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A spring stretches by 0.020m when a 1.5 ...

A spring stretches by `0.020m` when a 1.5 kg object is suspended from its end . How much mass should be attached to the spring so that its frequency of vibration is `f=3.1Hz` ?

Text Solution

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`sumF_(y)=ma_(y)=0`
`kx-mg=0impliesk=(mg)/(x)`
The force constant is,
`k=((1.5)(9.8))/((0.020))=14.7/0.020=735N//m.`
Frequency `=1/(2pi) sqrt((k)/(m))`
`implies4pi^(2)f^(2)=k/m`
`m=(k)/(4pi^(2)f^(2))impliesm=((735))/(4pi^(2)(3.0)^(2))`
`=((735))/(4xx(3.75)^(2)xx(9.0))`
`=735/(4xx9.85xx9.0)`
`=(735)/(354.6)=2.07kg`
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