A 2.30-kg bucket containing 14.0 kg of water is hanging from a vertical ideal sp

Question

A 2.30-kg bucket containing 14.0 kg of water is hanging from a vertical ideal spring of force constant 450 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.

Part A

When the bucket is half full, find the period of oscillation.

Express your answer with the appropriate units.

Part B

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

Part C

Is the period getting longer or shorter?

Part D

What is the shortest period this system can have?

Explanation / Answer

A.

Time period is given by:

T = 2*pi*sqrt (m/k)

mass of bucket when it's half full = 2.3 kg + 7 = 9.3 kg

T = 2*pi*sqrt (9.3/450) = 0.903 sec

B.

T = sqrt (m/k)

dT/dt = [2*pi/(2*sqrt (mk))]*(dm/dt)

dT/dt = [pi/sqrt (mk)]*(dm/dt)

dT/dt = [pi/sqrt (9.3*450)]*(-2*10^-3) = -9.71*10^-5 sec per sec

C.

Since dT/dt is negative period will be getting shorter

D.

Shortest time period will be when all water is leaked out, then

T = 2*pi*sqrt (2.3/450) = 0.45 sec

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