/*! 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 62 The spring system is connected t... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The spring system is connected to a crosshead that oscillates vertically when the wheel rotates with a constant angular velocity of \(\omega\). If the amplitude of the steady-state vibration is observed to be \(400 \mathrm{~mm}\), and the springs each have a stiffness of \(k=2500 \mathrm{~N} / \mathrm{m},\) determine the two possible values of \(\boldsymbol{\omega}\) at which the wheel must rotate. The block has a mass of \(50 \mathrm{~kg}\).

Short Answer

Expert verified
The two possible values of the angular velocity \(\omega\) at which the wheel must rotate are \(\pm 10 \, \mathrm{rad/s}\)

Step by step solution

01

Calculate the Natural Frequency

The first step is to calculate the natural frequency of the spring-system. This can be achieved using the formula for the natural frequency of a simple harmonic oscillator, \(\omega = \sqrt{\frac{k}{m}}\). Given that the spring constant \(k = 2500 \, \mathrm{N/m}\) and the mass \(m = 50 \, \mathrm{kg}\), we can substitute these values into the formula to obtain the natural frequency.
02

Calculate the Angular Velocity

The second step is to calculate the angular velocity. This can be done by rearranging the formula for natural frequency and taking the square root. Thus the angular velocity \(\omega\) can be calculated by the formula \(\omega = \pm \sqrt{\frac{k}{m}}\).
03

Solve for the Angular Velocity

After substituting the given values of \(k\) and \(m\) into the formula, we get two possible values of \(\omega\), a positive and a negative value. This is because the vibration can occur in two directions, upwards or downwards.

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.

Natural Frequency
The natural frequency of a system is a fundamental property that determines the rate at which the system tends to oscillate when disturbed from its equilibrium position. In the context of the spring-block system described in the exercise, the natural frequency represents how quickly the block will bounce up and down on the springs without any external force causing further motion.

Natural frequency is crucial because it defines the system's response to vibrations. If an external force, such as the rotation of a wheel, matches the natural frequency of the system, a phenomenon called resonance occurs, which can greatly amplify the vibrations and can potentially cause damage. The formula used to calculate natural frequency is \[\begin{equation}\omega_0 = \sqrt{\frac{k}{m}}\end{equation}\]\, where \(\omega_0\) is the natural frequency, \(k\) is the spring stiffness, and \(m\) is the mass of the object attached to the spring. Importantly, the natural frequency is measured in radians per second and signifies the intrinsic rate of undriven, undamped oscillations.
Simple Harmonic Oscillator
A simple harmonic oscillator (SHO) is a mechanical model that depicts motion featuring a restoring force proportional to displacement from equilibrium. This type of motion, known as simple harmonic motion (SHM), reflects the behavior of systems like springs and pendulums under ideal conditions—absent of external forces and damping. In SHM, the acceleration of the system is directly proportional to its displacement and is always directed towards the equilibrium position.

An object undergoing SHM will oscillate back and forth over the same path, within a fixed period that is determined by the system's natural frequency. The SHO is characterized by the formula for displacement \( x(t) = A \cos(\omega_0 t + \phi)\), where \(A\) is the amplitude, \(\omega_0\) the natural frequency, \(t\) the time, and \(\phi\) the phase angle. When dealing with such systems, understanding their natural frequency is essential because it allows us to predict their behavior and avoid resonance in structural applications.
Spring Stiffness
Spring stiffness, symbolized by the letter \(k\), is a measure of how resistant a spring is to deformation. In simpler terms, it indicates how 'stiff' or 'flexible' a spring is. The value of \(k\) dictates how much force is needed to stretch or compress the spring by a certain amount. This is epitomized by Hooke's Law, stated as \(F = -kx\), where \(F\) is the force applied, \(x\) is the displacement of the spring from its rest position, and the negative sign indicates that the force is restorative (i.e., it acts in the opposite direction of displacement).

The stiffness of the springs in the given spring-block system directly affects both the natural frequency of the system and the amplitude of oscillations under a force. A higher spring stiffness means that a greater force is needed to produce the same displacement, leading to a higher natural frequency, and thus, a faster oscillation rate. Understanding spring stiffness is crucial for designing systems that must absorb or control energy, such as vehicle suspensions or seismic isolators for buildings.

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 4-kg block is suspended from a spring that has a stiffness of \(k=600 \mathrm{~N} / \mathrm{m}\). The block is drawn downward \(50 \mathrm{~mm}\) from the equilibrium position and released from rest when \(t=0\). If the support moves with an impressed displacement of \(\delta=(10 \sin 4 t) \mathrm{mm},\) where \(t\) is in seconds, determine the equation that describes the vertical motion of the block. Assume positive displacement is downward.

A block having a mass of \(7 \mathrm{~kg}\) is suspended from a spring that has a stiffness \(k=600 \mathrm{~N} / \mathrm{m}\). If the block is given an upward velocity of \(0.6 \mathrm{~m} / \mathrm{s}\) from its equilibrium position at \(t=0,\) determine its position as a function of time. Assume that positive displacement of the block is downward and that motion takes place in a medium which furnishes a damping force \(F=(50|v|) \mathrm{N},\) where \(v\) is in \(\mathrm{m} / \mathrm{s}\)

The uniform beam is supported at its ends by two springs \(A\) and \(B\), each having the same stiffness \(k\). When nothing is supported on the beam, it has a period of vertical vibration of \(0.83 \mathrm{~s}\). If a 50 -kg mass is placed at its center, the period of vertical vibration is \(1.52 \mathrm{~s}\). Compute the stiffness of each spring and the mass of the beam.

The 40 -kg block is attached to a spring having a stiffness of \(800 \mathrm{~N} / \mathrm{m}\). A force \(F=(100 \cos 2 t) \mathrm{N},\) where \(t\) is in seconds is applied to the block. Determine the maximum speed of the block for the steady-state vibration.

The \(450-\mathrm{kg}\) trailer is pulled with a constant speed over the surface of a bumpy road, which may be approximated by a cosine curve having an amplitude of \(50 \mathrm{~mm}\) and wave length of \(4 \mathrm{~m}\). If the two springs \(s\) which support the trailer each have a stiffness of \(800 \mathrm{~N} / \mathrm{m},\) determine the speed \(v\) which will cause the greatest vibration (resonance) of the trailer. Neglect the weight of the wheels.

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.