Physics - SHM Question with Solution | TestHub

Consider mass $\mathrm{m}$ at $\mathrm{A}$. The forces exerted on it by the other two masses are given by ${\mathrm{F}}_1=\mathrm{G}\frac{{\mathrm{m}}^2}{{\mathrm{L}}^2}={\mathrm{F}}_2$ The angel between the two forces is $60^{\circ }$. Hence the resultant force The angel between the two forces is $60^{\circ }$. Hence the resultant force $\begin{array}{rl}& \mathrm{F}=\sqrt{{\mathrm{F}}_1^2+{\mathrm{F}}_1^2+2{\mathrm{F}}_1^2\cos 60^{\circ }}=\sqrt{3}{\mathrm{F}}_1\\ & \mathrm{F}=\sqrt{3}\cdot \mathrm{G}\frac{{\mathrm{m}}^2}{{\mathrm{L}}^2}\end{array}$ radius $r=\frac{L}{\sqrt{3}}$ For uniform circular motion, the gravitational force provides the centripetal force. $\begin{array}{rl}& \therefore \mathrm{mr}{\omega }^2=\mathrm{F}\\ & \therefore \mathrm{m}\frac{\mathrm{L}}{\sqrt{3}}\cdot {\omega }^2=\sqrt{3}\mathrm{G}\frac{{\mathrm{m}}^2}{{\mathrm{L}}^2}\\ & \therefore {\omega }^2=3\mathrm{G}\frac{\mathrm{m}}{{\mathrm{L}}^3}\end{array}$ $\begin{array}{rl}& \therefore \omega ={\left(3G\frac{\mathrm{m}}{{\mathrm{L}}^3}\right)}^{1/2}\\ & \therefore \frac{2\pi }{\mathrm{T}}={\left(\frac{3\mathrm{Gm}}{{\mathrm{L}}^3}\right)}^{1/2}\end{array}$ $\therefore \mathrm{T}=2\pi {\left(\frac{{\mathrm{L}}^3}{3Gm}\right)}^{1/2}$ $\therefore \mathrm{T}\propto {\mathrm{L}}^{3/2}$