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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