Question

A heavy & big sphere is hang with a string of length $\ell$, this sphere moves in a horizontal circular path making an angle $\theta$ with vertical then its time period is:

• A. $T=2\pi \sqrt{\frac{\ell }{g}}$
• B. $T=2\pi \sqrt{\frac{\ell \sin \theta }{g}}$
• C. $T=2\pi \sqrt{\frac{\ell \cos \theta }{g}}$
• D. $T=2\pi \sqrt{\frac{\ell }{g\cos \theta }}$

Answer 1

The correct option is C $T=2\pi \sqrt{\frac{\ell \cos \theta }{g}}$

According to diagram

$T\cos \theta =mg....\left(I\right)$

$Tsin\theta =mr{\omega }^2$

$T\sin \theta =ml\sin \theta {\omega }^2$

$T=ml{\omega }^2$

Now, divided equation (II) by (I)

$\frac{1}{\cos \theta }=\frac{l{\omega }^2}{g}$

${\omega }^2=\frac{g}{l\cos \theta }$

Now, the time period is

$T=\frac{2\pi }{\omega }$

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

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

Hence, the time period is $2\pi \sqrt{\frac{l\cos \theta }{g}}$