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?

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    <h3> Statement </h3>
    <p>
        $3.1.3.$ 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.</p><p>

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?

    <h3>Solution</h3>
    <p>
        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:
$$\boxed{U=\frac{kx^2}{2}}$$

    </p>

    <h4>Answer</h4>
    <p>
        a. $k=mv_0^2/x_0^2$<br>

б. $F=-kx, ~U=kx^2/2, ~k=m(v_0/x_0)^2$

No, it doesn't depend.

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