USAPhO 1998, Четвъртфинал — Задача 2

• (a, 5) 49 N/m
• (b, 10) 0.3 m
• (c, 10) 1.27 s

(a, 5) $\mathbf F=m\mathbf a$ becomes $kx_0-mg=0$ (3)

so that $k=mg/x_0=(1\ kg)(9.8\ N/kg)/(0.2\ m)=49\ N/m$. (2)

(b, 10) One may use the work-energy theorem,

$W_{nc}=\Delta E$ (3)

where $W_{nc}$ is the work done by non-conservative forces and $E$ is the total mechanical energy. With the zero of gravitational P.E. at the initial position of the hanging mass, $W_{nc}=\Delta E$ becomes

$-\mu mgx=\frac12kx^2-mgx$ (4)

The $x\ne0$ solution gives $x=2(mg/k)(1-\mu)=0.3\ m$. (3)

(c, 10) The system is equivalent to a particle of mass $2m$ being pulled in one direction by the force $mg$, in the opposite direction by the force $kx$. Thus, Newton’s second law gives

$mg-kx=2ma$ (3)

which says $a+(k/2m)x=\frac12g$, which is the equation of motion of a simple harmonic oscillator (in a uniform gravitational field), with angular frequency $\omega=\sqrt{k/2m}$. (4)

[Alternatively, if the student notes that the angular frequency squared equals $k$ divided by the mass that is oscillating, and notes that $2m$ is the mass that oscillates, give full credit for part (c) up to here.]

Next, convert to period: $T=2\pi/\omega$ gives $T=2\pi\sqrt{2m/k}=1.27\ s$. (3)