A particle of mass m is under the influence of a force F which varies with the displacement x according to the relation F = -kx + F₀ in which k and F₀ are constants. The particle when disturbed will oscillate

To find the point about which the particle will oscillate, we first need to determine the equilibrium position where the net force acting on the particle is zero.
Let $F=0$ at the equilibrium position $x=x_{\mathrm{eq}}$.

Given the force relation:
$F=-kx+F_0$

Setting $F=0$:
$-k{x}_{\mathrm{eq}}+F_0=0$
$k{x}_{\mathrm{eq}}=F_0$
$x_{\mathrm{eq}}=\frac{F_0}{k}$

Thus, the equilibrium position about which the particle will oscillate is $x=\frac{F_0}{k}$.

Now, let us analyze the restoring nature of the force. Let the displacement from the equilibrium position be $y$ such that:
$x=x_{\mathrm{eq}}+y=\frac{F_0}{k}+y$

Substituting this expression for $x$ back into the force equation:
$F=-k\left(\frac{F_0}{k}+y\right)+F_0$
$F=-F_0-ky+F_0$
$F=-ky$

Using Newton's second law, $F=ma=m\frac{d^{2}y}{d{t}^2}$, we have:
$m\frac{d^{2}y}{d{t}^2}=-ky$
$\frac{d^{2}y}{d{t}^2}+\left(\frac{k}{m}\right)y=0$

This is the standard equation of simple harmonic motion:
$\frac{d^{2}y}{d{t}^2}+{\omega }^{2}y=0$

Comparing the coefficients, we find the angular frequency $\omega$ of the oscillation is:
${\omega }^2=\frac{k}{m}$
$\omega =\sqrt{\frac{k}{m}}$

Therefore, the particle will oscillate about the point $x=\frac{F_0}{k}$ with an angular frequency $\omega =\sqrt{\frac{k}{m}}$.