Two particles, each of mass m, are connected by a light string of length `2L` as shown in A continuous force F is applied at the mid point of the string `(x=0)` at right angles to the initial position of the string. Show 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 distance 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
For motion of point C,
`F-2T cos theta=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 of motion will be `T cos (90^(@)-theta)=T sin thetha` , so if `a_(x)` is the acceleration of m along x-axis, then from `F=ma`
`T sin theta=ma_(x) or a_(x)=T sin theta//m` ....(ii)
Substituting the value of T from Eqn (i) in (ii)
`a_(x)=F/(2 cos theta)xx(sin theta)/m=F/(2m)tan theta=F/(2m) x/sqrt(L^(2)-x^(2))`
This is the required result From this, it is clear that if `x=L, a_(x)=infty` i.e, this situation cannot be realised in pratice 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.
Note: (i) Acceleration of particle in the direction of force will be,
`a_(y)=(T cos theta)/m=F/(2m)="constant"`
(ii) As particles move towards each other x or `theta` will decrease and so also the acceleration `a_(x)` and for `x=0,(theta=0^(@))a_(x)=0`