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Chapter 13: Problem 45

An apple weighs 1.00 \(\mathrm{N}\) . When you hang it from the end of a long spring of force constant 1.50 \(\mathrm{N} / \mathrm{m}\) 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)?

Short Answer

Expert verified
The unstretched length of the spring is effectively zero.

Step by step solution

01

Understand the Problem

We are given an apple that weighs 1.00 \( \mathrm{N} \). This apple causes a spring (with a constant of 1.50 \( \mathrm{N/m} \)) to bounce in simple harmonic motion (SHM). When stopped, the apple acts as a simple pendulum and has a frequency half that of the bouncing frequency. We need to find the unstretched length of the spring.
02

Calculate the Bounce (SHM) Frequency

The frequency of SHM for the spring can be determined using the formula for the frequency of a mass-spring system: \[ f_s = \frac{1}{2\pi} \sqrt{\frac{k}{m}} \] where \( k = 1.50 \ \mathrm{N/m} \) is the spring constant, and \( m \) is the mass of the apple. First, we find the mass \( m \) using the weight of the apple: \( W = mg = 1.00 \mathrm{N} \implies m = \frac{1.00}{9.81} \mathrm{kg} \). Substituting in, we get \[ f_s = \frac{1}{2\pi} \sqrt{\frac{1.50}{0.102}}. \]
03

Calculate the Pendulum Frequency

We know that the pendulum frequency \( f_p \) is half of the bounce frequency: \( f_p = \frac{1}{2} f_s \). The frequency of a simple pendulum is given by: \[ f_p = \frac{1}{2\pi} \sqrt{\frac{g}{L}}, \] where \( L \) is the effective length of the pendulum (which includes the stretch in spring) and \( g \approx 9.81 \ \mathrm{m/s^2} \).
04

Relate Frequencies to Solve for L

Using the fact that \( f_p = \frac{1}{2} f_s \), equate the expression for \( f_p \) to half the expression for \( f_s \) to solve for \( L \). This gives: \[ \frac{1}{2\pi} \sqrt{\frac{g}{L}} = \frac{1}{2} \left( \frac{1}{2\pi} \sqrt{\frac{k}{m}} \right). \] Simplify to find \[ L = \frac{g}{k} m = \frac{9.81}{1.50} \times 0.102 \approx 0.667 \ \mathrm{m}. \] This \( L \) represents the total length when the apple is suspended.
05

Find the Unstretched Length

From the total length, subtract the stretch caused by the weight of the apple to find the unstretched length. The stretch \( x \) can be found using Hooke's law: \( kx = mg \), which means \( x = \frac{mg}{k} = \frac{1.00}{1.50} \approx 0.667 \ \mathrm{m}. \) Then, the unstretched length \( L_0 \) is: \[ L_0 = L - x = 0.667 - 0.667 = 0 \ \mathrm{m}. \]

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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 simple harmonic motion (SHM) in physics. In this scenario, a mass is attached to a spring, and when displaced, it moves back and forth. Its motion can be described as oscillating around an equilibrium position.
The importance of such a system lies in its ability to show how forces like tension in the spring seek to restore the mass to its equilibrium point.

Key aspects include:
  • Equilibrium Position: The point where the forces acting on the mass are balanced and the system is at rest.
  • Amplitude: The maximum distance from the equilibrium position.
  • Frequency: How often the mass oscillates in one second.
  • Period: The time it takes to complete one full cycle of motion.
The mass-spring system's frequency is determined by the mass of the object and the spring's stiffness or spring constant.
Spring Constant
The spring constant, denoted as \( k \), is a crucial property of any spring. It quantifies how stiff or resilient a spring is, essentially representing the force required to extend or compress it by a unit distance.

In the equation of Hooke's Law, \( F = kx \), the force \( F \) applied to the spring is directly proportional to the displacement \( x \) from its natural or unstretched length.
  • Units: The spring constant is measured in Newtons per meter (N/m).
  • High \( k \): A high spring constant means the spring is stiff and requires more force to stretch it.
  • Low \( k \): A low spring constant indicates a less stiff spring that can be stretched easily with less force.
For a mass-spring system engaged in simple harmonic motion (SHM), the spring constant plays a vital role in determining the system's frequency and behavior.
Pendulum Frequency
The pendulum frequency is another fascinating example of simple harmonic motion but it involves a mass swinging back and forth due to gravity. In a simple pendulum, the mass (or bob) is attached to a fixed point by a string or rod of length \( L \).

This frequency depends essentially on two factors only:
  • Acceleration due to Gravity \( g \): affects how fast or slow the pendulum swings.
  • Length \( L \): the effective length of the pendulum, which affects the time period of its swing.
Given the equation for the pendulum's frequency, \( f = \frac{1}{2\pi} \sqrt{\frac{g}{L}} \), we can observe that the frequency is inversely related to the square root of the pendulum's length.
This means that longer pendulums swing with a lower frequency, while shorter pendulums swing with higher frequencies. It's interesting to note that for small angles, the pendulum frequency in this context becomes half of the bouncing frequency of the mass-spring system.

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Most popular questions from this chapter

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.

A harmonic oscillator has angular frequency \(\omega\) and amplitude \(A\) . (a) What are the magnitudes of the displacement and velocity when the elastic potential energy is equal to the kinetic energy? (Assume that \(U=0\) at equilibrium, (b) How often does this occur in each cycle? What is the time between occurrences? (c) At an instant when the displacement is equal to \(A / 2,\) what fraction of the total energy of the system is kinetic and what fraction is potential?

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 \(175-\mathrm{kg}\) sack of gravel sits on the middle of it. The beam is oscillating in SHM, with an amplitude of 40.0 \(\mathrm{cm}\) and a frequency of 0.600 cycles \(/ \mathrm{s}\). (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?

A \(1.50-\mathrm{kg}\) mass on a spring has displacement as a function of time given by the equation $$x(t)=(7.40 \mathrm{cm}) \cos \left[\left(4.16 \mathrm{s}^{-1}\right) t-2.42\right]$$ Find (a) the time for one complete vibration; (b) the force constant of the spring; (c) the maximum speed of the mass; (d) the maximum force on the mass; (e) the position, speed, and acceleration of the mass at \(t=1.00 \mathrm{s} ;(\mathrm{f})\) the force on the mass at that time.

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.
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