Answer
(a) $T = \frac{mg}{cos~\phi}$
(b) The period of the pendulum is $2\pi~\sqrt{\frac{L~cos~\phi}{g}}$
Work Step by Step
(a) The vertical component of the string's tension is equal in magnitude to the bob's weight. We can find an expression for the tension $T$:
$T~cos~\phi = mg$
$T = \frac{mg}{cos~\phi}$
(b) The horizontal component of the tension provides the centripetal force to keep the ball moving around in a circle. We can find the ball's angular speed:
$T~sin~\phi =m~\omega^2~r$
$(\frac{mg}{cos~\phi})~sin~\phi =m~\omega^2~L~sin~\phi$
$\frac{mg}{cos~\phi} =m~\omega^2~L$
$\frac{g}{L~cos~\phi} =\omega^2$
$\omega = \sqrt{\frac{g}{L~cos~\phi}}$
We can find the period $P$ of the pendulum:
$P = \frac{2\pi}{\omega}$
$P = \frac{2\pi}{\sqrt{\frac{g}{L~cos~\phi}}}$
$P = 2\pi~\sqrt{\frac{L~cos~\phi}{g}}$
The period of the pendulum is $2\pi~\sqrt{\frac{L~cos~\phi}{g}}$
