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Chapter 13: Problem 7

When a body of unknown mass is attached to an ideal spring with force constant \(120 \mathrm{N} / \mathrm{m},\) it is found to vibrate with a frequency of 6.00 Ha. Find (a) the period of the motion; \((b)\) the angular frequency; (c) the mass of the body.

Short Answer

Expert verified
(a) Period: 0.1667 s; (b) Angular frequency: 37.70 rad/s; (c) Mass: 0.0843 kg.

Step by step solution

01

Understand the given data and what is required

We know that the spring has a force constant \(k = 120 \, \text{N/m}\) and the frequency of the vibration is \(f = 6.00 \, \text{Hz}\). We need to find (a) the period \(T\), (b) the angular frequency \(\omega\), and (c) the mass \(m\) of the body.
02

Find the period of the motion (T)

The period \(T\) is the reciprocal of the frequency \(f\). Use the formula:\[T = \frac{1}{f}\]Substitute \(f = 6.00 \, \text{Hz}\):\[T = \frac{1}{6.00} = 0.1667 \, \text{seconds}\]
03

Find the angular frequency (蠅)

The angular frequency \(\omega\) can be found using the relation:\[\omega = 2\pi f\]Substitute \(f = 6.00\):\[\omega = 2 \times 3.1416 \times 6.00 \approx 37.70 \, \text{rad/s}\]
04

Find the mass of the body (m)

We can find the mass using the formula relating the spring constant \(k\), angular frequency \(\omega\), and mass \(m\):\[\omega = \sqrt{\frac{k}{m}}\]Rearrange the formula to solve for \(m\):\[m = \frac{k}{\omega^2}\]Substitute \(k = 120 \, \text{N/m}\) and \(\omega = 37.70\):\[m = \frac{120}{(37.70)^2} \approx 0.0843 \, \text{kg}\]

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Key Concepts

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

Spring Constant
The spring constant, denoted as \(k\), is a measure of the stiffness of a spring. It defines how much force is needed to stretch or compress the spring by a unit distance. Units of \(k\) are typically in Newtons per meter (N/m).
  • The higher the spring constant, the stiffer the spring is, meaning more force is needed to produce a given amount of stretch or compression.
  • In simple harmonic motion (SHM), the spring constant plays a crucial role in determining the characteristics of the motion.
  • The force in a spring is mathematically expressed as \(F = -kx\), where \(x\) is the displacement from equilibrium.
A spring with \(k = 120 \, \text{N/m}\) is relatively stiff, which affects the frequency and vibrational motion of the mass attached to it.
Frequency and Period
Frequency and period are interrelated concepts in the study of oscillations, including simple harmonic motion. Frequency \(f\) refers to how often the motion repeats itself in one second, and it is measured in hertz (Hz).
  • The period \(T\) is the time it takes for one complete cycle of motion. It is measured in seconds.
  • They are inverses of one another, as seen in the formula: \(T = \frac{1}{f}\).
For example, in the exercise given, with a frequency of \(6.00 \, \text{Hz}\), the period \(T\) can be easily calculated as \(0.1667 \, \text{seconds}\). Understanding these concepts helps in predicting the timing of subsequent motion cycles in vibrational systems.
Angular Frequency
Angular frequency \(\omega\) is a vector quantity that describes how quickly an object undergoes rotation or oscillation per unit time. It is directly related to the frequency of the motion but is expressed in terms of radians per second (rad/s).
  • It is given by the equation \(\omega = 2\pi f\), where \(f\) is the frequency.
  • Angular frequency is essential in analyzing oscillations because it integrates the cyclical nature represented by the radians of a circle.
In the original problem, substituting \(f = 6.00\) into the formula provides \(\omega \approx 37.70 \, \text{rad/s}\), offering insight into how fast the system cycles through its motion.
Mass-Spring System
A mass-spring system is a simple mechanical model used to analyze vibrations and oscillations. It consists of a mass attached to a spring, which can either compress or extend.
  • This system is a classic example of simple harmonic motion.
  • The properties of the spring (like the spring constant \(k\)) and the mass determine the motion's characteristics, such as frequency and amplitude.
  • The mass in a mass-spring system can be found using the formula \(m = \frac{k}{\omega^2}\).
In our exercise, with \(k = 120 \, \text{N/m}\) and \(\omega = 37.70 \, \text{rad/s}\), the mass is calculated to be approximately \(0.0843 \, \text{kg}\), demonstrating the direct relationship between the system's parameters and the dynamics of its motion.
Vibrational Motion
Vibrational motion is an essential concept in physics that describes the oscillation of components around an equilibrium position. It is seen in many physical systems, including mass-spring models and molecular structures.
  • In simple harmonic motion, the object moves back and forth over the same path with predictable periodicity.
  • Factors like mass, spring constant, and damping (if any) affect the motion's frequency and amplitude.
  • This type of motion is characterized by sinusoidal patterns when graphed over time.
Understanding vibrational motion aids in predicting how different systems will respond when displaced and provides foundational knowledge for more complex oscillatory dynamics, like those found in mechanical and acoustic systems.

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Most popular questions from this chapter

A child with poor table manners is sliding his 250 -g dinner plate back and forth in SHM with an amplitude of 0.100 \(\mathrm{m}\) on a horizontal surface. At a point 0.060 \(\mathrm{m}\) away from equilibrium, the speed of the plate is 0.300 \(\mathrm{m} / \mathrm{s}\) . (a) What is the period? (b) What is the displacement when the speed is 0.160 \(\mathrm{m} / \mathrm{s} ?(\mathrm{c})\) In the center of the dinner plate is a \(10.0-\mathrm{g}\) carrot slice. If the carrot slice is just on the verge of slipping at the endpoint of the path, what is the coefficient of static friction between the carrot slice and the plate?

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)?

A \(1.50-\mathrm{kg}\) ball and a \(2.00-\mathrm{kg}\) ball are glued together with the lighter one below the heavier one. The upper ball is attached to a vertical ideal spring of force constant \(165 \mathrm{N} / \mathrm{m},\) and the system is vibrating vertically with amplitude 15.0 \(\mathrm{cm} .\) The gine connecting the balls is old and weak, and it suddenly comes loose when the balls are at the lowest position in their motion. (a) Why is the glue more likely to fail at the lowest point than at any other point in the motion? (b) Find the amplitude and frequency of the vibrations after the lower ball has come loose.

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 \(50.0-\mathrm{g}\) 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 \(\mathrm{m} . \mathrm{A}\) 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 \(b\).
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