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Chapter 14: Problem 75

A 2.00-kg bucket containing 10.0 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. 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) 1.57 s; (b) Period decreases at \( -3.57 \times 10^{-6} \text{ s/s} \); (c) Shortest period is 0.42 s.

Step by step solution

01

Understand the Problem

We are dealing with an oscillating spring-mass system. The total initial mass (bucket + water) is 2 kg + 10 kg = 12 kg. Water leaks at 2 g/s, so when the bucket is half full, it will have 5 kg of water left. We need to find (a) the period of oscillation when the bucket is half full, (b) how the period changes over time, and (c) the minimum possible period given the current rate of water loss.
02

Formulate the Period of Oscillation

The period of oscillation for a spring-mass system is given by:\[T = 2\pi \sqrt{\frac{m}{k}}\]where \(m\) is the total mass of the system and \(k\) is the spring constant (450 N/m). Initially, when the bucket is half full, \(m = 2 \text{ kg (bucket)} + 5 \text{ kg (water)} = 7 \text{ kg}\).
03

Calculate the Period (a)

Substitute \(m = 7 \text{ kg}\) and \(k = 450 \text{ N/m}\) into the equation:\[T = 2\pi \sqrt{\frac{7}{450}}\]Calculate to find:\[T \approx 1.57 \text{ seconds}\]
04

Analyze the Change in Period Over Time (b)

Given the mass changes with time because of the water leaking, we take the derivative of the period formula with respect to time. The derivative of \(T\) with respect to \(m\) is:\[\frac{dT}{dm} = \pi \frac{1}{\sqrt{km}}\]Convert the leak rate to kg/s: \(2 \text{ g/s} = 0.002 \text{ kg/s}\). The rate of change of the period w.r.t time \(t\) is given by:\[\frac{dT}{dt} = \frac{dT}{dm} \times \frac{dm}{dt} = \pi \frac{1}{\sqrt{450 \times 7}} \times (-0.002).\]
05

Calculate Change in Period Over Time

Calculate the actual rate of change:\[\frac{dT}{dt} = \pi \times \frac{1}{\sqrt{3150}} \times (-0.002) \approx -3.57 \times 10^{-6}\text{ s/s (seconds per second)}\]This shows that the period is decreasing, indicating it is getting shorter over time.
06

Determine the Shortest Period (c)

The shortest period occurs when the system has the smallest mass, which is just the bucket (2 kg, no water). Substitute into the period formula:\[T = 2\pi \sqrt{\frac{2}{450}} \approx 0.42 \text{ seconds.}\]

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Key Concepts

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

Period of Oscillation
The period of oscillation is the time it takes for a mass-spring system to complete one full cycle of movement. This cycle comprises a movement from the starting position back to the same position after passing through an entire wavelength. In a series of processes where the mass changes, like in this example of a leaking bucket, understanding the period becomes crucial for analysis.
The formula commonly used to calculate the period (\[T\]) is based on the mass (\[m\]) attached to a spring and the spring constant (\[k\]):
\[T = 2\pi \sqrt{\frac{m}{k}}\]
For the given problem: when the bucket is half full, the oscillating mass is 7 kg, and with the spring constant of 450 N/m, the calculated period is approximately 1.57 seconds. This formula emphasizes that as the mass decreases, the period also changes, illustrating the sensitivity of the system to variations in mass.
Spring Constant
The spring constant is a measure of a spring's stiffness. This fundamental property within a mass-spring system plays an essential role in determining the frequency and period of oscillations.
The spring constant is denoted by the symbol (\[k\]), and it defines how resistant the spring is to being compressed or extended. In mathematical terms, the spring constant appears in Hooke's Law, which states that the force (\[F\]) needed to extend or compress a spring by a distance (\[x\]) is directly proportional to that distance:
\[F = kx\]
Within the context of the period of oscillation, a high spring constant results in a quicker oscillation period, suggesting a stiffer spring responds more rapidly to changes in mass, as evident in the formula \[T = 2\pi \sqrt{\frac{m}{k}}\]. In the given exercise, a spring constant of 450 N/m indicates a moderately stiff spring.
Mass-Spring System
A mass-spring system is a foundational model in physics used to understand oscillatory motion. This system consists of a mass attached to a spring that can stretch and compress. When examining its dynamics, one observes that the mass, when displaced and released, will oscillate due to the restoring force of the spring.
Key characteristics of a mass-spring system include:
  • Mass (\[m\]): Represents the weight attached to the spring, crucial for calculating oscillation period.
  • Spring Constant (\[k\]): As discussed, it depicts the spring's stiffness and influences how rapidly the system oscillates.
In the given scenario, varying the bucket's mass (due to leaking water) impacts the system's dynamics, causing the period of oscillation to change over time. Understanding this model helps appreciate how mass and spring interplay shapes overall system behavior.
Dynamics of Oscillating Systems
Oscillating systems are prevalent in many scientific and engineering applications. Dynamics of such systems refer to how these systems evolve over time, especially in response to external changes or initial conditions.
In a mass-spring system:
  • The dynamics include the interaction between mass, spring constant, and external forces, such as gravity or friction.
  • This particular problem showcases how a changing mass influences oscillatory behavior, with water leaking out resulting in mass reduction over time.
As water progressively leaks, the mass-spring system transitions, displaying changed oscillation characteristics, mainly a decreasing period. The analysis of these dynamics is crucial for understanding behavior in real-life applications, like automotive suspension systems or earthquake-resistant buildings. Mastery over these principles allows for designing systems that better manage oscillatory motions.

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Most popular questions from this chapter

A 50.0-g hard-boiled egg moves on the end of a spring with force constant \(k =\) 25.0 N/m. Its initial displacement is 0.300 m. A damping force \(F_x = -bv_x\) acts on the egg, and the amplitude of the motion decreases to 0.100 m in 5.00 s. Calculate the magnitude of the damping constant \(b\).

While on a visit to Minnesota ("Land of 10,000 Lakes"), you sign up to take an excursion around one of the larger lakes. When you go to the dock where the 1500-kg boat is tied, you find that the boat is bobbing up and down in the waves, executing simple harmonic motion with amplitude 20 cm. The boat takes 3.5 s to make one complete up-and-down cycle. When the boat is at its highest point, its deck is at the same height as the stationary dock. As you watch the boat bob up and down, you (mass 60 kg) begin to feel a bit woozy, due in part to the previous night's dinner of lutefisk. As a result, you refuse to board the boat unless the level of the boat's deck is within 10 cm of the dock level. How much time do you have to board the boat comfortably during each cycle of up-and-down motion?

A large, 34.0-kg bell is hung from a wooden beam so it can swing back and forth with negligible friction. The bell's center of mass is 0.60 m below the pivot. The bell's moment of inertia about an axis at the pivot is 18.0 kg \(\cdot\) m\(^2\). The clapper is a small, 1.8-kg mass attached to one end of a slender rod of length \(L\) and negligible mass. The other end of the rod is attached to the inside of the bell; the rod can swing freely about the same axis as the bell. What should be the length \(L\) of the clapper rod for the bell to ring silently\(-\)that is, for the period of oscillation for the bell to equal that of the clapper?

A cheerleader waves her pom-pom in SHM with an amplitude of 18.0 cm and a frequency of 0.850 Hz. Find (a) the maximum magnitude of the acceleration and of the velocity; (b) the acceleration and speed when the pom-pom's coordinate is \(x = +\)9.0 cm; (c) the time required to move from the equilibrium position directly to a point 12.0 cm away. (d) Which of the quantities asked for in parts (a), (b), and (c) can be found by using the energy approach used in Section 14.3, and which cannot? Explain.

An object with mass 0.200 kg is acted on by an elastic restoring force with force constant 10.0 N/m. (a) Graph elastic potential energy \(U\) as a function of displacement \(x\) over a range of \(x\) from \(-\)0.300 m to \(+\)0.300 m. On your graph, let 1 cm \(=\) 0.05 J vertically and 1 cm \(=\) 0.05 m horizontally. The object is set into oscillation with an initial potential energy of 0.140 J and an initial kinetic energy of 0.060 J. Answer the following questions by referring to the graph. (b) What is the amplitude of oscillation? (c) What is the potential energy when the displacement is one-half the amplitude? (d) At what displacement are the kinetic and potential energies equal? (e) What is the value of the phase angle \(\phi\) if the initial velocity is positive and the initial displacement is negative?
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