As shown in the figure,two particles each of | vedclass

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(B) Let $T$ be the tension in the string and $	heta$ be the angle the string makes with the horizontal surface. When the separation between the parti

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

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(B) Let $T$ be the tension in the string and $	heta$ be the angle the string makes with the horizontal surface. When the separation between the particles is $2 x$,the distance from the midpoint $P$ to each particle is $x$. The length of the string segment from the midpoint to each particle is $a$. Thus,$\cos 	heta = \frac{x}{a}$ and $\sin 	heta = \frac{\sqrt{a^2-x^2}}{a}$.For the vertical equilibrium of the midpoint $P$,the upward force $F$ is balanced by the vertical components of the tension $T$ from both sides:$2 T \sin 	heta = F \implies T = \frac{F}{2 \sin 	heta}$.For each particle of mass $m$,the horizontal force causing acceleration $a'$ is the horizontal component of the tension $T$:$T \cos 	heta = m a'$.Substituting $T$ into the equation for $a'$:$a' = \frac{T \cos 	heta}{m} = \frac{F \cos 	heta}{2 m \sin 	heta} = \frac{F}{2 m 	an 	heta}$.Since $	an 	heta = \frac{\sin 	heta}{\cos 	heta} = \frac{\sqrt{a^2-x^2}/a}{x/a} = \frac{\sqrt{a^2-x^2}}{x}$,we have:$a' = \frac{F}{2 m} \frac{x}{\sqrt{a^2-x^2}}$.

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