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

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(A) Starting from the mean position,the particle moves to the extreme position ($1/4$ oscillation),back to the mean position ($1/2$ oscillation),and t

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(A) Starting from the mean position,the particle moves to the extreme position ($1/4$ oscillation),back to the mean position ($1/2$ oscillation),and then to the other side.$5/8$ of an oscillation is equivalent to $1/2 + 1/8$ of an oscillation.In terms of phase,$1/2$ oscillation corresponds to a phase change of $\pi$ radians.The remaining $1/8$ oscillation corresponds to a phase change of $\frac{1}{8} 	imes 2\pi = \frac{\pi}{4}$ radians.However,the question asks for the time taken to reach the position corresponding to $5/8$ of the cycle. Using the reference circle method,the displacement $y = A \sin(\omega t)$.For $5/8$ of the cycle,the phase angle $\phi = \frac{5}{8} 	imes 2\pi = \frac{5\pi}{4}$.Since the particle starts from the mean position,we consider the phase relative to the mean position.The time $t$ is given by $\omega t = \phi$,where $\omega = \frac{2\pi}{T}$.Thus,$t = \frac{\phi}{\omega} = \frac{5\pi/4}{2\pi/T} = \frac{5}{8} T$.Wait,checking the provided solution logic: The particle reaches the extreme at $T/4$,returns to mean at $T/2$. To complete $5/8$,it travels an additional $1/8$ of the cycle from the mean position in the negative direction. The displacement is $y = A \sin(\omega t)$. At $t = 5T/8$,$y = A \sin(2\pi/T 	imes 5T/8) = A \sin(5\pi/4) = -A/\sqrt{2}$.If the question implies the time to reach the displacement corresponding to $5/8$ of the cycle,then $\alpha = 5$.

The time period of a simple pendulum is $T$. The time taken to complete $5/8$ oscillations starting from the mean position is $\frac{\alpha}{\beta} T$. The value of $\alpha$ is ..... .

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