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A simple pendulum of length $l$ and having a bob of mass $\mathrm{M}$ is suspended in a car. The car is moving on a circular track of radius $R$ with a uniform speed $v$. If the pendulum makes small oscillations in a radial direction about its equilibrium position, what will be its time period?
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Verified Answer
In this case, the bob of the pendulum is under the action of two accelerations.
(i) Acceleration due to gravity ' $g$ ' acting vertically downwards.
(ii) Centripetal acceleration $a_c=\frac{v^2}{R}$ acting along the horizontal direction.
$\therefore$ Effective acceleration, $g^{\prime}=\sqrt{g^2+a_c^2}$
$$
\text { or } g^{\prime}=\sqrt{g^2+\frac{v^4}{R^2}}
$$
Now time period,
$$
T^{\prime}=2 \pi \sqrt{\frac{l}{g}}=2 \pi \sqrt{\frac{l}{\sqrt{g^2+\frac{v^4}{R^2}}}}
$$
(i) Acceleration due to gravity ' $g$ ' acting vertically downwards.
(ii) Centripetal acceleration $a_c=\frac{v^2}{R}$ acting along the horizontal direction.
$\therefore$ Effective acceleration, $g^{\prime}=\sqrt{g^2+a_c^2}$
$$
\text { or } g^{\prime}=\sqrt{g^2+\frac{v^4}{R^2}}
$$
Now time period,
$$
T^{\prime}=2 \pi \sqrt{\frac{l}{g}}=2 \pi \sqrt{\frac{l}{\sqrt{g^2+\frac{v^4}{R^2}}}}
$$
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