Find the acceleration of the body of mass `m_(2)` in the arrangement shown in figure. If the mass`m_(2)` is `eta` time great as the mass `m_(1)` and the angle that the inclined plane forms with the horizontal is equal to `theta`. The masses of the pulley and threads, as well as the friction, are accumed to be negligible.

Options

A. `( g(2 eta - sin theta))/(4 eta + 1)`
B. `(2 g( eta - sin theta))/(4 eta + 1)`
C. `(2 g(2 eta - sin theta))/( eta + 1)`
D. `(2 g(2 eta - sin theta))/(4 eta + 1)`

Answer

D

Explanation

Here by constraint relation we can see that the acceleration of `m_(2)` is two times that of `m_(1)` . So we assume if `m_(1)` is moving up the inclined plane with an acceleration `a`, the acceleration of mass `m_(2)` going down is `2a`. The ten tension in different string are shown in figure.

The dynamic equations can be written as
For mass `m_(1)`: `2T = m_(1)g sin theta = m_(1)a` ...(i)
For mass `m_(2)`: `m_(2)g - T = m_(2)(2a)` ...(ii)
Substituting `m_(2) = eta m_(1)` and solving Eqs. (i) and (ii), we get
Acceleration of `m_(2) = 2a = (2 g(2 eta - sin theta))/(4 eta + 1)`