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Chapter 13: Problem 6

In a physics lab, you attach a \(0.200-\mathrm{kg}\) air-track glider to the end of an ideal spring of negligible mass and start it oscillating. The elapsed time from when the glider first moves through the equilibrium point to the second time it moves through that point is 2.60 s. Find the spring's force constant.

Short Answer

Expert verified
The spring force constant is approximately 0.029 N/m.

Step by step solution

01

Understand the Problem

The problem involves a glider attached to a spring, and we need to determine the spring's force constant. We are given the mass of the glider and the oscillation time from equilibrium point to equilibrium point.
02

Recognize the Physics Principle

This problem relates to simple harmonic motion (SHM), where the time between successive passages through the equilibrium point is half the period of oscillation.
03

Determine Time Period

Calculate the full period from the given time. The full period, \(T\), is twice the elapsed time since the glider moves from equilibrium to the next equilibrium in half a period. So, \(T = 2 \times 2.60 \, \text{s} = 5.20 \, \text{s}\).
04

Use the Formula for the Period of SHM

The period of a mass-spring system is given by \( T = 2\pi \sqrt{\frac{m}{k}} \), where \(m\) is the mass and \(k\) is the spring constant.
05

Solve for Spring Constant

Rearrange the formula \( T = 2\pi \sqrt{\frac{m}{k}} \) to solve for \(k\):\[ k = \frac{4\pi^2 m}{T^2} \]Substitute \(m = 0.200 \, \text{kg}\) and \(T = 5.20 \, \text{s}\) into the equation:\[ k = \frac{4\pi^2 \times 0.200}{(5.20)^2} \approx 0.029 \, \text{N/m} \]
06

Double Check Calculations

Ensure that the substitution and calculations used are correct. Check each step, including the formula rearrangement and arithmetic operation, to confirm accuracy.

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Key Concepts

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

Simple Harmonic Motion
At the heart of the problem is simple harmonic motion (SHM), which is a type of periodic motion where an object oscillates about an equilibrium position. In SHM, the motion is sinusoidal in time, which means it follows a smooth repetitive oscillation. A common example of SHM is a mass-spring system, like the glider and spring in the exercise. This motion is characterized by a few important properties:
  • The object oscillates back and forth over the same path.
  • The maximum displacement from the equilibrium position is known as the amplitude.
  • The motion repeats itself at regular intervals, known as the period.
The forces responsible for this type of motion are "restorative forces," meaning they work to return the object to its equilibrium point. This is why springs, with their ability to pull or push back to their natural length, are classic examples of restorative forces in SHM.
Mass-Spring System
A mass-spring system serves as a fundamental example of SHM in physics. It combines a mass (the glider, in this case) attached to a spring to demonstrate the principles of oscillation. When the mass is displaced from its equilibrium position, the spring force acts to restore the mass back to equilibrium. Here are some key features:
  • The spring force follows Hooke's Law: \( F = -kx \), where \( k \) is the spring constant and \( x \) is the displacement from equilibrium.
  • Energy is conserved鈥攐scillation transforms potential energy stored in the spring to kinetic energy of the mass, and back again.
  • The spring constant \( k \) measures the stiffness of the spring: a stiffer spring has a larger \( k \).
This system is widely used in classrooms to help students comprehend the dynamic relationship between force, mass, and oscillation, offering an intuitive way to explore mechanical principles.
Period of Oscillation
The period of oscillation \( T \) is a crucial measure in SHM. It represents the time for one complete cycle of motion. For a mass-spring system, the period depends on two primary factors: the mass \( m \) and the spring constant \( k \). The formula that links these variables is:
\[ T = 2\pi \sqrt{\frac{m}{k}} \]This equation shows that the period is:
  • Directly proportional to the square root of the mass \( m \). A heavier mass lengthens the period, resulting in slower oscillations.
  • Inversely proportional to the square root of the spring constant \( k \). A stiffer spring (higher \( k \)) shortens the period, resulting in faster oscillations.
Understanding this relationship helps in determining how changes to the system affect its oscillatory behavior, echoing the importance of accurate calculations in experimental physics.
Physics Lab Experiment
Physics lab experiments, like the one described, provide a hands-on opportunity to explore theoretical concepts. In this experiment, a mass-spring setup offers a straightforward yet effective method to observe SHM and understand fundamental physics principles. Here's how an experiment around this would typically proceed:
  • Set up the air-track glider and spring to ensure minimal friction.
  • Initiate oscillation by slightly displacing the glider and releasing it.
  • Use a stopwatch to measure the time between sequential passes through the equilibrium point. Double this time for the complete period.
  • Apply the formula \[ k = \frac{4\pi^2 m}{T^2} \] to calculate the spring constant accurately, as shown in the solution.
This experiment not only solidifies classroom learning but also develops crucial lab skills like precise measurement and data analysis, providing a strong foundation in scientific investigation.

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Most popular questions from this chapter

A holiday ornament in the shape of a hollow sphere with mass \(M=0.015 \mathrm{kg}\) and radius \(R=0.050 \mathrm{m}\) is hung from a tree limb by a small loop of wire attached to the surface of the sphere. If the ornament is displaced a small distance and reased, it swings back and forth as a physical pendulum with negligible friction. Calculate its period. (Hint: Use the parallel-axis theorem to find the moment of inertia of the sphere about the pivot at the tree limb.)

A cheerleader waves her pom-pom in SHM with an amplitude of 18.0 \(\mathrm{cm}\) and a frequency of 0.850 \(\mathrm{Hz}\) . Find (a) the maximum magnitude of the acceleration and of the velocity; \((6)\) the acceleration and speed when the poin-poin's coordinate is \(x=+9.0 \mathrm{cm}\); (c) the time required to move from the equilibrium position directly to a point 12.0 \(\mathrm{cm}\) away. (d) Which of the quantities asproach used parts \((\mathrm{a}),(\mathrm{b}),\) and \((\mathrm{c})\) can be found using the energy approach used in Section \(13.3,\) and which cannot? Explain.

When a 0.750 -kg mass oscillates on an ideal spring, the frequency is 1.33 Hz. What will the frequency be if 0.220 kg are added to the original mass, and (b) subtracted from the original mass? Try to solve this problem without finding the force constant of the spring.

A uniform rod of length \(L\) oscillates through small angles about a point a distance \(x\) from its center. (a) Prove that its angular frequency is \(\sqrt{g x /\left[\left(L^{2} / 12\right)+x^{2}\right] .(b)}\) Show that its maximum angular frequency occurs when \(x=L / \sqrt{12}\) . (c) What is the length of the rod if the maximum angular frequency is 2\(\pi \mathrm{rad} / \mathrm{s} ?\)

A harmonic oscillator consists of a \(0.500-\mathrm{kg}\) mass attached to an ideal spring with force constant 140 \(\mathrm{N} / \mathrm{m}\) . Find (a) the period; (b) the frequency; (c) the angular frequency of the oscillations.
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