A 2.40 bucket containing 11.5 of water is hanging from a vertical ideal spring o

Question

A 2.40 bucket containing 11.5 of water is hanging from a vertical ideal spring of force constant 115 and oscillating up and down with an amplitude of 3.00 . Suddenly the bucket springs a leak in the bottom such that water drops out at a steady rate of 2.00 .
When the bucket is half full, find the period of oscillation.
When the bucket is half full, find the rate at which the period is changing with respect to time.
Is the period getting longer or shorter?
What is the shortest period this system can have?

Explanation / Answer

T = 2(pi) Sqrt(m/K)
and we known that dm/dt = -2x10^-3 kg/s, and the rate of change of the period is dT/dt

A) when the bucket is half full, m = 5.5kg (half the water is empty) + 2.2 (weight of the bucket) = 7.7kg

Therefore, T = 2(pi) x Sqrt(7.7/120)

B) DT/dt = pi/Sqrt(mK) dm/dt

Therefore, DT/dt = pi / (Sqrt(7.7x120)) x -2x10^-3

C) as DT/dt is negative, the period is getting shorter

D) the bucket is empty, so m=2.2kg. Therefore, T = 2(pi) x Sqrt(2.4/120)

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