Two particles of mass m each are tied at the ends of a light string of length 2a. The whole system is kept on a frictionless horizontal surface with the string held tight so that each mass is at a distance 'a' from the center P (as shown in figure). Now, the mid-point of the string is pulled vertically upwards with a small but constant force F. As a result, the particles move towards each other on the surface. The magnitude of acceleration, when the separation between them becomes 2x, is :

Options

A. `(F)/(2m)(a)/(sqrt(a^(2)-x^(2))`
B. `(F)/(2m)(a)/(sqrt(x-x^(2))`
C. `(F)/(2m)(x)/(a)`
D. `(F)/(2m)sqrt(a^(2)-x^(2))/(x)`

Answer

b

Explanation

As shown in fig. (i) the separation between the twomasses is 2x. Each mass will move in the horizontal direction towards each other as the force .F. acts vertically. Let the tension in each part of the string be T. The state of forces at the point P and on the mass A are shown in fig. (ii) and (ii) above. When the separation x is reached net force at the point P is given by:

Fig. (ii) 2T sin theta=F…(i)
Also for the mass A, Fig. ..(ii)
`R+T sin theta= mg...(ii)`
` T cos theta= mf...(iii)`
where .f is the acceleration with which each ball moves. From (1) and (iii)
Acceleration, `f =(F cos theta)/(2m)=(F)/(2m)xx(x)/sqrt(a^(2)x^(2))`
Hence (b) is the correct choice