Question
A simple pendulum is set into oscillations in a uniformly travelling car along a horizontal road. What happens to its period if the car takes a sudden turn towards the left ?

Answer


The equilibrium position of the string makes an angle $\theta=\tan ^{-1}\left( a _{ c } / g \right)$ with the vertical due to the centrifugal force to the right.
The centripetal acceleration, $a_c$, is horizontal and towards the left. The acceleration due to gravity is vertically downward.
$
\therefore g _{\text {eff }}=\sqrt{g^2+a_{ c }^2}
$
so that the period of oscillation $T=2 \pi \sqrt{L / g_{\text {eff }}}$
$\therefore$ As the car takes a sudden left turn, the period of oscillation decreases.

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