {\mathrm{E}}_2=\frac{1}{2}{\mathrm{m}}_2{{\mathrm{\omega }}_2}^2{{\mathrm{r}}_2}^2=\frac{1}{2}\mathrm{m}\times \frac{4}{9}{\mathrm{\pi }}^2\times {\mathrm{L}}^2 \\ \frac{{\mathrm{E}}_1}{{\mathrm{E}}_2}=\frac{\frac{1}{2}{\mathrm{m\pi }}^2{\mathrm{L}}^2}{\frac{1}{2}\mathrm{m}\times \frac{4}{9}{\mathrm{\pi }}^2\times {\mathrm{L}}^2}=\frac{9}{4} \end{gathered}$$