In a deep vertical shaft underground there are two weights of equal mass connected by a weightless non-stretchable thread, which is thrown over a fixed block. Find the period of vertical oscillation of the weights if the radius of the Earth is 6400 km, the acceleration of free fall at the surface is $10 m/s^{2}$.
Decision:
At the surface of the Earth, the force of gravity acting on a body of mass $m$ is the force of universal gravitation between that body and the Earth. If we consider the Earth to be a homogeneous ball with radius $R$, then
$mg= G \frac{m M}{R^{2}}$, (1)
Where $M$ is the mass of the Earth; $G$ is the gravitational constant. In a mine, when both weights are at a depth $h$ underground, a force acts on each of them from the Earth's side
$F= G \frac{m \rho V}{(R-h)^{2}}= \frac{4}{3} \pi G \rho m (R-h)$,
where $\rho$ is the density of the Earth, $V=\frac{4}{3} \pi (R-h)^{3}$ is the volume of a ball of radius
$R-h$.
If now one of the weights (for example, the right one) is lifted to the height $x$ from the equilibrium position, it will be at the depth $h-x$ and the thread tension force $T$ and the gravitational force $F_{1}(x)$, equal to
$F_{1}(x)= \frac{4}{3} \pi G \rho m (R-(h-x))$.
The equation of motion of the right weight in projection on the upward coordinate axis Oh will be as follows
$ma_{1}=- \frac{4}{3} \pi G \rho m (R-h+x) + T$, (2)
where $a_{1}$ is the projection of the acceleration of the right weight on the x-axis. Since the thread is inextensible, the left weight will fall to a distance $x$ from the equilibrium position. The left weight will be acted on by the vertically upward force of the thread tension $T$, since the thread is weightless, and the gravitational force $F_{2}(x)$ equal to
$F_{2}(x)=\frac{4}{3} \pi G \rho m (R- (h+x))$.
The equation of motion of the left weight in projection on the x-axis has the form
$-ma_{2} = - \frac{4}{3} \pi G \rho m (R-h-x) + T$. (3)
Tick as the thread is inextensible, then the moduli of the accelerations of the right and left weights are equal and their directions are opposite ($a_{1}=a_{2} \equiv a$).
Subtracting equation (3) from equation (2), we obtain
$2ma=- \frac{8}{3} \pi G \rho mx$,
or
$ma+ \omega^{2} x = 0$, (4)
where the notation
$\omega^{2}= \frac{4}{3} \pi G \rho$.(5)
It follows from equation (4) that the weights will make vertical harmonic oscillations, the period of which is equal to
$T= \frac{2 \pi }{ \omega}$. (6)
Let us substitute in formula (1) the mass of the Earth equal to $M= \rho V= \frac{4}{3} \pi \rho R^{3}$. Thus we can get an expression for the acceleration of free fall in the form of
$g = \frac{4}{3} \pi G \rho R$. (7)
From (5) and (7) we obtain the frequency
$\omega = \sqrt{ \frac{g}{R}}$
and oscillation period
$T=2 \pi \sqrt{ \frac{R}{g}} \approx 80 min$.
Decision:
At the surface of the Earth, the force of gravity acting on a body of mass $m$ is the force of universal gravitation between that body and the Earth. If we consider the Earth to be a homogeneous ball with radius $R$, then
$mg= G \frac{m M}{R^{2}}$, (1)
Where $M$ is the mass of the Earth; $G$ is the gravitational constant. In a mine, when both weights are at a depth $h$ underground, a force acts on each of them from the Earth's side
$F= G \frac{m \rho V}{(R-h)^{2}}= \frac{4}{3} \pi G \rho m (R-h)$,
where $\rho$ is the density of the Earth, $V=\frac{4}{3} \pi (R-h)^{3}$ is the volume of a ball of radius
$R-h$.
If now one of the weights (for example, the right one) is lifted to the height $x$ from the equilibrium position, it will be at the depth $h-x$ and the thread tension force $T$ and the gravitational force $F_{1}(x)$, equal to
$F_{1}(x)= \frac{4}{3} \pi G \rho m (R-(h-x))$.
The equation of motion of the right weight in projection on the upward coordinate axis Oh will be as follows
$ma_{1}=- \frac{4}{3} \pi G \rho m (R-h+x) + T$, (2)
where $a_{1}$ is the projection of the acceleration of the right weight on the x-axis. Since the thread is inextensible, the left weight will fall to a distance $x$ from the equilibrium position. The left weight will be acted on by the vertically upward force of the thread tension $T$, since the thread is weightless, and the gravitational force $F_{2}(x)$ equal to
$F_{2}(x)=\frac{4}{3} \pi G \rho m (R- (h+x))$.
The equation of motion of the left weight in projection on the x-axis has the form
$-ma_{2} = - \frac{4}{3} \pi G \rho m (R-h-x) + T$. (3)
Tick as the thread is inextensible, then the moduli of the accelerations of the right and left weights are equal and their directions are opposite ($a_{1}=a_{2} \equiv a$).
Subtracting equation (3) from equation (2), we obtain
$2ma=- \frac{8}{3} \pi G \rho mx$,
or
$ma+ \omega^{2} x = 0$, (4)
where the notation
$\omega^{2}= \frac{4}{3} \pi G \rho$.(5)
It follows from equation (4) that the weights will make vertical harmonic oscillations, the period of which is equal to
$T= \frac{2 \pi }{ \omega}$. (6)
Let us substitute in formula (1) the mass of the Earth equal to $M= \rho V= \frac{4}{3} \pi \rho R^{3}$. Thus we can get an expression for the acceleration of free fall in the form of
$g = \frac{4}{3} \pi G \rho R$. (7)
From (5) and (7) we obtain the frequency
$\omega = \sqrt{ \frac{g}{R}}$
and oscillation period
$T=2 \pi \sqrt{ \frac{R}{g}} \approx 80 min$.