/*! 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 42 A block which has a mass \(m\) i... [FREE SOLUTION] | 91Ó°ÊÓ

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A block which has a mass \(m\) is suspended from a spring having a stiffness \(k\). If an impressed downward vertical force \(F=F_{O}\) acts on the weight, determine the equation which describes the position of the block as a function of time.

Short Answer

Expert verified
The equation which describes the position of the block as a function of time is \( y(t) = A \cos(\sqrt{\frac{k}{m}}\cdot t) + B \sin(\sqrt{\frac{k}{m}}\cdot t) + \frac{F_{O}}{k} \), where \( A \) and \( B \) are constants that depend on initial conditions.

Step by step solution

01

Setting Up the Problem

Let's assume the equilibrium position of the block is \( y=0 \), and any displacement downward from this position will be considered positive. According to Hooke's Law, this spring force exerted on the block is proportional to its displacement from the equilibrium. The equation for Hooke's law can be represented as \( -ky = ma \), where \( y \) is the displacement, \( a \) is the acceleration, and the negative sign indicates that the force exerted by the spring opposes the displacement. Therefore, the equation of motion considering an additional external impressed force \( F=F_{O} \) becomes \( F - ky = ma \).
02

Formulating the Differential Equation

The acceleration \( a \) can also be described as second-derivative of displacement \( y \) with respect to time \( t \). Therefore, the equation of motion can be written as a differential equation: \( F_{O} - ky = m \cdot \frac{d^2y}{dt^2} \).
03

Solving the Differential Equation

Rearranging the terms and dividing by \( m \) gives \( \frac{d^2y}{dt^2} + \frac{k}{m}y = \frac{F_{O}}{m} \). This is a linear differential equation representing simple harmonic motion with a constant source term. The solution will take the form \( y(t) = A \cos(\omega t) + B \sin(\omega t) + \frac{F_{O}}{k} \), where \( A \) and \( B \) are constants that depend on initial conditions, and \( \omega = \sqrt{\frac{k}{m}} \) is the natural frequency of the system.

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Most popular questions from this chapter

The \(40-\mathrm{kg}\) 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.

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 \(\mathrm{m} / \mathrm{s}\)

If the lower end of the 6 -kg slender rod is displaced a small amount and released from rest, determine the natural frequency of vibration. Each spring has a stiffness of \(k=200 \mathrm{~N} / \mathrm{m}\) and is unstretched when the rod is hanging vertically.

When a \(3-\mathrm{kg}\) block is suspended from a spring, the spring is stretched a distance of \(60 \mathrm{~mm}\). Determine the natural frequency and the period of vibration for a \(0.2-\mathrm{kg}\) block attached to the same spring.

An \(8-\mathrm{kg}\) block is suspended from a spring having a stiffness \(k=80 \mathrm{~N} / \mathrm{m} .\) If the block is given an upward velocity of \(0.4 \mathrm{~m} / \mathrm{s}\) when it is \(90 \mathrm{~mm}\) above its equilibrium position, determine the equation which describes the motion and the maximum upward displacement of the block measured from the equilibrium position. Assume that positive displacement is measured downward.

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