Question

A 2.50-kg bucket containing 12.0 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.
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?
Is the period getting longer or shorter?
Part D
What is the shortest period this system can have?

Answer 1

The period of oscillation when the bucket is half full is 1.03 s. The rate at which the period is changing with respect to time is 2.00 g/s. The period is getting shorter as the mass of the system decreases.

Step-by-step explanation:

Part A:

When the bucket is half full, the mass of water in it is 12.0 kg / 2 = 6.0 kg.

The effective mass of the system is the sum of the mass of the bucket and the mass of the water, which is 2.50 kg + 6.0 kg = 8.50 kg.

The period of oscillation can be calculated using the formula:

T = 2π√(m/k)

Where T is the period, m is the effective mass of the system, and k is the force constant of the spring.

T = 2π√(8.50 kg / 450 N/m) = 1.03 s

Part B:

To find the rate at which the period is changing concerning time, we need to find the derivative of the period concerning time.

The rate at which the period is changing concerning time is equal to the rate at which the effective mass is changing concerning time.

Since water is leaking out of the bucket at a steady rate of 2.00 g/s, the rate at which the effective mass is changing is also 2.00 g/s.

Part C:

The period is getting shorter as the mass of the system decreases due to the water leaking out of the bucket.

Part D:

The shortest period this system can have is when the bucket is empty and the effective mass of the system is equal to the mass of the bucket, which is 2.50 kg. Using the same formula as in part A, the period can be calculated as:

T = 2π√(2.50 kg / 450 N/m) = 0.31 s

Answer 2

The problem involves calculating the period of oscillation for a spring-bucket system, the rate at which this period changes, the direction of change in period, and the shortest possible period during a water leakage scenario. The key concept is simple harmonic motion in physics.

Step-by-step explanation:

The student is working on a problem related to simple harmonic motion (SHM) in a physics class. They are given a system consisting of a spring with a known force constant and a bucket with water oscillating vertically. We can derive the period of oscillation for the half-full bucket, the rate at which this period changes over time due to a steady rate of water leaking, the direction of change in period, and the shortest possible period of the system.

Part A: The period of oscillation when the bucket is half full is found using the formula T = 2π√(m/k) where m is the mass of the bucket plus half the initial water mass, and k is the spring constant.

Part B: The rate at which the period is changing is found by differentiating the period with respect to mass, dT/dm, and then multiplying that by the rate at which mass is changing, dm/dt, due to the water leaking.

Part C: As mass decreases due to the water leaking, the period gets shorter since the period is directly proportional to the square root of the mass.

Part D: The shortest period this system can have would occur when only the bucket remains, without any water inside. This is calculated using the period formula with minimal mass.