Two particles of mass m each are tied at the | vedclass

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(B) Let the angle made by the string with the horizontal be $	heta$. The length of each half of the string is $a$. When the separation between the pa

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$\frac{F}{2m} \frac{x}{\sqrt{a^2-x^2}}$

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(B) Let the angle made by the string with the horizontal be $	heta$. The length of each half of the string is $a$. When the separation between the particles is $2x$,the horizontal distance of each particle from the center $P$ is $x$. From the geometry,$\cos 	heta = \frac{x}{a}$ and $\sin 	heta = \frac{\sqrt{a^2-x^2}}{a}$. Considering the vertical equilibrium of the midpoint of the string: $2T \sin 	heta = F$,so $T = \frac{F}{2 \sin 	heta}$. The horizontal force on each particle is $T \cos 	heta = ma$,where $a$ is the acceleration of the particle. Substituting $T$: $ma = \left( \frac{F}{2 \sin 	heta} ight) \cos 	heta = \frac{F}{2} \cot 	heta$. Since $\cot 	heta = \frac{\cos 	heta}{\sin 	heta} = \frac{x/a}{\sqrt{a^2-x^2}/a} = \frac{x}{\sqrt{a^2-x^2}}$,we get $ma = \frac{F}{2} \left( \frac{x}{\sqrt{a^2-x^2}} ight)$,which gives the acceleration $a = \frac{F}{2m} \left( \frac{x}{\sqrt{a^2-x^2}} ight)$.

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