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

A 0.2 kg of mass hangs at the end of a spring. When 0.02 kg 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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When 0.02 kg mass is added, the spring stretches by 7 cm
As `mg=kx`
`therefore k=(mg)/(x)=(0.02xx10)/(7xx10^(-2))=(20)/(7)Nm^(-1)`
When 0.02 kg mass is removed, the period of vibration will be
`T=2xsqrt((m)/(k))=2pisqrt((0.2)/(20//7))=2pisqrt((7)/(100))=(2pixx2.645)/(10)=1.66s`
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