Simple pendulums for small oscillations

The string is inextensible and of negligible mass and the amplitude is small, so the system is a simple pendulum and the formula for the period for small oscillations applies, which depends only on the length of the string and on $g$. The period of the pendulum is denoted by $T_0$, because the letter $T$ denotes the magnitude of the tension in the string. The frequency is the reciprocal of the period. The number of oscillations in one minute is the ratio of $60\mathrm{s}$ to the period; since the ratio equals $27.3$, the complete oscillations are $27$.

Squaring the period formula isolates the length, so $\ell =\frac{g{T}_0^2}{4{\pi }^2}$, slightly less than one meter. On the Moon the length of the string is the same and only the gravitational acceleration changes. The period is inversely proportional to $\sqrt{g}$, therefore on a celestial body with weaker gravity the pendulum oscillates more slowly, with a period about $2.5$ times that on Earth. The ratio of the periods is obtained without knowing $\ell$, and it provides a check of the second result, since $2.00\cdot 2.46\mathrm{s}=4.92\mathrm{s}$.

With the elevator at rest the elevator’s frame is inertial and the period formula with $g$ applies. When the elevator accelerates upward, its frame is not inertial, and in the elevator’s frame the particle is also acted on by the fictitious force $-m\vec{a}$, pointing downward. This force adds to the weight, so the pendulum behaves as in a gravitational field of strength $g+a$. The period $T_a$ therefore becomes shorter, because $g_{\mathrm{eff}}$ is greater than $g$. In the limiting case $a=0$ we recover $T_0$.

The amplitude of $8.0°$ corresponds to $0.140\mathrm{rad}$, an angle for which $\sin \theta \approx \theta$ with excellent accuracy, so the period is that of small oscillations. For the speed we use the conservation of mechanical energy, which holds because the tension is always perpendicular to the velocity and does no work and air resistance is negligible, so the only force that does work is the weight. The particle drops by the height $\ell (1-\cos {\theta }_0)$, and the potential energy lost becomes kinetic energy. At the lowest point the tension and the weight are vertical and the particle travels along a circle, therefore their resultant is the centripetal force; from this we obtain the tension, which exceeds the weight by $0.048\mathrm{N}$. As a check, harmonic motion gives $v_{\max }=\omega \ell {\theta }_0=3.50\cdot 0.80\cdot 0.140m/s=0.391m/s$, in agreement with the exact computation.