Explanation

(a) [image:5_image226.jpeg]
From force diagram,
$T_{1} = m_{1}a_{1}$ …(i)
$T_{2} = 2T_{1}\ \ \ \ \ \ \ \ \ \ \ \ $…(ii)
$m_{2}g - T_{2} = m_{2}a_{1}$
$m_{2}g - 2T_{2} = m_{2}a_{1}$ …(iii)
Total work done by tensions should be zero,
$\therefore$ $T_{1}x_{1} - T_{2}x_{2} = 0$
or $T_{1}x_{1} = T_{2}x_{2}$ or $T_{1}x_{2} = 2T_{1}x_{2}$
or $x_{1} = 2x_{2}$ or $\frac{d^{2}x_{1}}{dt^{2}} = \frac{2d^{2}x_{2}}{dt^{2}}$
$\therefore a_{1} = 2a_{2}\ \ \ \ \ \ \ \ $…(iv)
After solving Eqs. (i), (iii) and (iv),
$a_{2} = \frac{m_{2}g}{4m_{1} + m_{2}}$