What is the ratio between the distance travel | vedclass

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(C) In one time period $(T)$,an oscillator performing Simple Harmonic Motion $(SHM)$ moves from the mean position to the extreme position $(+A)$,then

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$4$

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(C) In one time period $(T)$,an oscillator performing Simple Harmonic Motion $(SHM)$ moves from the mean position to the extreme position $(+A)$,then back to the mean position,then to the other extreme position $(-A)$,and finally returns to the mean position.The total distance travelled $(d)$ is the sum of the magnitudes of these displacements:$d = |+A| + |-A| + |-A| + |+A| = 4A$Here,$A$ is the amplitude of the oscillation.The ratio of the distance travelled $(d)$ to the amplitude $(A)$ is:$	ext{Ratio} = \frac{d}{A} = \frac{4A}{A} = 4$

What is the ratio between the distance travelled by the oscillator in one time period and the amplitude?

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