Two identical particles A and B, each of mass $m$ are interconnected by spring of stiffness $k$ . If the particle B is applied a force (directed to outside) $F$ & the elongation of the spring is $x$ , the relative acceleration between the particle is equal to:

\frac{F}{2 m}

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\frac{F - k x}{2 m}

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\frac{F - 2 k x}{m}

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\frac{k x}{m}

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Solution

The correct option is C $\frac{F - 2 k x}{m}$
As seen in the FBD of given system, block A experiences force $k x$ in the right direction and block B experiences a force $F - k x$ in the right direction. As we know relative accelerations of two particles moving in same direction is found out by subtracting from each other.
So,
$a_{b a} = a_{b} - a_{a}$
$⟹ a_{b a} = \frac{F - 2 k x}{m}$

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Two identical particles A and B, each of mass m are interconnected by spring of stiffness k. If the particle B is applied a force (directed to outside) F & the elongation of the spring is x, the relative acceleration between the particle is equal to:

Two identical particles A and B, each of mass $m$ are interconnected by spring of stiffness $k$ . If the particle B is applied a force (directed to outside) $F$ & the elongation of the spring is $x$ , the relative acceleration between the particle is equal to: