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A 0.2kg of mass hangs at the end of a sp...

A 0.2kg of mass hangs at the end of a spring. When 0.02kg more mass is added to the end of the spring, it stretches 7 cm more. If the 0.02 kg mass is removed what will be the period of vibration of the system?

Text Solution

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The correct Answer is:
`1.66`

When 0.02 kg mass is added, the spring stretches by 7 cm
As mg = Kx
` :. K = (mg)/( x) = (0.02 xx 10)/( 7 xx 10^(-2)) = (20)/( 7) Nm^(-1)`
When 0.02 kg mass is removed the period of vibration will be
`T = 2 pi sqrt((m.)/(K)) = 2 pi sqrt(( 0.2)/(20//7))`
`= 2 pi sqrt((7)/(100)) = (2 pi xx 2.645)/(10) = 1.66` s
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