/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 6 In a physics lab, you attach a \... [FREE SOLUTION] | 91Ó°ÊÓ

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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's force constant is approximately 0.0583 N/m.

Step by step solution

01

Understanding the Problem

We have a glider of mass \( m = 0.200 \) kg attached to an ideal spring, and it oscillates through the equilibrium point. The time from the first pass through the equilibrium point to the second pass given is 2.60 s. We need to find the spring's force constant \( k \). This problem deals with harmonic oscillations.
02

Determine the Time Period

The time given (2.60 s) is for the glider to pass through the equilibrium point twice. This represents half of the complete oscillation period (from equilibrium to maximum displacement and back again to equilibrium). So, the full time period \( T \) is two times the given time: \( T = 2 \times 2.60 \) s = 5.20 s.
03

Use the Formula for Time Period of a Mass-Spring System

The time period \( T \) of a mass-spring system is given by the formula: \[ T = 2 \pi \sqrt{\frac{m}{k}} \]where \( m \) is mass and \( k \) is the spring constant. We can rearrange this equation to solve for \( k \): \[ k = \frac{4 \pi^2 m}{T^2} \]
04

Substitute Known Values

Substitute \( m = 0.200 \) kg and \( T = 5.20 \) s into the rearranged formula to find \( k \):\[ k = \frac{4 \pi^2 \times 0.200}{5.20^2} \]
05

Calculate the Spring Constant

Compute the value using a calculator:\[ k = \frac{4 \times (3.1416)^2 \times 0.200}{27.04} = \frac{7.8954 \times 0.200}{27.04} \approx 0.0583 \text{ N/m} \] Thus, the spring force constant \( k \) is approximately 0.0583 N/m.

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Key Concepts

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

Harmonic Oscillation
Harmonic oscillation refers to a type of periodic motion where an object moves back and forth over the same path, about an equilibrium position. This motion is driven by a restoring force proportional to the displacement. In simpler terms, when you pull back an object attached to a spring and let go, the object will move back and forth repeating its motion in a regular way.
The motion is considered harmonic because it can be described by sinusoidal functions, making predictions about the motion at any point in time possible. An ideal spring, like the one in our problem, follows Hooke's Law. This law states that the force needed to extend or compress a spring by some distance is proportional to that distance: \[ F = -kx \]where \(F\) is the force applied, \(k\) is the spring constant, and \(x\) is the displacement from the equilibrium position. The negative sign indicates that the force is always directed towards the equilibrium position.
Mass-Spring System
A mass-spring system is a common example used to understand harmonic oscillation. It generally consists of a mass attached to a spring, which can oscillate back and forth when set in motion. This system showcases the essential features of harmonic oscillation.
In this setup, the mass moves due to the spring’s restoring force, which aims to return the mass to its equilibrium position. The oscillation results from the interplay between kinetic and potential energy:
  • Kinetic energy dominates when the mass passes through the equilibrium point at maximum speed.
  • Potential energy peaks when the spring is at its maximum displacement, either compressed or stretched.
Understanding the dynamics of a mass-spring system helps to learn about real-world applications like shock absorbers in vehicles or even the tiny movements in your car’s suspension system.
Period of Oscillation
The period of oscillation, often represented by \(T\), is the time it takes for one complete cycle of motion in a harmonic oscillator, such as a mass-spring system, to occur. It is a crucial feature as it offers insights into the system's dynamics.
In our example, we found that the time it took for the mass to move through the equilibrium point twice was the same as half the period. By doubling this observed time, we determine the full period. The formula for the period \(T\) in a mass-spring system is:\[ T = 2\pi\sqrt{\frac{m}{k}} \]where \(m\) is the mass and \(k\) is the spring constant.
This formula, derived from the laws of physics, implies that heavier masses and stiffer springs affect the period. Larger masses lead to longer periods, while stiffer springs, or higher spring constants, produce shorter periods. This relationship is fundamental in designing systems that rely on predictable and stable oscillations, ensuring they operate effectively in devices like clocks or even your smartphone's gyroscope.

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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 \(k=25.0 \mathrm{N} / \mathrm{m} .\) Its initial displacement is 0.300 \(\mathrm{m} .\) A damping force \(F_{x}=-b v_{x}\) acts on the egg, and the amplitude of the motion decreases to 0.100 \(\mathrm{m}\) in 5.00 \(\mathrm{s}\) . Calculate the magnitude of the damping constant \(b\) .

We want to hang a thin hoop on a horizontal nail and have the hoop make one complete small-angle oscillation each 2.0 \(\mathrm{s}\) What must the hoop's radius be?

A tuning fork labeled 392 Hz has the tip of each of its two prongs vibrating with an amplitude of 0.600 \(\mathrm{mm} .\) (a) What is the maximum speed of the tip of a prong? (b) A housefly (Musca domestica) with mass 0.0270 \(\mathrm{g}\) is holding onto the tip of one of the prongs. As the prong vibrates, what is the fly's maximum kinetic energy? Assume that the fly's mass has a negligible effect on the frequency of oscillation.

CALC Jerk. A guitar string vibrates at a frequency of 440 Hz. A point at its center moves in SHM with an amplitude of 3.0 \(\mathrm{mm}\) and a phase angle of zero. (a) Write an equation for the position of the center of the string as a function of time. (b) What are the maximum values of the magnitudes of the velocity and acceleration of the center of the string? (c) The derivative of the acceleration with respect to time is a quantity called the jerk. Write an equation for the jerk of the center of the string as a function of time, and find the maximum value of the magnitude of the jerk.

A 1.80-kg monkey wrench is pivoted 0.250 \(\mathrm{m}\) from its center of mass and allowed to swing as a physical pendulum. The period for small-angle oscillations is 0.940 s. (a) What is the moment of inertia of the wrench about an axis through the pivot? (b) If the wrench is initially displaced 0.400 rad from its equilibrium position, what is the angular speed of the wrench as it passes through the equilibrium position?

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