$3.1.3.$ a. A body of mass $m$ suspended on a spring oscillates such that the greatest value of velocity is $v_0$, and the greatest deviation from the equilibrium position is $x_0$. Determine the stiffness of the spring.
b. The velocity of a body of mass $m$ suspended on a spring and oscillating depends on the coordinate of the body $x$ according to the law $v = v_0 \sqrt{ 1 - (\frac{x}{x_0})^2}$. Find the relation between the force acting on the body and the potential energy of the body and the coordinate $x$. Does the result depend on the nature of the force that makes the body move according to the above law?
Since the oscillations are harmonic, the greatest velocity $v_0$ can be found through the greatest deviation from the equilibrium position $x_0$. To do this, we take the derivative of the solution to the equation of harmonic oscillations $$ v=\frac{dx}{dt}=\frac{d}{dt}\left(x_0\sin\omega t\right)=x_0\omega\cos\omega t $$ From which it follows that the maximum speed $v_0$ is equal to $$ v_0=x_0\omega$$ Substituting the value for the cyclic frequency $\omega$ of the spring pendulum $\left(\omega = \sqrt{\frac{k}{m}}\right)$ obtained earlier from 3.1.2 $$ v_0=x_0\sqrt{\frac{k}{m}}\Rightarrow\boxed{k=\frac{v_0^2m}{x_0^2}} $$ b) The restoring force in these harmonic oscillations will be the elastic force of the spring $F=-kx$, since in any case in a conventional spring pendulum the restoring force is equal to the elastic force of the spring. Then we find the work of this force as in 3.1.2: $$ \boxed{U=\frac{kx^2}{2}} $$
a. $k=mv_0^2/x_0^2$ б. $F=-kx, ~U=kx^2/2, ~k=m(v_0/x_0)^2$ No, it doesn't depend.