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Angle of Repose

Two blocks 1 ...

Question

Two blocks (1 and 2) of equal mass m are connected by an ideal string (see figure below) over a frictionless pulley. The blocks arc 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. The block 1 is slowly pulled down a distance x and released. Just after the release the possible values of the magnitudes of the acceleration of the blocks $a_{1}$ and $a_{2}$ can be.

A

either

(a_{1} = a_{2} = \frac{(k_{1} + k_{2}) x}{2 m})

or

(a_{1} = \frac{k_{1} x}{m} - g a n d a_{2} = \frac{k_{2} x}{m} + g)

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B

(a_{1} = a_{2} = \frac{(k_{1} + k_{2}) x}{2 m})

only

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C

(a_{1} = a_{2} = \frac{(k_{1} - k_{2}) x}{2 m})

only

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D

either

(a_{1} = a_{2} = \frac{(k_{1} - k_{2}) x}{2 m})

or

(a_{1} = a_{2} \frac{(k_{1} k_{2}) x}{(k_{1} + k_{2}) m} - g)

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Solution

The correct option is A either $(a_{1} = a_{2} = \frac{(k_{1} + k_{2}) x}{2 m})$ or
$(a_{1} = \frac{k_{1} x}{m} - g a n d a_{2} = \frac{k_{2} x}{m} + g)$
For Mass 1
$T + k_{1} - m g = m a_{1}$
For mass 2
$k_{2} x + m g - T = m a_{2}$
if after release the string remains tight then $a_{1} = a_{2}$
so $a_{1} = a_{2} = \frac{(k_{1} + k_{2}) x}{2 m}$
Now if string becomes loose the T=0
So $a_{1} = \frac{k_{1} x}{m} - g$
$a_{2} = \frac{k_{2} x}{m} + g$

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