Question

Figure shows a particle of mass m attached with 4 identical springs each of spring constant K and each of which are initially in their natural length L. The gravitational force is neglected. If the mass is slightly displaced by distance $x$ along a line perpendicular to the plane of the figure (passing through itself) and released then the force acting on particle just when it is released is proportional to $x^n$, then $n$ is

Answer 1

$Fnet=4F\cos \theta =4K\left(\sqrt{L^2+X^2}-L\right].\frac{X}{\sqrt{L^2+X^2}}=4KX\left(1-\frac{L}{\sqrt{L^2+X^2}}\right]=4KX\left(\frac{X^2}{2L^2}\right)=\frac{2K}{L^2}{X}^3$
Hence, value of $n=3$