The time period of a simple pendulum is T. Th | vedclass

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(C) The angular displacement of a simple pendulum performing simple harmonic motion can be represented as $	heta(t) = \beta \sin(\omega t)$,where $\b

Vedclass · Accepted Answer

$\frac{T}{2\pi} \sin^{-1} \left( \frac{\alpha}{\beta} \right)$

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(C) The angular displacement of a simple pendulum performing simple harmonic motion can be represented as $	heta(t) = \beta \sin(\omega t)$,where $\beta$ is the angular amplitude and $\omega$ is the angular frequency.Given that the time period is $T$,the angular frequency is $\omega = \frac{2\pi}{T}$.We want to find the time $t$ taken to reach an angular displacement $	heta = \alpha$ starting from the equilibrium position $O$ (where $	heta = 0$ at $t = 0$).Substituting the values into the equation: $\alpha = \beta \sin(\omega t)$.Rearranging for $t$: $\sin(\omega t) = \frac{\alpha}{\beta}$.$\omega t = \sin^{-1} \left( \frac{\alpha}{\beta} ight)$.Substituting $\omega = \frac{2\pi}{T}$: $\left( \frac{2\pi}{T} ight) t = \sin^{-1} \left( \frac{\alpha}{\beta} ight)$.Therefore,$t = \frac{T}{2\pi} \sin^{-1} \left( \frac{\alpha}{\beta} ight)$.

The time period of a simple pendulum is $T$. The angular amplitude is $\beta$. How much time will the bob of the pendulum take to move from the equilibrium position $O$ to position $A$,where the string makes an angle $\alpha$ with the vertical?

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