/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 10 An oscillating block-spring syst... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

An oscillating block-spring system takes \(0.75 \mathrm{~s}\) to begin re- peating its motion. Find (a) the period, (b) the frequency in hertz, and (c) the angular frequency in radians per second.

Short Answer

Expert verified
(a) 0.75 s, (b) 1.333 Hz, (c) 8.377 rad/s

Step by step solution

01

Understanding the System

This system involves a block attached to a spring that moves back and forth in a regular periodic motion. The time it takes to repeat this motion is given as 0.75 seconds. This time is known as the period of the motion.
02

Determine the Period (T)

The period, represented by \(T\), is the time taken for one complete cycle of oscillation. It is given directly in the problem as \(T = 0.75\) seconds.
03

Calculate the Frequency

Frequency, denoted by \(f\), is defined as the number of cycles per second. The relationship between period and frequency is given by \(f = \frac{1}{T}\). Using the period \(T = 0.75\,\text{s}\), calculate the frequency:\[ f = \frac{1}{0.75} = 1.333\, \text{Hz} \]
04

Calculate the Angular Frequency

Angular frequency, denoted by \(\omega\), is related to the frequency by the formula \(\omega = 2\pi f\). Using the frequency \(f = 1.333\,\text{Hz}\), calculate the angular frequency:\[ \omega = 2\pi \times 1.333 = 8.377\, \text{rad/s} \]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Oscillation Period
The oscillation period is a fundamental concept in the study of harmonic motion, especially observable in block-spring systems. The period, symbolized as \( T \), is the duration required to complete one full cycle of motion. It's quite telling of an oscillator's dynamics, essentially giving you a sense of how quickly or slowly the system oscillates. In the provided example, the oscillation period is directly given as \( T = 0.75 \) seconds. This number indicates that every 0.75 seconds, the oscillating block returns to its initial position, beginning another cycle. The concept of period is essential as it serves as a stepping stone for calculating other related properties like frequency and angular frequency. Understanding the period can help you predict the behavior of various oscillatory systems, whether they be mechanical systems like springs or even waves.
Frequency Calculation
Frequency is another key parameter to grasp when dealing with oscillations. It answers the question: "How many cycles occur in one second?" Represented by \( f \), frequency is measured in hertz (Hz), which simply means cycles per second. In a harmonic oscillator, once you know the period, you can easily find the frequency using the formula \( f = \frac{1}{T} \). It's as simple as dividing one by the period. For our block-spring system, given that the period \( T \) is 0.75 seconds, the frequency is \[ f = \frac{1}{0.75} = 1.333 \text{ Hz} \]. This tells us that the block completes approximately 1.333 cycles each second.
  • Frequency allows us to understand the rate of oscillation without directly observing it over a longer period.
  • Higher frequency indicates more cycles per second, whereas lower frequency suggests fewer cycles.
It is widely used in physics and engineering to describe repetitive events, from the ticking of a clock to the vibrations of a guitar string.
Angular Frequency
The concept of angular frequency extends the idea of frequency into a form that’s directly proportional to the circular path or angle in radians. Denoted as \( \omega \), this parameter describes how fast an object moves through its cycle and is measured in radians per second. Angular frequency is calculated using the formula \( \omega = 2\pi f \), where \( f \) is the frequency in hertz. With our example, and a frequency of \( 1.333 \text{ Hz} \), the angular frequency comes out to be \( \omega = 2\pi \times 1.333 = 8.377 \text{ rad/s} \).
  • Angular frequency links the linear aspects of frequency with the rotational context, useful in systems involving rotational motion or wave mechanics.
  • Together with period and frequency, it helps in understanding the dynamics of oscillating systems and more complex mechanical scenarios where angular displacement needs to be considered.
In essence, while frequency tells you how often a cycle completes per second, angular frequency shows how much angle is covered in that period, thereby bridging linear and rotational perspectives.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A massless spring with spring constant \(19 \mathrm{~N} / \mathrm{m}\) hangs vertically. A body of mass \(0.20 \mathrm{~kg}\) is attached to its free end and then released. Assume that the spring was unstretched before the body was released. Find (a) how far below the initial position the body descends, and the (b) frequency and (c) amplitude of the resulting SHM.

A \(50.0 \mathrm{~g}\) stone is attached to the bottom of a vertical spring and set vibrating. If the maximum speed of the stone is \(15.0 \mathrm{~cm} / \mathrm{s}\) and the period is \(0.500 \mathrm{~s}\), find the (a) spring constant of the spring, (b) amplitude of the motion, and (c) frequency of oscillation.

A \(10 \mathrm{~g}\) particle undergoes SHM with an amplitude of \(2.0 \mathrm{~mm}\), a maximum acceleration of magnitude \(8.0 \times 10^{3} \mathrm{~m} / \mathrm{s}^{2}\), and an unknown phase constant \(\phi\). What are (a) the period of the motion, (b) the maximum speed of the particle, and (c) the total mechanical energy of the oscillator? What is the magnitude of the force on the particle when the particle is at (d) its maximum displacement and (e) half its maximum displacement?

Two particles execute simple harmonic motion of the same amplitude and frequency along close parallel lines. They pass each other moving in opposite directions each time their displacement is half their amplitude. What is their phase difference?

The vibration frequencies of atoms in solids at normal temperatures are of the order of \(10^{13} \mathrm{~Hz}\). Imagine the atoms to be connected to one another by springs. Suppose that a single silver atom in a solid vibrates with this frequency and that all the other atoms are at rest. Compute the effective spring constant. One mole of silver \((6.02 \times\) \(10^{23}\) atoms) has a mass of \(108 \mathrm{~g}\).

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.