Two particles, each of mass m, are connected by a light string of length 2L as shown in figure. Continous force F is applied· at the midpoint of the string (x = 0) at right angles to the initial position of the strir that acceleration of m in the direction at right angles to F is given by
`a_x=F/(2m)(x)/(sqrt(L^2-x^2))`
where x-is the perpendicular ~stance of one of the particles from the line of action of F. Discuss the situation when x=L.

Explanation

Let the mid point of the string O be displaced by y downwards along y-axis, so that at a given instant each string makes an angle `theta` with the y-axis as shown in figure. For motion of point C,
`F=2Tcostheta=0xxa_y`, i.e., `T=F/((2costheta))....(i)`

Now consider the motion of mass m at A towards B or vice-versa. Then as component of T in the direction motion will be `Tcos(90^@-theta)=Tsintheta`, so if `a_x` is the acceleration of m along x-axis, then from F=ma
`Tsintheta=ma_x` or `a_x=(Tsintheta)/m` .....(ii)
Substituting the value of T from Eqn. (i) in (ii),
`a_x=F/(2costheta)xx(sintheta)/m=F/(2m)tantheta=F/(2m)x/(sqrt(L^2-x^2))`
This is the required result. From this, it is clear that if x = L, a. = `oo` , i.e., this situation cannot be realised in practice as when F is applied, the mid point of string will be depressed in the y direction and so x cannot remain equal to L.