Question

A heavy particles is tied to end A of string of length 1.6 M its other end is fixed its revolves as a conical pendulum with the string making 60° degree with the vertical then the time period of the revolution of the particle is

Answer 1

Final Answer : $\frac{2\sqrt{2}\pi}{5}s$

Steps:
1) Since, particle revolves with uniform velocity.
=> Radial Force is provided by horizontal component of Tension.
$Tsin(60\degree) = m\omega^2r \\ \\ => Tsin(60\degree) = m\omega^2l sin(60\degree) \\ \\ => T = m\omega^2l$

2) Also,
Vertical Component of Tension is balanced by 'mg'.

$Tcos(60\degree) = mg \\ \\ => m\omega^2l cos(60\degree) = mg \\ \\ => \omega =\sqrt{ \frac{2g}{l}}$

3) Then, we have
Time Period,
$T = \frac{2\pi}{\omega} \\ \\ = 2\pi \sqrt{\frac{l}{2g}} = 2\pi \sqrt{\frac{1.6}{2*10}} \\ \\ = \frac{2\sqrt{2}\pi}{5}\: s$