The bob moves along a circular arc of radius \(l\). The forces acting along the tangential direction (restoring forces) are:

• Component of gravity: \(mg \sin \theta\) (towards equilibrium)
• Applied force: \(F = mg\) (towards equilibrium)

Using Newton’s Second Law for tangential motion (\(F_t = m a_t = m l \alpha\)):

Assuming the angle \(\theta\) is small, we can approximate \(\sin \theta \approx \theta\).

Let \(\phi = \theta + 1\). Then \(\frac{d^2\phi}{dt^2} = \frac{d^2\theta}{dt^2}\). The equation becomes \(\frac{d^2\phi}{dt^2} = – \omega^2 \phi\), where \(\omega = \sqrt{\frac{g}{l}}\). This represents Simple Harmonic Motion (SHM) about the mean position \(\phi = 0\) (which corresponds to \(\theta = -1\) rad).

The general solution is \(\phi(t) = A \cos(\omega t)\).
At \(t=0\), the bob is at rest at \(\theta = \theta_0\). \[ \phi(0) = \theta_0 + 1 \implies A = \theta_0 + 1 \] So, the equation of motion is: \[ \theta(t) + 1 = (\theta_0 + 1) \cos(\omega t) \] We need to find the time \(t\) when the thread becomes vertical, i.e., \(\theta = 0\).

Given \(\theta_0 = \frac{2}{\sqrt{3}} – 1\): \[ 1 + \theta_0 = \frac{2}{\sqrt{3}} \] \[ \cos(\omega t) = \frac{1}{2/\sqrt{3}} = \frac{\sqrt{3}}{2} \] \[ \omega t = \frac{\pi}{6} \]

Substituting \(\omega = \sqrt{\frac{g}{l}}\): \[ t = \frac{\pi}{6 \omega} = \frac{\pi}{6} \sqrt{\frac{l}{g}} \]