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Chapter 10: Problem 86

When an object of mass \(m_{1}\) is hung on a vertical spring and set into vertical simple harmonic motion, it oscillates at a frequency of 12.0 \(\mathrm{Hz}\) . When another object of mass \(m_{2}\) is hung on the spring along with the first object, the frequency of the motion is 4.00 \(\mathrm{Hz}\) . Find the ratio \(m_{2} / m_{1}\) of the mases.

Short Answer

Expert verified
The ratio \( \frac{m_2}{m_1} \) is 35.

Step by step solution

01

Understand Frequency and Mass Relationship

The frequency of a simple harmonic oscillator is given by the formula \( f = \frac{1}{2\pi} \sqrt{\frac{k}{m}} \), where \( f \) is the frequency, \( k \) is the spring constant, and \( m \) is the mass attached to the spring.
02

Set Up Equations for Given Frequencies

For the first case with mass \( m_1 \): \( 12 = \frac{1}{2\pi}\sqrt{\frac{k}{m_1}} \).For the second case with mass \( m_1 + m_2 \): \( 4 = \frac{1}{2\pi}\sqrt{\frac{k}{m_1 + m_2}} \).
03

Solve for Spring Constant and Substitute

From the first equation, solve for the spring constant \( k \):\[ k = (2\pi \times 12)^2 m_1 \].Substitute this into the second equation:\[ 16\pi^2 (m_1 + m_2) = 576\pi^2 m_1 \].
04

Simplify and Solve for Ratio

Cancel out \( \pi^2 \) and simplify:\[ 16(m_1 + m_2) = 576m_1 \].Expand and re-arrange to find \( m_2 \):\[ 16m_1 + 16m_2 = 576m_1 \].\[ 16m_2 = 560m_1 \].\[ \frac{m_2}{m_1} = \frac{560}{16} = 35 \].

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Key Concepts

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

Understanding Frequency in Simple Harmonic Motion
Frequency in the context of simple harmonic motion refers to how often an oscillating object completes one cycle of motion in a specific time, usually one second. It's measured in Hertz (Hz). In simple harmonic motion like the one involving springs, the frequency is determined by the characteristics of the system. The mathematical formula that relates frequency (\( f \)) to the mass (\( m \)) and the spring constant (\( k \)) is given by:
  • \( f = \frac{1}{2\pi} \sqrt{\frac{k}{m}}\)
This equation shows that frequency is
  • Inversely proportional to the square root of the mass.
  • Directly proportional to the square root of the spring constant.
When you increase the mass, the frequency decreases, meaning the object takes longer to complete one cycle if it's heavier.
The Role of Mass in Oscillation
Mass is a crucial factor that affects how a spring oscillator behaves. It determines how easily the spring can move and how quickly it can return to its original position during oscillation. When a spring is paired with a mass, the oscillation becomes smoother as the system's inertia comes into play. A simple rule to remember for mass in oscillation is:
  • The larger the mass, the slower the oscillation.
  • The smaller the mass, the quicker the oscillation.
This concept is vital in experiments involving different masses on the same spring. Heavier masses reduce the frequency of oscillation because they require more force to return to their equilibrium position.
Spring Constant's Impact on Oscillation
The spring constant, denoted by \( k \), is a measure of a spring's stiffness. It describes how much force is needed to compress or stretch the spring by a unit distance. A high spring constant means the spring is very stiff and requires more force to deform, whereas a low spring constant indicates a more flexible spring. In the equation \( f = \frac{1}{2\pi} \sqrt{\frac{k}{m}} \), the spring constant has a significant role:
  • A larger spring constant \( k \) increases the frequency \( f \), making oscillations quicker.
  • A smaller \( k \) decreases the frequency, resulting in slower oscillations.
Thus, by changing the spring constant, you can directly influence the dynamics of an oscillating system.
Basic Principles of Oscillation
Oscillation refers to the repeated back and forth movement of an object from its equilibrium position. In simple harmonic motion, such as an object on a spring, oscillations are regular and predictable. Key characteristics of oscillation in this context include:
  • The motion is periodic, which means it repeats after equal intervals of time.
  • The motion consists of a harmonic pattern, usually a sine or cosine wave when graphed.
Understanding these core aspects can help you predict and calculate the behavior of oscillating systems. In the problem of determining the ratio of masses using oscillation frequencies, the basic principles of oscillation help set up the equations necessary to find the solution.

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Most popular questions from this chapter

A spring \((k=830 \mathrm{N} / \mathrm{m})\) is hanging from the ceiling of an elevator, and a 5.0 -kg object is attached to the lower end. By how much does the spring stretch (relative to its unstrained length) when the elevator is accelerating upward at \(a=0.60 \mathrm{m} / \mathrm{s}^{2} ?\)

A hand exerciser utilizes a coiled spring. A force of 89.0 \(\mathrm{N}\) is required to compress the spring by 0.0191 \(\mathrm{m} .\) Determine the force needed to compress the spring by 0.0508 \(\mathrm{m} .\)

A horizontal spring is lying on a frictionless surface. One end of the spring is attached to a wall, and the other end is connected to a movable object. The spring and object are compressed by \(0.065 \mathrm{m},\) released from rest, and subsequently oscillate back and forth with an angular frequency of 11.3 \(\mathrm{rad} / \mathrm{s}\) . What is the speed of the object at the instant when the spring is stretched by 0.048 \(\mathrm{m}\) relative to its unstrained length?

The drawing shows a top view of a frictionless horizontal surface, where there are two springs with particles of mass \(m_{1}\) and \(m_{2}\) attached to them. Each spring has a spring constant of 120 \(\mathrm{N} / \mathrm{m}\) . The particles are pulled to the right and then released from the positions shown in the drawing. How much time passes before the particles are side by side for the first time at \(x=0 \mathrm{m}\) if \(\quad\) (a) \(m_{1}=m_{2}=\) 3.0 \(\mathrm{kg}\) and \(\quad(\mathrm{b}) m_{1}=3.0 \mathrm{kg}\) and \(m_{2}=27 \mathrm{kg} ?\)

One end of a piano wire is wrapped around a cylindrical tuning peg and the other end is fixed in place. The tuning peg is turned so as to stretch the wire. The piano wire is made from steel \(\left(Y=2.0 \times 10^{11} \mathrm{N} / \mathrm{m}^{2}\right) .\) It has a radius of 0.80 \(\mathrm{mm}\) and an unstrained length of 0.76 \(\mathrm{m}\) . The radius of the tuning peg is 1.8 \(\mathrm{mm}\) . Initially, there is no tension in the wire, but when the tuning peg is turned, tension develops. Find the tension in the wire when the tuning peg is turned through two revolutions. Ignore the radius of the wire compared to the radius of the tuning peg.
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