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## Homework Statement

A 2.10 kg bucket containing 13.0 kg of water is hanging from a vertical ideal spring of force constant 130 N/m and oscillating up and down with an amplitude of 3.00 cm. Suddenly the bucket springs a leak in the bottom such that water drops out at a steady rate of 2.00 g/s.

When the bucket is half full, find the rate at which the period is changing with respect to time.

## Homework Equations

T=2[itex]\pi[/itex]sqrt(Ʃm/k)

## The Attempt at a Solution

I know that I need to find T as a function of t, then take the derivative wrt t and evaluate it at the time when the bucket is half full.

But I'm not sure how to set this up.. I tried:

T=2[itex]\pi[/itex]sqrt[(m

_{1}+m

_{2}+Δmt)/k]

where m

_{1}=2.1 kg, m

_{2}=13 kg, and Δm=0.02 kg/s

but it seems to be the wrong set-up. Anyone have any ideas?

Also, when the bucket is half full, is t=(half the mass of water)/(0.02 kg/s)=6.5/0.02=325 seconds? Is this the t that we should evaluate the derivative at? (assuming we figure it out first :tongue2:)

Thanks for helping!