91Ó°ÊÓ

\(\begin{array}{l}{\rm{force}}\;{\rm{constant}} = k\\{\rm{mass}} = M\\{\rm{Angle}} = \theta \end{array}\)

The time required to complete one cycle is known as the period.

The motion is simple harmonic if the equation of motion for the angular oscillations is of the form \(\frac{{{d^2}\theta }}{{d{t^2}}} = - \frac{K}{I}\theta \) , and in this case the period is,\(T = 2\pi \sqrt {\frac{I}{K}} \) .

For a slender rod pivoted about tis center \(I = \frac{1}{{12}}M{L^2}\)

The torque on the rod about the pivot is \(\tau = \left( {k\frac{L}{2}\theta } \right)\frac{L}{2} = I\alpha = I\frac{{{d^2}\theta }}{{d{t^2}}}\) gives

Is proportional to \(\theta \) and the motion is angular SHM. \(\frac{K}{I} = \frac{{3k}}{M} = {\omega ^2}\)

\(T = 2\pi \sqrt {\frac{M}{{3k}}} \)

The expression we used for the torque \(\tau = - \left( {k\frac{L}{2}\theta } \right)\frac{L}{2}\) is valid only when \(\theta \)is small enough for \(\sin \;\theta \approx \theta \;{\rm{and}}\;\cos \;\theta \approx 1\)

Hence the period is \(T = 2\pi \sqrt {\frac{M}{{3k}}} \)