Question

A string of length $0.5m$ carries a bob with a period $2\pi s$. Calculate the angle of inclination of string with vertical and tension in the string.

Answer 1

For equilibrium , $m{\omega }^{2}r=mg\cos \theta$

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

$m(\frac{2\pi }{T}{)}^{2}r=mg\cos \theta$

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

Putting $r=0.5,T=2\pi$

$2\pi =2\pi \sqrt{\frac{0.5}{10\cos A}}$

$1=\frac{1}{20\cos \theta }$

$\cos \theta =\frac{1}{20}$

Hence angle of inclination $={\cos }^{-1}\frac{1}{20}$