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A simple pendulum of length $1 \mathrm{~m}$ and having a bob of mass $100 \mathrm{~g}$ is suspended in a car, moving on a circular track of radius $100 \mathrm{~m}$ with uniform speed $10 \mathrm{~m} / \mathrm{s}$. If the pendulum makes small oscillation in a radial direction about its equilibrium position, then its time period can be given by $\mathrm{T}=2 \pi / \alpha^{1 / 4}$. The value of $\alpha$ is [Take $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2$ ]
[Take g = 10 m/s2]
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[Take g = 10 m/s2]
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Verified Answer
The correct answer is:
101
The bob of the simple pendulum will experience the acceleration due to gravity and the centripetal acceleration provided by the circular motion of the car. Acceleration due to gravity $=\mathrm{g}$ Centripetal acceleration $=\mathrm{v}^2 / \mathrm{R}$ Effective acceleration will be
$$
g^{\prime}=\sqrt{g^2+\frac{v^4}{R^2}}
$$
Time period, $T=2 \pi \sqrt{\frac{\mathrm{L}}{\mathrm{g}^{\prime}}}$
$$
\begin{aligned}
& \Rightarrow \quad \frac{2 \pi}{2 / 4}=2 \pi \sqrt{\frac{1}{\sqrt{g^2+\frac{\mathrm{v}^4}{\mathrm{R}^2}}}} \\
& \Rightarrow \quad \alpha=\mathrm{g}^2+\frac{\mathrm{v}^4}{\mathrm{R}^2}=(10)^2+\frac{(10)^4}{(100)^2}=101
\end{aligned}
$$
$$
g^{\prime}=\sqrt{g^2+\frac{v^4}{R^2}}
$$
Time period, $T=2 \pi \sqrt{\frac{\mathrm{L}}{\mathrm{g}^{\prime}}}$
$$
\begin{aligned}
& \Rightarrow \quad \frac{2 \pi}{2 / 4}=2 \pi \sqrt{\frac{1}{\sqrt{g^2+\frac{\mathrm{v}^4}{\mathrm{R}^2}}}} \\
& \Rightarrow \quad \alpha=\mathrm{g}^2+\frac{\mathrm{v}^4}{\mathrm{R}^2}=(10)^2+\frac{(10)^4}{(100)^2}=101
\end{aligned}
$$
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