A 2.00kg bucket containing 11.5kg of water is hanging from a vertical ideal spri

Question

A 2.00kg bucket containing 11.5kg of water is hanging from a vertical ideal spring of force constant 140N/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.

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

Given that

The mass of bucket is (m) =2kg

The mass of water contained in bucket is (M) =11.5kg

The vertical ideal spring of force constant(k) = 140N/m

The amplitude of oscillations(A) = 3.00 cm =0.03m

Suddenly the bucket springs a leak in the bottom such that water drops out at a steady rate of (dm/dt) = 2.00*10-3kg/s.

We know that

The time period is given by T =2piSqrt(m/k)

a)

when the bucket is half full, m = 5.75kg (half the water is empty) + 2.0 (weight of the bucket) = 7.75kg

Now the period of oscillation is given by

T= 2*3.14Sqrt(7.75/140) =1.477s

b)

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

dT/dt =- pi/Sqrt(mK) dm/dt =-3.14/Sqrt(7.75*140)* 2x10-3=-47.66s

c)

The rate at which the period is changing with respect to time is negative in above case. So, the period gets shorter.

d)

If the bucket becomes empty then m =2kg then the shortest time period isgiven by

T =2pi*Sqrt(m/k) =2*3.14*Sqrt(2/140) =0.7506s

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