Question

A small sphere of mass 200 gm is attached to an inextensible string of length 130 cm whose upper end is fixed to the ceiling. The sphere is made to describe a horizontal circle of radius 50 cm. Calculate the time period of this conical pendulum and the tension in the string. $({\pi }^2=10)$

Answer 1

The angle is given as,

$\cos \theta =\frac{\sqrt{l^2-r^2}}{l}$

The time period of the pendulum is given as,

$t=2\pi \sqrt{\frac{l\cos \theta }{g}}$

$t=2\pi \sqrt{\frac{\sqrt{l^2-r^2}}{g}}$

$t=2\pi \sqrt{\frac{\sqrt{{\left(130\times {10}^{-2}\right)}^2-{\left(50\times {10}^{-2}\right)}^2}}{9.8}}$

$t^2=4{\pi }^2\times \frac{\sqrt{{\left(130\times {10}^{-2}\right)}^2-{\left(50\times {10}^{-2}\right)}^2}}{9.8}$

$t=2.21\sec$

The tension on the string is given as,

$T\cos \theta =mg$

$T=\frac{200\times {10}^{-3}\times 9.8}{\frac{\sqrt{{\left(130\times {10}^{-2}\right)}^2-{\left(50\times {10}^{-2}\right)}^2}}{130\times {10}^{-2}}}$

$=2.123N$

Thus, the time period of the pendulum is $2.21\sec$ and the tension in the string is $2.123N$.