- Home
- Standard 11
- Physics
13.Oscillations
medium
The bob of simple pendulum having length $l$, is displaced from mean position to an angular position $\theta$ with respect to vertical. If it is released, then velocity of bob at lowest position
A
$\sqrt{2 g l \cos \theta}$
B
$\sqrt {2gl(1 + \cos \theta )} $
C
$\sqrt {2gl(1 - \cos \theta )} $
D
$\sqrt{2 gl}$
(AIPMT-2000)
Solution
(c)If suppose bob rises up to a height h as shown then after releasing potential energy at extreme position becomes kinetic energy of mean position
$ \Rightarrow mgh = \frac{1}{2}mv_{\max }^2$
$ \Rightarrow {v_{\max }} = \sqrt {2gh} $
Also, from figure $\cos \theta = \frac{{l – h}}{l}$
$ \Rightarrow h = l(1 – \cos \theta )$
So, ${v_{\max }} = \sqrt {2gl(1 – \cos \theta )} $
Standard 11
Physics
