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(D) From the graph,the maximum distance from the ceiling is $100\ cm$ and the minimum distance is $30\ cm$.These represent the two extreme positions o

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The speed is maximum at $t = \frac{1}{4}\ T$

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(D) From the graph,the maximum distance from the ceiling is $100\ cm$ and the minimum distance is $30\ cm$.These represent the two extreme positions of the oscillation.The total distance between the extreme positions is $100\ cm - 30\ cm = 70\ cm$.Since the total distance between extreme positions is $2A$,we have $2A = 70\ cm$,which gives the amplitude $A = 35\ cm$.The mean position is at a distance of $30\ cm + 35\ cm = 65\ cm$ from the ceiling.At $t = 0$,the mass is at the upper extreme position $(100\ cm)$.At $t = \frac{1}{4}\ T$,the mass reaches the mean position $(65\ cm)$.At $t = \frac{1}{2}\ T$,the mass reaches the lower extreme position $(30\ cm)$.In simple harmonic motion,the speed is maximum at the mean position.Therefore,the speed is maximum at $t = \frac{1}{4}\ T$.

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