NBPhO 2023, Теория и експеримент — Задача 6. String-coupled masses

i) (2 т.)

For in-phase oscillations, the string connecting the masses always remains parallel to the $x$-axis. In that case, the motion is confined to the $yz$-plane and is identical to a pendulum with length $r\sin\theta$ hung from the axis of rotation:

Grading:

• Stating $\sum F_x = 0$, or stating that the connecting string remains parallel to the $x$-axis (0.6 pts)
• Realising the effective pendulum length is $r\sin\theta$, i.e. the axis of rotation is the line connecting the wall points (0.8 pts)
• Correct expression for $\omega_1$ (0.6 pts)

ii) (3 т.)

Now suppose one of the balls moves along the $y$ axis to a distance $ym$ and the other ball — by $-y$. The length of the horizontal string remains constant, hence $y^2 + (r - x)^2 = r^2$, where $x$ denotes the displacement of the ball in $x$-direction. For small displacements, we can neglect the smallest term $x^2$, hence $x = y^2/2r$. The length of the slanted string remains also constant, hence $(r\sin\theta - z)^2 + (r\cos\theta + x)^2 + y^2 = r^2$, where $z$ denotes the vertical displacement. Neglecting here the smallest terms $z^2$ and $x^2$, we obtain $2rz\sin\theta = 2rx\cos\theta + y^2 = y^2(1 + \cos\theta)$, hence

Now we can write down energy conservation law: $\frac{1}{2}\dot{y}^2 + gz = const$, where dot denotes the time derivative. Differentiating this over time we obtain

so that

Grading:

• Equation for a cylinder surface (0.4 pts)
• Equation for a sphere surface (0.4 pts)
• Both surface equations correct and coordinates consistent (0.2 pts)
• Realising $x^2$- and $z^2$-terms are negligible, and not discarding $y^2$ (0.2 pts)
• Solving $z$ in terms of $y$ (0.5 pts)
• Energy conservation between $z$ and $\dot{y}$ (0.3 pts)
• Solving relation between $\ddot{y}$ and $y$ (0.6 pts)
• Correct expression for $\omega_2$ (0.3 pts)
• $\omega_2$ correctly in terms of $\omega_1$ (0.1 pts)

Alternative solution: Let one of the masses (mass A) be displaced by $y$ and the other by $-y$. Let $T'$ be the tension in the two outermost strings (equal magnitudes by symmetry) and $T$ in the connecting string. Newton’s law along the $y$-axis for mass A then yields

i.e.

Without oscillations, $T_0' = mg/\sin\theta$ and $T_0 = T_0'\cos\theta$. Let us write $T' = T_0' + \Delta T'$ and $T = T_0 + \Delta T$, i.e.

Let us make some estimates. Clearly, $|\Delta T| \sim |\Delta T'|$. The maximum of $\Delta T'$ occurs when the mass swings by its lowest point with velocity $\dot{y}$, where its trajectory has a radius of curvature $R \sim r\sin\theta$. Moreover, small normal mode oscillations imply $y = A\cos\omega_2 t$ for some amplitude $A \ll r\sin\theta$, i.e. $|\dot{y}| \le A\omega_2$. Finally, assume $\omega_2 = k\omega_1$ holds for some $k \sim 1$. We have

i.e. quadratically smaller in $(A/r\sin\theta)$ than $T_0'$. Thus, $T' + T \approx mg\frac{1 + \cos\theta}{\sin\theta}$ and

(The result shows $k \le \sqrt{2}$, i.e. the assumption holds.) Grading:

• Force equation, either along $y$ or full vector form (0.4 pts)
• Solving relation between $\ddot{y}$ and $y$ in terms of tension forces (0.4 pts)
• Unperturbed values for $T'$ and $T$ (0.4 pts)
• Reasonable estimate of $|\Delta T'|$ (0.6 pts)
• Showing $|\Delta T'|$ is quadratically smaller than $T'$ (0.8 pts)
• Correct expression for $\omega_2$ (0.3 pts)
• $\omega_2$ correctly in terms of $\omega_1$ (0.1 pts)
• Remark: Accept solution with assumption $\Delta T'/T' \ll 1$ without estimates