Question

Two masses each equal to $m$ are lying on $x$-axis at $(-a,0)$ and $(+a,0)$ respectively. They are connected by a light string and are constrained to move along the line joining their centre. A force $F$ is applied at the origin along $y$-axis resulting into motion of masses towards each other. The acceleration of each mass when positions of masses at any instant becomes $(-x,0)$ and $(+x,0)$ is given by:

• A. $\frac{F}{m}\frac{x}{\sqrt{a^2-x^2}}$
• B. $\frac{F}{m}\frac{\sqrt{a^2-x^2}}{x}$
• C. $\frac{Fx}{2m\sqrt{a^2-x^2}}$
• D. $\frac{F}{2m}\sqrt{\frac{a^2-x^2}{x}}$

Answer 1

The correct option is C $\frac{Fx}{2m\sqrt{a^2-x^2}}$

For vertical force, $F=T\cos \theta +T\cos \theta =2T\cos \theta$
or $T=\frac{F}{2\cos \theta }....\left(1\right)$
The force causing the motion is given by $F_m=T\sin \theta =\frac{F}{2\cos \theta }\sin \theta$
or $F_m=\frac{F}{2}\tan \theta =\frac{F}{2}.\frac{x}{\sqrt{a^2-x^2}}$
Acceleration $=\frac{F_m}{m}=\frac{Fx}{2m\sqrt{a^2-x^2}}$