SHM in Pendulums

TYPES OF PENDULUM

COMPOUND PENDULUM

When a rigid body is suspended from an axis and made to oscillate about that then it is called compound pendulum.

C = Initial position of center of mass

C' = Position of center of mass after time t

S = Point of suspension

$\ell$ = Distance between point of suspension and center of mass (it remains constant during motion)

$T=2\pi \sqrt{\frac{I}{mg\ell }}$

$I=I_{CM}+m{\ell }^2$

where $I_{CM}$= moment of inertia relative to the axis which passes from the center of mass & parallel to the axis of oscillation.

$T=2\pi \sqrt{\frac{I_{CM}+m{\ell }^2}{mg\ell }}$

where $I_{CM}=m{k}^2$

k = gyration radius (axis passes from centre of mass)

$T=2\pi \sqrt{\frac{m{k}^2+m{\ell }^2}{mg\ell }}$

$T=2\pi \sqrt{\frac{k^2+{\ell }^2}{\ell g}}=2\pi \sqrt{\frac{L_{eq}}{g}}$

$L_{eq}=\frac{k^2}{\ell }+\ell$= equivalent length of simple pendulum;

T is minimum when $\ell =k.$

TORSIONAL PENDULUM

Consider a body, such as a disc or a rod, suspended at the end of a wire, as shown in the fig.

Fig. Torsional Pendulum

When the end of the wire is twisted by an angle $\theta$, the restoring torque is given by

$\tau =-k\theta$

where k is called the torsional constant.

Using Newton's law

$I\frac{d^2\theta }{d{t}^2}=-k\theta$

or $I\frac{d^2\theta }{d{t}^2}+\frac{k}{I}\theta =0$

Thus, the angular frequency and the time period of oscillation are given by

$\omega =\sqrt{\frac{k}{I}}$

and $T=2\pi \sqrt{\frac{I}{k}}$

PHYSICAL PENDULUM

Any rigid body suspended from a fixed support constitutes a physical pendulum. Consider the situation when the body is displaced through a small angle $\theta$. Torque on the body about o is given by

$\tau =mg\ell \sin \theta$ (1)

where $\ell$ = distance between point of suspension and centre of mass of the body.

If I be the M.I. of the body about O.

then $\tau =I\alpha$ (2)

From (1) and (2), we get

$I\frac{d^2\theta }{d{t}^2}=-mg\ell \sin \theta$ as $\theta$ and $\frac{d^2\theta }{d{t}^2}$ are oppositely directed.

$\Rightarrow \frac{d^2\theta }{d{t}^2}=-\frac{mg\ell }{I}\theta$ since $\theta$ is very small.

Comparing with the equation $\frac{d^2\theta }{d{t}^2}=-{\omega }^2\theta$, we get

$\omega =\sqrt{\frac{mg\ell }{I}}$

$\Rightarrow T=2\pi \sqrt{\frac{I}{mg\ell }}$

Exmaple : A rod of mass M and length L is pivoted about its end O as shown in the figure(10.19). Find the period of SHM.

Fig. A rod suspended about its end acts like a physical pendulum

Solution: Restoring torque is

${\tau }_0=-Mg\frac{L}{2}\sin \theta$

Using Newton's Second Law and the small angle approximation, we get

$I_0\frac{d^2\theta }{d{t}^2}+Mg\frac{L}{2}\theta =0$

Using parallel-axes theorem

$I_0=I_c+M{\left(\frac{L}{2}\right)}^2=\frac{M{L}^2}{12}+\frac{M{L}^2}{4}=\frac{M{L}^2}{3}$

$\therefore \frac{d^2\theta }{d{t}^2}+\left(\frac{3}{2}\frac{g}{L}\right)\theta =0$

Thus, $T=2\pi \sqrt{\frac{2L}{3g}}$