Two identical point masses each of mass $m$ are connected one another by a massless string of length $L$ . A constant force $F$ is applied at the midpoint of the string. If $l$ is the instantaneous distance between the two masses, What will be the acceleration of each mass.

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None of the above

$\frac{F}{m} [\frac{l}{\sqrt{L^{2} - l^{2}}}]$

$\frac{F}{2 m} [\frac{l}{\sqrt{L^{2} - l^{2}}}]$

$\frac{F}{2 m} [\frac{1}{\sqrt{L^{2} - l^{2}}}]$

answer is A.

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Detailed Solution

$T h e s i t u a t i o n i s s h o w n i n f i g u r e L e t T b e t h e t e n s i o n i n t h e s t r i n g . F o r p o i n t O, w e h a v e 2 T \sin θ = F o r T = \frac{F}{2 \sin θ}$

$N o w a c c e l e r a t i o n o f e a c h m a s s; a_{x} = \frac{F_{x}}{m}, o r a_{x} = \frac{T \cos θ}{m} = \frac{(F / 2 \sin θ) \cos θ}{m} = \frac{F c o t θ}{2 m} = \frac{F}{2 m} (\frac{l / 2}{\sqrt{{(L / 2)}^{2} - {(l / 2)}^{2}}}) = \frac{F}{2 m} [\frac{l}{\sqrt{L^{2} - l^{2}}}]$