The bob of a simple pendulum of length L is r | vedclass

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(A) For a simple pendulum with small angular displacement $	heta$,the angular position $	heta(t)$ at time $t$ is given by the equation of Simple Har

Vedclass · Accepted Answer

$L 	heta \cos \left[\sqrt{\frac{g}{L}} \cdot tight]$

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(A) For a simple pendulum with small angular displacement $	heta$,the angular position $	heta(t)$ at time $t$ is given by the equation of Simple Harmonic Motion $(SHM)$: $	heta(t) = 	heta_0 \cos(\omega t)$.Here,the initial angular displacement is $	heta$,so $	heta(t) = 	heta \cos(\omega t)$.The angular frequency $\omega$ of a simple pendulum is given by $\omega = \sqrt{\frac{g}{L}}$.Therefore,the angular displacement at time $t$ is $	heta(t) = 	heta \cos\left(\sqrt{\frac{g}{L}} \cdot tight)$.The linear displacement $s$ of the bob along the arc is given by the product of the length of the pendulum and the angular displacement: $s = L \cdot 	heta(t)$.Substituting the expression for $	heta(t)$,we get $s = L 	heta \cos\left(\sqrt{\frac{g}{L}} \cdot tight)$.

The bob of a simple pendulum of length $L$ is released from a position of small angular displacement $\theta$. Its linear displacement at time $t$ is ($g =$ acceleration due to gravity).

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