Question

A pendulum of mass $m$ and length $l$ is suspended from the ceiling of a trolley which has a constant acceleration $a$. The maximum deflection $\theta$ of the pendulum from the vertical is

• A. ${\tan }^{-1}\left(\frac{a}{g}\right)$
• B. ${\tan }^{-1}\left(\frac{2a}{g}\right)$
• C. $2{\tan }^{-1}\left(\frac{a}{g}\right)$
• D. ${\tan }^{-1}\left(\frac{a}{g}\right)$

Answer 1

The correct option is C $2{\tan }^{-1}\left(\frac{a}{g}\right)$

The free-body diagram of the pendulum with respect to the trolley is shown in figure.


The forces acting on the bob are:
Force due to gravity, $mg$
Pseudo force, $ma$
Tension in the string, $T$

For the maximum deflection from vertical,
Work done by gravity is $W_g=-mgl(1-\cos \theta )$
$=-mgl\left(2{\sin }^2\frac{\theta }{2}\right)$
Work done by pseudo force, $W_{ps}=mal\sin \theta$
Work done by tension $W_T=0$

At the position of maximum deflection, the velocity of the bob is zero, $\Delta KE=0$.
Applying work-energy theorem, we get
$W_g+W_{ps}+W_T=\Delta K$
$\Rightarrow -mgl\left[2{\sin }^2\frac{\theta }{2}\right]+mal\left[2\sin \frac{\theta }{2}\cos \frac{\theta }{2}\right]=0$
$\therefore \tan \frac{\theta }{2}=\frac{a}{g}$ or $\theta =2{\tan }^{-1}\left(\frac{a}{g}\right)$