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Chapter 14: Problem 81

CALC A 2.00 -kg bucket containing 10.0 \(\mathrm{kg}\) of water is hanging from a vertical ideal spring of force constant 125 \(\mathrm{N} / \mathrm{m}\) and oscillating up and down with an amplitude of 3.00 \(\mathrm{cm} .\) Suddenly the bucket springs a leak in the bottom such that water drops out at a steady rate of 2.00 \(\mathrm{g} / \mathrm{s}\) . When the bucket is half full, find (a) the period of oscillation and (b) the rate at which the period is changing with respect to time. Is the period getting longer or shorter? (c) What is the shortest period this system can have?

Short Answer

Expert verified
(a) 1.49 s; (b) Period is decreasing at 0.0004 s/s; (c) Shortest period is 1.00 s.

Step by step solution

01

Determine Initial Mass and Parameters

Initially, the bucket has a mass of \(2.00\, \mathrm{kg}\) and contains \(10.0\, \mathrm{kg}\) of water. Therefore, the total mass \(m = 12.0\, \mathrm{kg}\). The force constant of the spring is \(k = 125\, \mathrm{N/m}\). The amplitude \(A = 0.03\, \mathrm{m}\) (converted from \(3.00\, \mathrm{cm}\)). The rate of water leakage is \(2.00\, \mathrm{g/s} = 0.002\, \mathrm{kg/s}\).
02

Calculate Period of Oscillation When Half Full

The period of oscillation for a mass-spring system is given by \(T = 2\pi \sqrt{\frac{m}{k}}\). When the bucket is half full, the water mass is \(5.0\, \mathrm{kg}\), so the total mass \(m = 2.0 + 5.0 = 7.0\, \mathrm{kg}\). Thus, the period \(T = 2\pi \sqrt{\frac{7.0}{125}} \approx 1.49\, \mathrm{s}\).
03

Determine Rate of Change of Period

The rate at which the period changes with respect to time is given by differentiating the period formula: \(\frac{dT}{dt} = \frac{d}{dt}\left(2\pi \sqrt{\frac{m(t)}{k}}\right)\). Since \(m(t)=7-0.002t\) (since we're losing 0.002 kg/s), we use the chain rule: \(\frac{dT}{dt} = \pi \cdot \frac{1}{\sqrt{\frac{m}{k}}} \cdot \frac{dm}{dt}\). Solving this gives \(\frac{dT}{dt} = -0.0004\, \mathrm{s/s}\), indicating the period is decreasing.
04

Find the Shortest Period

The shortest period occurs when the mass is minimized, which is when all the water has leaked out. Here, only the bucket remains, so \(m = 2.0\, \mathrm{kg}\). Substituting into the period equation, \(T = 2\pi \sqrt{\frac{2.0}{125}} \approx 1.00\, \mathrm{s}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Mass-Spring System
A mass-spring system is a classic example of harmonic motion that is often used to describe physical systems. In this type of system, a mass is attached to a spring, and when displaced, it oscillates back and forth.
In our scenario, the bucket and the water together act as the mass, while the spring is characterized by its spring constant.
As the bucket containing water leaks, the mass of this system changes over time, which in turn impacts its oscillation dynamics.
The mass determines how much force the spring exerts back in response to the displacement, making it a crucial factor in understanding the overall movement of the system.
Overall, knowing how the mass-spring system operates can provide a better grasp of harmonic motion scenarios in various forms, from a simple classroom setup to complex engineering designs.
Spring Constant
The spring constant, denoted as \( k \), is a measure of a spring鈥檚 stiffness and is defined as the force required to compress or extend the spring by a unit length.
In this exercise, the spring constant is given as 125 N/m. This value indicates how resistant the spring is to being stretched or compressed.
A higher spring constant means a stiffer spring, which impacts the oscillation period.
In harmonic motion, the spring constant works together with the mass to determine how quickly an oscillation cycle completes.
The rigidity that the spring constant provides ensures that the system returns to equilibrium quickly, which is why it plays a vital role in the calculation of the period of oscillation.
Period of Oscillation
The period of oscillation is the time it takes for a mass-spring system to complete one full cycle of motion.
Mathematically, it is given by the equation: equation: \( T = 2\pi \sqrt{\frac{m}{k}} \)
where \( m \) is the mass and \( k \) is the spring constant.
In this problem, when the bucket is half-full, the total mass is reduced. As the mass decreases, the velocity of oscillations increases, leading to a shorter period.
  • Initial state with full bucket: larger mass, longer period.
  • Half-full bucket: mass decrease, shorter period of about 1.49 seconds.
  • Empty bucket: minimum mass, shortest period of around 1 second.
Understanding how mass affects the period helps in analyzing real-world systems where mass can vary over time, such as with the leaking bucket.
Rate of Change
The rate of change in the context of this exercise refers to how the period of oscillation changes as the mass of the system decreases.
Here, water leaks at a continuous rate, causing a decrease in mass, which directly affects the period.
To determine the rate of change, we differentiate the formula for the period with respect to time:\( \frac{dT}{dt} = \pi \cdot \frac{1}{\sqrt{\frac{m}{k}}} \cdot \frac{dm}{dt} \).
This rate calculated as \(-0.0004 \, \mathrm{s/s}\) implies the period is getting shorter as the water leaks out.
The steady reduction in mass results in a faster oscillation period, illustrating a practical application of calculus in assessing dynamic changes in physical systems.
By understanding the rate of change, one can predict how the system will behave over time, which is critical in engineering and physics applications.

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Most popular questions from this chapter

A uniform beam is suspended horizontally by two identical vertical springs that are attached between the ceiling and each end of the beam. The beam has mass 225 \(\mathrm{kg}\) , and a 175 -kg sack of gravel sits on the middle of it. The beam is oscillating in \(\mathrm{SHM}\) , with an amplitude of 40.0 \(\mathrm{cm}\) and a frequency of 0.600 cycle/s. (a) The sack of gravel falls off the beam when the beam has its maximum upward displacement. What are the frequency and amplitude of the subsequent SHM of the beam? (b) If the gravel instead falls off when the beam has its maximum speed, what are the frequency and amplitude of the subsequent SHM of the beam?

You are watching an object that is moving in SHM. When the object is displaced 0.600 m to the right of its equilibrium position, it has a velocity of 2.20 \(\mathrm{m} / \mathrm{s}\) to the right and an acceleration of 8.40 \(\mathrm{m} / \mathrm{s}^{2}\) to the left. How much farther from this point will the object move before it stops momentarily and then starts to move back to the left?

A 0.500 -kg glider, attached to the end of an ideal spring with force constant \(k=450 \mathrm{N} / \mathrm{m},\) undergoes \(\mathrm{SHM}\) with an amplitude of 0.040 \(\mathrm{m} .\) Compute (a) the maximum speed of the glider; (b) the speed of the glider when it is at \(x=-0.015 \mathrm{m} ;\) (c) the magnitude of the maximum acceleration of the glider; (d) the acceleration of the glider at \(x=-0.015 \mathrm{m} ;\) (e) the total mechanical energy of the glider at any point in its motion.

A 0.400 -kg object undergoing SHM has \(a_{x}=-2.70 \mathrm{m} / \mathrm{s}^{2}\) when \(x=0.300 \mathrm{m} .\) What is the time for one oscillation?

A Pendulum on Mars. A certain simple pendulum has a period on the earth of 1.60 s. What is period on the surface of Mars, where \(g=3.71 \mathrm{m} / \mathrm{s}^{2} ?\)
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