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Chapter 14: Problem 7

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

Short Answer

Expert verified
(a) 0.167 s, (b) 37.70 rad/s, (c) 0.0844 kg.

Step by step solution

01

Find the period of the motion

The period of oscillation \(T\) is the reciprocal of the frequency \(f\). Given the frequency \(f = 6.00\, \text{Hz}\), we can find the period using the formula: \(T = \frac{1}{f}\). Substituting the value, we get \(T = \frac{1}{6.00} \approx 0.167\, \text{s}\).
02

Calculate the angular frequency

Angular frequency \(\omega\) is related to frequency by the formula \(\omega = 2\pi f\). Given \(f = 6.00\, \text{Hz}\), substituting in the values yields \(\omega = 2\pi \times 6.00 \approx 37.70\, \text{rad/s}\).
03

Calculate the mass of the body

The angular frequency \(\omega\) is also related to the mass \(m\) and the spring constant \(k\) via the formula \(\omega = \sqrt{\frac{k}{m}}\). Rearranging for \(m\), we get \(m = \frac{k}{\omega^2}\). Given \(k = 120\, \text{N/m}\) and \(\omega \approx 37.70\, \text{rad/s}\), substituting in these values gives \(m = \frac{120}{(37.70)^2} \approx 0.0844\, \text{kg}\).

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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, often symbolized as \( k \), is a measure of a spring's stiffness. It tells us how much force is needed to stretch or compress a spring by a certain amount. The higher the spring constant, the stiffer the spring is and the more force it takes to change its length.
To understand it better:
  • Units: The unit of spring constant is Newton per meter (N/m).
  • Hooke's Law: The relationship between force and spring constant is expressed by Hooke's Law \( F = kx \), where \( F \) is the force applied, and \( x \) is the displacement from its equilibrium position.
  • Application: Knowing the spring constant helps in designing systems with desired elasticity and in predicting how a system will respond to an external force.
Oscillation Period
The oscillation period, termed as \( T \), is the time taken for one complete cycle of vibration to pass a given point. In this exercise, it is calculated as the reciprocal of the frequency of the vibrations.
  • Calculation Formula: The period \( T \) is given by \( T = \frac{1}{f} \), where \( f \) is the frequency.
  • Recurring Motion: In oscillatory motion, such as spring vibration, the period remains constant as long as the system is undisturbed.
  • Units: The unit for the period is seconds (s).
By knowing how long one cycle takes, you can determine the pace of any periodic motion.
Angular Frequency
Angular frequency, denoted by \( \omega \), tells us how quickly an object oscillates in a circular path. Despite being called 'angular', it applies even in linear systems, like the vibrations of a spring.
  • Relationship to Frequency: It's related to frequency \( f \) through the formula \( \omega = 2\pi f \). This shows the link between rotational and linear perspectives of motion.
  • Units: Angular frequency is measured in radians per second (rad/s).
  • Comprehensive Understanding: Knowing \( \omega \) helps in analyzing systems subject to harmonic motion. It is crucial for solutions involving trigonometric functions like sine and cosine.
Mass Calculation
Calculating the mass of an oscillating body in a spring-mass system can be intriguing. Given is the relationship between mass \( m \), spring constant \( k \), and angular frequency \( \omega \).
  • Key Formula: The mass is derived from \( m = \frac{k}{\omega^2} \).
  • Spring Influence: The stiffness of the spring affects how the system oscillates. Thus, \( k \) is a part of the mass calculation.
  • Understanding Oscillation: Mass affects how quickly or slowly a system responds to forces鈥攅ssential for things like shock absorbers or measuring devices.
  • Units: Mass is measured in kilograms (kg).
This in-depth look into the calculations helps ensure that the components of any oscillating system are correctly designed for their purposes.

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Most popular questions from this chapter

A small block is attached to an ideal spring and is moving in SHM on a horizontal, frictionless surface. The amplitude of the motion is 0.250 \(\mathrm{m}\) and the period is 3.20 \(\mathrm{s} .\) What are the speed and acceleration of the block when \(x=0.160 \mathrm{m} ?\)

A thin metal disk with mass \(2.00 \times 10^{-3} \mathrm{kg}\) and radius 2.20 \(\mathrm{cm}\) is attached at its center to a long fiber (Fig. E14.42). The disk, when twisted and released, oscillates with a period of 1.00 s. Find the torsion constant of the fiber.

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

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.

CP On the planet Newtonia, a simple pendulum having a bob with mass 1.25 \(\mathrm{kg}\) and a length of 185.0 \(\mathrm{cm}\) takes 1.42 \(\mathrm{s}\) . when released from rest, to swing through an angle of \(12.5^{\circ}\) , where it again has zero speed. The circumference of Newtonia is measured to be \(51,400 \mathrm{km}\) . What is the mass of the planet Newtonia?
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