The path length of oscillation of a simple pe | vedclass

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(B) The path length of oscillation is the total distance between the two extreme positions, which is equal to $2a$, where $a$ is the amplitude.Given,

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$8 \pi \,cm/s$

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(B) The path length of oscillation is the total distance between the two extreme positions, which is equal to $2a$, where $a$ is the amplitude.Given, $2a = 16 \,cm$, so the amplitude $a = 8 \,cm$.The length of the pendulum is $l = 1 \,m$.The angular frequency of a simple pendulum is given by $\omega = \sqrt{\frac{g}{l}}$.Substituting $g = \pi^2 \,m/s^2$ and $l = 1 \,m$, we get $\omega = \sqrt{\frac{\pi^2}{1}} = \pi \,rad/s$.The maximum velocity $v_{max}$ in simple harmonic motion is given by $v_{max} = a\omega$.Substituting the values, $v_{max} = 8 \,cm 	imes \pi \,rad/s = 8\pi \,cm/s$.

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

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