A string of length $l>d$ is fixed at points lying on a horizontal plane at a distance $d$ from each other. A weightless small pulley can roll on the string, to which a body of weight $G$ is attached. What is the period of oscillation of the body for small displacements? (See the article published on page 129 of volume XXVIII of KML!)
I. Solution. The body moves along an elliptical arc. In the case of a small displacement, the elliptical arc can be approximated by the arc of a circle tangent to the end of the minor axis. The radius of this circle is $\varrho=a^{2} / b$, (K. M. L. Vol. XXVIII, p. 129), where $a$ and $b$ are the semi-major and semi-minor axes of the ellipse, respectively.

![](https://cdn.mathpix.com/cropped/2024_05_02_3a0790e5a42f51850553g-1.jpg?height=466&width=442&top_left_y=325&top_left_x=820)

Figure 1

The pendulum described in the problem can be replaced by a pendulum with a radius of $\varrho$ (Figure 1), and its period of oscillation is

$$
T=2 \pi \sqrt{\frac{\underline{\varrho}}{g}}=2 \pi \sqrt{\frac{a^{2}}{b g}}=2 \pi a \sqrt{\frac{1}{b g}}=2 \pi \frac{l}{2} \sqrt{\frac{1}{g \sqrt{\frac{l^{2}-d^{2}}{2}}}}=\pi l \sqrt{\frac{\sqrt{2}}{g \sqrt{l^{2}-d^{2}}}}
$$

János Végh (Debrecen, KLTE Gyak. Gimn., III. o. t.)

II. Solution. The equation of the ellipse is

$$
\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 . \quad \text { Here } \quad a=\frac{l}{2}, \quad b=\frac{\sqrt{l^{2}-d^{2}}}{2}
$$

The coordinates of the body at the lowest point of the trajectory are $\left(0, y_{0}\right)$. If the body is displaced horizontally by a distance $A$, its ordinate changes to $y_{1}$ (Figure 2):

![](https://cdn.mathpix.com/cropped/2024_05_02_3a0790e5a42f51850553g-1.jpg?height=341&width=545&top_left_y=1537&top_left_x=774)

Figure 2

$$
\frac{A^{2}}{a^{2}}+\frac{y_{1}^{2}}{b^{2}}=1, \quad y_{1}=b \sqrt{1-\frac{A^{2}}{a^{2}}}
$$

The elevation of the body is $\Delta h=y_{0}-y_{1}=b\left(1-\sqrt{1-\frac{A^{2}}{a^{2}}}\right)$.

Using the principle of conservation of energy, we know that at the lowest point, the velocity is maximal, and the potential energy is 0, while at the position displaced horizontally by $A$, the potential energy is $m g \Delta h$, and the kinetic energy is 0.

$$
(1 / 2) m v^{2}=m g \Delta h
$$

In the case of a small displacement, the motion of the body can be considered as a harmonic oscillation with amplitude $A$, and the velocity amplitude is $v=A 2 \pi / T$. Substituting this and the value of $\Delta h$ into (2), and solving for $T$, while neglecting the term containing $A^{2}$, we obtain the same result as in the first solution.

József Klebniczki (Szeged, JATE Ságvári E. Gyak. Gimn., III. o. t.)