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A 0.950 kg mass hangs vertically from a ...

A 0.950 kg mass hangs vertically from a spring that he a spring constant of 8.50 N /m . The mass is set into vertical oscillations and after 600 s , you find tht the amplitude of the oscillation is `1/10` That of initial amplitude.
What is the damping constant associated with this motion?

Text Solution

Verified by Experts

The amplitude is `A=A_(0) e^(-alphat)`
Divide by `A_(0))` take natural logarithms and solve for `alpha`
`A/(A_(0))=e^(-alphat)impliesl_(n)((A)/(A_(0)))=-alphat`
`impliesalpha=-1/tl_(n)((A)/(A_(0)))`
Since `alpha=(beta)/(2m), "the "beta=2malpha.` Thus,
`beta=-(2m)/(t)l_(n)((A)/(A_(0)))`
Now substituting `0.950kg " for"m,600s.` for t and 0.10 for `((A)/(A_(0)))`
`beta=-(2m)/(t)l_(n)((A)/(A_(0)))=-(2(0.950))/(600)l_(n)0.10`
`=7.3xx10^(-3)kgs^(-1)`
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