Question

A 2.20 kg bucket containing 11.0 kg of water is hanging from a vertical ideal spring of force constant 120 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. A)When the bucket is half full, find the rate at which the period is changing with respect to time. B)What is the shortest period this system can have?

Answer 1

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)

T = 1.59

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

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

= - 6197.9

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)

T =0.888