In the shown figure masses of the pulleys and strings as well as friction between the string and pulley is negligible. Find the ratio of accelerations of the masses $m_{1} a n d m_{2}$ $\frac{a_{2}}{a_{1}} =$

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answer is 2.

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Detailed Solution

Let the lengths of the strings passes over A is $l_{1}$ and of that passes over B is $l_{2}$ $y_{A} a n d y_{B}$ be distance of pulleys. From ground while $y_{1} a n d y_{2}$ are distances of $m_{1} a n d m_{2}$ from ground respectively.
$∵ (y_{A} - y_{1}) + (y_{A} - y_{B}) = l_{1}$
$\Rightarrow 2 y_{A} - y_{1} - y_{B} = l_{1}$
$∵$ $y_{A}$ is constant $∵ y_{1} + y_{B} = 2 y_{A} - l_{1} =$ constant
Differentiating twice this equation w. r. t. time we get
$\frac{d^{2} y_{1}}{d t^{2}} + \frac{d^{2} y_{B}}{d t^{2}} = 0 \Rightarrow a_{1} = - a_{B} ..... (i)$
Similarly $y_{B} + y_{B} - y_{2} = l_{2}$
$\Rightarrow 2 y_{B} - y_{2} =$ constant
$\Rightarrow 2 \frac{d^{2} y_{B}}{d t^{2}} - \frac{d^{2} y_{2}}{d t^{2}} = 0 \Rightarrow 2 a_{B} = a_{2} = 2 a_{1} ....... (i i)$