A classical mechanics problem by Abdullah Faysal

$T=2π \sqrt{\frac{ L \cos \theta}{g}}=2(3.14) \sqrt{\frac{ 0.23 \cos 27^\circ}{9.81}}=0.908\text{ [s]}.$

Let $l$ be the length of the wire and $\theta$ the angle it does with the vertical. The trajectory of the pendulum describes a circle or radius $r=lsin(\theta)$ . We can define $e_r$ to be the unit vector directed along the radius and $e_z$ the vertical component, perpendicular to $e_r$ .

First let's project the forces on the axes. There's $T$ the tension of the wire and $W=mg$ the weight of the pendulum. The $z$ -component of the tension exactly compensates the weight (since the pendulum doesn't move along the z-axis to keep $\theta$ constant), so we get $T= mg/cos(\theta)$ . Hence, the $r$ -component of $T$ is $T_r= -mgtan(\theta)$ .

We then apply Newton's second law: $m$ a = $mgtan(\theta)$ $e_r$ . Since the speed in polar coordinate is $v = r d \theta /dt$ and $r$ is constant, $a = v^2/r$ . Thus we have $v= \sqrt{grtan(\theta)}$ . The oscillation period is given by the perimeter of the circle divided by the speed: $T=2\pi r/v$ .

With $\theta = 27°$ and $l = 0.23 m$ we have $T \approx 0.9085 s$