As shown in the figure, two particle each of mass m tied at the ends of a light string of length 2a are kept on a frictionless horizontal surface. When the mid-point (P) of the string is pulled vertically upwards with a small but constant force F, the particles move towards each other on the surface. Magnitude of acceleration of each particle, when the separation between them becomes 2x is

Options

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

Answer

B

Explanation

Given length =a
Equating the vertical components of force,
`2T sin theta = F`…(i)
Equating the horizontal forces,
`T cos theta = ma.` …(ii)

Dividing Eq (i) by `F_(1)`, (ii), we get
`2 tan theta = (F)/(ma.)`
or `a. = (F)/(2m tan theta) = (F)/(2m((sqrt(a^(2) - x^(2)))/(x))) = (Fx)/(2m sqrt(a^(2) - x^(2))) " " ( :. tan theta = ("perpendicular")/("base") = (sqrt(a^(2)-x^(2)))/(x))`