Two particles of mass $m$ each are tied at the ends of a light string of length $2a$. The whole system is kept on a frictionless horizontal surface with the string held tight so that each mass is at a distance $a$ from the centre $P$ (as shown in the figure). Now, the mid-point of the string is pulled vertically upwards with a small but constant force $F$. As a result, the particles move towards each other on the surface. The magnitude of acceleration, when the separation between them becomes $2x$, is

Step 1: Set up the geometry of the system.

Let $h$ be the vertical height of the midpoint $P$ above the horizontal surface when the separation between the particles is $2x$.

$$x^2 + h^2 = a^2$$ $$h = \sqrt{a^2 - x^2}$$

Let $\theta$ be the angle that each half of the string makes with the horizontal.

$$\sin\theta = \frac{h}{a}$$ $$\cos\theta = \frac{x}{a}$$

Step 2: Analyze the forces acting at the midpoint $P$.

Since the string is light, the midpoint $P$ is massless, meaning the net force on it must be zero.

$$F = 2T\sin\theta$$ $$T = \frac{F}{2\sin\theta}$$

Substitute $\sin\theta = \frac{h}{a}$ into the tension equation.

$$T = \frac{Fa}{2h}$$

Step 3: Analyze the horizontal motion of one particle.

The horizontal component of the tension force drives the acceleration $a_{acc}$ of each particle of mass $m$.

$$m a_{acc} = T\cos\theta$$

Substitute the expressions for $T$ and $\cos\theta$.

$$m a_{acc} = \left(\frac{Fa}{2h}\right)\left(\frac{x}{a}\right)$$ $$m a_{acc} = \frac{Fx}{2h}$$

Step 4: Solve for the magnitude of acceleration.

Isolate the acceleration $a_{acc}$ and substitute $h = \sqrt{a^2 - x^2}$.

$$a_{acc} = \frac{Fx}{2mh}$$ $$a_{acc} = \frac{Fx}{2m\sqrt{a^2 - x^2}}$$

Answer: The magnitude of acceleration is $\dfrac{Fx}{2m\sqrt{a^2 - x^2}}$.