
As shown in the figure, two particle each of mass $m$ tied at the ends of a light string of length $2 a$ are…

- $\frac{F}{2 m} \frac{a}{\sqrt{a^2-x^2}}$
- $\frac{F}{2 m} \frac{x}{\sqrt{a^2-x^2}}$
- $\frac{F}{2 m} \frac{x}{a}$
- $\frac{F}{2 m} \frac{\sqrt{a^2-x^2}}{x}$
Solution

$ 2 \tan \theta=\frac{F}{m a^{\prime}} $ or $ \begin{array}{r} a^{\prime}=\frac{F}{2 m \tan \theta}=\frac{F}{2 m\left(\frac{\sqrt{a^2-x^2}}{x}\right)}=\frac{F x}{2 m \sqrt{a^2-x^2}} \\ \left(\because \tan \theta=\frac{\text { perpendicular }}{\text { base }}=\frac{\sqrt{a^2-x^2}}{x}\right) \end{array} $
Asked in: AP EAMCET 2018 (23 Apr Shift 1)