Question

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 center 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

• A. $\frac{F}{2m}\frac{a}{\sqrt{a^2-x^2}}$
• B. $\frac{F}{2m}\frac{x}{\sqrt{a^2-x^2}}$
• C. $\frac{F}{2m}\frac{x}{a}$
• D. $\frac{F}{2m}\frac{\sqrt{a^2-x^2}}{a}$

Answer 1

The correct option is B $\frac{F}{2m}\frac{x}{\sqrt{a^2-x^2}}$

$\because F=2Tcos\theta$
$\Rightarrow T=\frac{F}{2cos\theta }$

Also,
Acceleration, $f=\frac{Tsin\theta }{m}$
$\frac{F}{2m}tan\theta$
$=(\frac{F}{2m}\frac{1}{\sqrt{a^2-x^2}}$