The path length of oscillation of a simple pe | vedclass

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(C) Given: Length of pendulum $l = 1 \ m$,Path length $= 16 \ cm$,$g = \pi^2 \ m/s^2$. Since the path length is the total distance between the two ext

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$8 \pi \ cm/s$

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(C) Given: Length of pendulum $l = 1 \ m$,Path length $= 16 \ cm$,$g = \pi^2 \ m/s^2$. Since the path length is the total distance between the two extreme positions,the amplitude $A$ is half of the path length. $A = \frac{16 \ cm}{2} = 8 \ cm = 0.08 \ m$. The time period $T$ of a simple pendulum is given by $T = 2 \pi \sqrt{\frac{l}{g}}$. Substituting the values: $T = 2 \pi \sqrt{\frac{1}{\pi^2}} = 2 \pi 	imes \frac{1}{\pi} = 2 \ s$. The angular frequency $\omega$ is $\omega = \frac{2 \pi}{T} = \frac{2 \pi}{2} = \pi \ rad/s$. The maximum velocity $V_{max}$ is given by $V_{max} = A \omega$. $V_{max} = 8 \ cm 	imes \pi \ rad/s = 8 \pi \ cm/s$.

The path length of oscillation of a simple pendulum of length $1 \ m$ is $16 \ cm$. Its maximum velocity is $(g = \pi^2 \ m/s^2)$

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