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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

The length of a simple pendulum is 0.79 \(\mathrm{m}\) and the mass of the particle (the bob) at the end of the cable is 0.24 \(\mathrm{kg}\) . The pendulum is pulled away from its equilibrium position by an angle of \(8.50^{\circ}\) and released from rest. Assume that friction can be neglected and that the resulting oscillatory motion is simple harmonic motion. (a) What is the angular frequency of the motion? (b) Using the position of the bob at its lowest point as the reference level, determine the total mechanical energy of the pendulum as it swings back and forth. (c) What is the bob's speed as it passes through the lowest point of the swing?

The fan blades on a jet engine make one thousand revolutions in a time of 50.0 \(\mathrm{ms}\) . Determine (a) the period (in seconds) and (b) the frequency (in \(\mathrm{Hz} )\) of the rotational motion. (c) What is the angular frequency of the blades?

Between each pair of vertebrae in the spinal column is a cylindrical disc of cartilage. Typically, this disc has a radius of about \(3.0 \times 10^{-2} \mathrm{m}\) and a thickness of about \(7.0 \times 10^{-3} \mathrm{m} .\) The shear modulus of cartilage is \(1.2 \times 10^{7} \mathrm{N} / \mathrm{m}^{2}\) . Suppose that a shearing force of magnitude 11 \(\mathrm{N}\) is applied parallel to the top surface of the disc while the bottom surface remains fixed in place. How far does the top surface move relative to the bottom surface?

In a room that is 2.44 m high, a spring (unstrained length \(=0.30 \mathrm{m} )\) hangs from the ceiling. A board whose length is 1.98 \(\mathrm{m}\) is attached to the free end of the spring. The board hangs straight down, so that its \(1.98-\mathrm{m}\) length is perpendicular to the floor. The weight of the board (104 \(\mathrm{N} )\) stretches the spring so that the lower end of the board just extends to, but does not touch, the floor. What is the spring constant of the spring?

An 11.2 -kg block and a 21.7 -kg block are resting on a horizontal frictionless surface. Between the two is squeezed a spring (spring constant \(=1330 \mathrm{N} / \mathrm{m}\) ). The spring is compresed by 0.141 \(\mathrm{m}\) from its unstrained length and is not attached to either block. With what speed does each block move away after the mechanism keeping the spring squeezed is released and the spring falls away?
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