Two masses each equal to m are lying on x-axis at `(-a,0)(+a,0)` respectively as shown in figure They are connected by a light string A force F is applied at the origin along vertical direction As a result the masses move toward each other without loosing contact with ground What is the acceleration of each mass? Assume the instantanceous position of the masses as`(-x,0)`and `(x,0)`

Options

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

Answer

D

Explanation

From figure `F= 2T cos theta`
or `T=F/2 cos theta`
The force responsible for motion of masses on x-axis is `T sin theta`

`ma=T sin theta=F/(2cos theta)xxsin theta`
`=F/2 tan theta=F/2xx(OB)/(OA)=F/2xx x/sqrt((a^(2)-x^(2)))`
So, `a=F/(2m)xxF/sqrt((a^(2)-x^(2)))`