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A simple pendulum is released from rest at the horizontally stretched position. When the string makes an angle $\theta$ with the vertical, the angle $\phi$ which the acceleration vector of the bob makes with the string is given by
$\phi=0$
$\phi=\tan ^{-1}\left(\frac{\tan \theta}{2}\right)$
$\phi=\tan ^{-1}(2 \tan \theta)$
$\phi=\frac{\pi}{2}$
Solution
(b)
Let $v$ is the velocity of bob at position $\theta$ after being released from horizontal position.
From energy conservation, we have
$m g h=\frac{1}{2} m v^2$
$\Rightarrow m g(l \cos \theta)=\frac{1}{2} m v^2$
$\Rightarrow \quad \frac{v^2}{l}=2 g \cos \theta \quad \dots(i)$
Restoring force on bob is component of weight.
So, tangential component of acceleration is
$a_t=g \sin \theta \quad \dots(ii)$
If total acceleration a makes angle $\phi$ with string. Then,
Tangential component,
$a_t=a \sin \phi$
and radial component, $a_c=a \cos \phi$
So, $\quad \frac{a_t}{a_c}=\frac{a \sin \phi}{a \cos \phi}=\tan \phi$
$\Rightarrow \quad \tan \phi=\frac{a_i}{a_c} \quad \dots(iii)$
Substituting values of $a_t$ and $a_c$ from Eqs.
$(i)$ and $(ii)$ in Eq. $(iii)$, we get
$\tan \phi=\frac{g \sin \theta}{2 g \cos \theta}$
$\Rightarrow \quad \tan \phi=\frac{\tan \theta}{2}$
So, $\quad \phi=\tan ^{-1}\left(\frac{\tan \theta}{2}\right)$
