3.1.3 | до правки #19988 | Решения Савченко

Statement

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?

Solution

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:

$\boxed{U=\frac{kx^2}{2}}$

Answer

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.