MCQ
Two blocks $1$ and $2$ of equal mass $m$ are connected by an ideal string (see figure below) over a frictionless pulley. The blocks are attached to the ground by springs having spring constants $k_1$ and $k_2$ such that $k_1 > k_2$. Initially, both springs are unstretched. Block $1$ is slowly pulled down a distance $x$ and released. Just after the release the possible values of the magnitudes of the accelerations of the blocks $a_1$ and $a_2$ can be
  • A
    either $\left(a_1=a_2=\frac{\left(k_1+k_2\right) x}{2 m}\right)$ or $\left(a_1=\frac{k_1 x}{m}-g\right.$ and $\left.a_2=\frac{k_2 x}{m}+g\right)$
     
  • $\left(a_1=a_2=\frac{\left(k_1+k_2\right) x}{2 m}\right)$ only
  • C
    $\left(a_1=a_2=\frac{\left(k_1-k_2\right) x}{2 m}\right)$ only
  • D
    either $\left(a_1=a_2=\frac{\left(k_1-k_2\right) x}{2 m}\right)$ $\text { or }\left( \alpha _1=a_2=\frac{\left(k_1 k_2\right) x}{\left(k_1+k_2\right) m}-g\right)$

Answer

Correct option: B.
$\left(a_1=a_2=\frac{\left(k_1+k_2\right) x}{2 m}\right)$ only
b
(b)

Free body diagram for block $1$ and $2$ are as shown below.

So, equations of force balance are

$T+k_1 x-m g =m a_1$

$\text { and }$ $k_2 x+m g-T =m a_2$

As string is in extensible,

$a_1=a_2=a \text { (let) }$

Then adding both equations, we have

$\left(k_1+k_2\right) x=2 m a$

$\Rightarrow \quad a=\left(\frac{k_1+k_2}{2 m}\right) \cdot x$

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