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You hang various masses \(m\) from the end of a vertical, 0.250-kg spring that obeys Hooke's law and is tapered, which means the diameter changes along the length of the spring. Since the mass of the spring is not negligible, you must replace \(m\) in the equation \(T =\) 2\(\pi\sqrt{ m/k }\) with \(m + m_\mathrm{eff}\), where \(m_\mathrm{eff}\) is the effective mass of the oscillating spring. (See Challenge Problem 14.93.) You vary the mass m and measure the time for 10 complete oscillations, obtaining these data: (a) Graph the square of the period \(T\) versus the mass suspended from the spring, and find the straight line of best fit. (b) From the slope of that line, determine the force constant of the spring. (c) From the vertical intercept of the line, determine the spring's effective mass. (d) What fraction is \(m_\mathrm{eff}\) of the spring's mass? (e) If a 0.450-kg mass oscillates on the end of the spring, find its period, frequency, and angular frequency.

Short Answer

Expert verified
Graph \(T^2\) vs. mass to find the slope for \(k\); determine \(m_\mathrm{eff}\) from the intercept; calculate new oscillation parameters.

Step by step solution

01

Understand the System

In this problem, a spring of mass 0.250 kg, which obeys Hooke's Law, is used. The system's dynamics are described by the formula for the period of oscillation: \(T = 2\pi\sqrt{ \frac{m+m_\mathrm{eff}}{k} }\). The effective mass \(m_\mathrm{eff}\) accounts for the mass of the spring, making the physical scenario more accurate. We'll explore various components like graphing \(T^2\) vs. mass \(m\), the force constant \(k\), the effective mass \(m_\mathrm{eff}\), and then use them to calculate the specifications for a 0.450-kg mass.
02

Graphing and Analyzing Data

Graph \(T^2\) against the mass \(m\). This involves taking the measured periods for various masses, squaring these periods, and plotting them on a graph with the horizontal axis as the mass \(m\) and the vertical axis as \(T^2\). The result should be a linear relationship, with the slope giving information about the spring constant \(k\).
03

Determine the Force Constant from the Slope

The graph of \(T^2\) vs. \(m\) should yield a straight line. The slope \(b\) from the relationship \(T^2 = \frac{4\pi^2}{k}(m+m_\mathrm{eff})\) is calculated from the graph. Knowing the relationship, \(b = \frac{4\pi^2}{k}\), solve for \(k\): \(k = \frac{4\pi^2}{b}\).
04

Determine Effective Mass from the Y-intercept

Once you have the plot, the intercept \(a\) when \(m = 0\) corresponds to \(\frac{4\pi^2 m_\mathrm{eff}}{k}\). From this, extract \(m_\mathrm{eff}\) using the relationship \(m_\mathrm{eff} = \frac{a \cdot k}{4 \pi^2}\).
05

Calculate the Fraction of Unstrung Mass

The effective mass \(m_\mathrm{eff}\) is a fraction of the total spring mass. The fraction \(\frac{m_\mathrm{eff}}{m_\mathrm{spring}}\) is determined by dividing \(m_\mathrm{eff}\) by the mass of the spring itself, 0.250 kg.
06

Compute the Period, Frequency, and Angular Frequency

Solve for the new period \(T\) using the formula with the new mass 0.450 kg. The calculated period \(T = 2\pi \sqrt{ \frac{0.450 + m_\mathrm{eff}}{k} }\). Once \(T\) is known, find the frequency \(f\) using \(f = \frac{1}{T}\) and the angular frequency \(\omega = 2\pi f\).

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Key Concepts

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

Spring Constant
The spring constant, denoted as \( k \), is a crucial parameter in Hooke's Law, which describes how the force exerted by a spring is directly proportional to its extension or compression. It is a measure of the stiffness of the spring. The formula for Hooke’s Law is \( F = kx \), where \( F \) is the force applied to the spring, and \( x \) is the displacement from the spring’s equilibrium position.
  • A larger spring constant means a stiffer spring, requiring more force to achieve the same displacement.
  • A smaller spring constant indicates a more flexible spring, where less force is needed for the same displacement.
Understanding how to determine the spring constant involves graphing the square of the period \( T^2 \), across varying masses \( m \), and using the slope of this graph to find the spring constant employing the formula \( k = \frac{4\pi^2}{b} \) where \( b \) is the slope obtained from the graph.
Effective Mass
The effective mass \( m_\mathrm{eff} \) is another component that expands on Hooke's Law by accounting for the mass of the spring itself. This addition is particularly important in systems where the spring’s mass is not negligible, such as tapered springs where the mass distribution changes along the length of the spring. The formula for the period of oscillation, \( T = 2\pi\sqrt{ (m + m_\mathrm{eff})/k } \), includes the effective mass, allowing for more accurate predictions of the system's behavior.
  • The effective mass can be extracted from the y-intercept of the \( T^2 \) versus mass graph.
  • It is calculated using \( m_\mathrm{eff} = \frac{a \cdot k}{4 \pi^2} \), where \( a \) is the y-intercept and \( k \) is the previously determined spring constant.
Oscillations
Oscillations refer to the repeated and periodic motion of the system, where a mass on a spring moves back and forth through its equilibrium position. The time it takes to complete one full cycle of motion is called the period \( T \). Oscillations in a mass-spring system are a fundamental example of simple harmonic motion (SHM), where the restoring force is proportional to displacement.
  • As the mass is added or removed from the spring, the period of oscillation changes accordingly.
  • Understanding the nature of oscillations helps in predicting the system's response to external forces.
The linear relationship between \( T^2 \) and mass \( m \) helps visualize how oscillations are influenced by changes in mass, allowing clear observations of these relationships in practice.
Angular Frequency
Angular frequency \( \omega \) provides insight into how quickly an oscillation occurs in a system governed by simple harmonic motion. It is closely related to the frequency \( f \) and is given by \( \omega = 2\pi f \). Angular frequency is measured in radians per second, offering a different perspective from the regular frequency measured in cycles per second.
  • The angular frequency relates to the speed of oscillations and is key in defining various aspects of wave and oscillatory mechanics.
  • It is essential for calculating the dynamics of a mass-spring system, especially when analyzing the motion in terms of angles rather than periods or frequencies.
Once you have determined the period \( T \), it becomes straightforward to calculate both the frequency \( f = \frac{1}{T} \) and \( \omega = 2\pi f \), revealing crucial details about the motion's characteristics.

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Most popular questions from this chapter

A machine part is undergoing SHM with a frequency of 4.00 Hz and amplitude 1.80 cm. How long does it take the part to go from \(\chi =\) 0 to \(\chi = -1.80\) cm?

An apple weighs 1.00 N. When you hang it from the end of a long spring of force constant 1.50 N/m and negligible mass, it bounces up and down in SHM. If you stop the bouncing and let the apple swing from side to side through a small angle, the frequency of this simple pendulum is half the bounce frequency. (Because the angle is small, the back-and-forth swings do not cause any appreciable change in the length of the spring.) What is the unstretched length of the spring (with the apple removed)?

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An object is undergoing SHM with period 1.200 s and amplitude 0.600 m. At \(t =\) 0 the object is at \(x =\) 0 and is moving in the negative \(x\)-direction. How far is the object from the equilibrium position when \(t =\) 0.480 s?

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