# JEE Main & Advanced Physics Simple Harmonic Motion Some Other Types of Pendulum

Some Other Types of Pendulum

Category : JEE Main & Advanced

(1) Infinite length pendulum :  If the length of the pendulum is comparable to the radius of earth then

$T=2\pi \sqrt{\frac{1}{g\left[ \frac{1}{l}+\frac{1}{R} \right]}}$

(i) If $l<<R$, then $\frac{1}{l}>>\frac{1}{R}$  so  $T=2\pi \sqrt{\frac{l}{g}}$

(ii) If $l>>R(\to \infty )\,\text{then }\frac{1}{l}<\frac{1}{R}$      so $T=2\pi \sqrt{\frac{R}{g}}=2\pi \sqrt{\frac{6.4\times {{10}^{6}}}{10}}\cong 84.6$ minutes

and it is the maximum time period which an oscillating simple pendulum can have

(iii) If $l=R$   so  $T=2\pi \sqrt{\frac{R}{2g}}\cong 1hour$

(2) Second's Pendulum : It is that simple pendulum whose time period of vibrations is two seconds.

Putting $T=2\,\sec$ and $g=9.8m/{{\sec }^{2}}$ in $T=2\pi \sqrt{\frac{l}{g}}$ we get $l=\frac{4\times 9.8}{4{{\pi }^{2}}}=0.993$$m=99.3\,m$

Hence length of second?s pendulum is 99.3 cm or nearly 1 meter on earth surface.

For the moon the length of the second?s pendulum will be 1/6 meter [As ${{g}_{moon}}=\frac{{{g}_{\text{Earth}}}}{6}$]

(3) Compound 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 $\tau =mgl\ \sin \theta$                                                                                           ...(i)

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

If $l$ be the M.I. of the body about O. Then $\tau =I\alpha$ ...(ii)

From (i) and (ii), we get $I\frac{{{d}^{2}}\theta }{d{{t}^{2}}}=-mgl\sin \theta$ as $\theta$ and $\frac{{{d}^{2}}\theta }{d{{t}^{2}}}$ are oppositely directed $\Rightarrow$ $\frac{{{d}^{2}}\theta }{d{{t}^{2}}}=-\frac{mgl}{I}\theta$ since $\theta$is very small

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

$\omega =\sqrt{\frac{mgl}{I}} \Rightarrow T=2\pi \sqrt{\frac{I}{mgl}}$

Also $I={{I}_{cm}}+m{{l}^{2}}$                (Parallel axis theorem)

$=m{{k}^{2}}+m{{l}^{2}}$                                          (where $k=$ radius of gyration)

$\therefore$ $T=2\pi \sqrt{\frac{m{{K}^{2}}+m{{l}^{2}}}{mgl}}=2\pi \sqrt{\frac{\frac{{{K}^{2}}}{l}+l}{g}}$$=2\pi \sqrt{\frac{{{l}_{\text{eff}}}}{g}}$

${{l}_{eff}}=$  Effective length of pendulum = Distance between point of suspension and centre of mass.

Some common physical pendulum

 Body Time period Bar $T=2\pi \sqrt{\frac{2\,l}{3g}}$ Ring $T=2\pi \sqrt{\frac{2R}{g}}$ Disc $T=2\pi \sqrt{\frac{3R}{2g}}$

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