/*! 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 46 A \(5-\mathrm{kg}\) block is sus... [FREE SOLUTION] | 91Ó°ÊÓ

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A \(5-\mathrm{kg}\) block is suspended from a spring having a stiffness of \(300 \mathrm{~N} / \mathrm{m}\). If the block is acted upon by a vertical force \(F=(7 \sin 8 t) \mathrm{N}\), where \(t\) is in seconds, determine the equation which describes the motion of the block when it is pulled down \(100 \mathrm{~mm}\) from the equilibrium position and released from rest at \(t=0 .\) Assume that positive displacement is downward.

Short Answer

Expert verified
The equation of motion for the block is \( x(t) = -0.1 \cos(\sqrt{60}*t) - 0.044 \sin(8*t) \).

Step by step solution

01

Establish the Differential Equation

For a block suspended from a spring, the governing differential equation for the motion of the block can be written as \( m*x''(t) + k*x(t) = F(t) \), where \( m \) is the mass of the block, \( k \) is the spring constant, \( x(t) \) is the displacement of the block, \( F(t) \) is the force acting on the block, and \( x''(t) \) denotes the second derivative of \( x(t) \). Substituting the provided values for \( m, k, \) and \( F(t) \), we get \( 5*x''(t) + 300*x(t) = 7 \sin (8t) \).
02

Identify Initial Conditions

The initial conditions provided are that the block is released from a position \( x(0) = 0.1m \) (100mm) below the equilibrium, and it is released from rest which means the initial velocity \( x'(0) = 0 \). However, since the positive displacement is considered downward, the sign of the initial position must be negative: \( x(0) = -0.1m \).
03

Determine Homogenous and Particular Solutions

We need to find the homogeneous and particular solutions to the differential equation. The homogeneous equation is \( 5*x''(t) + 300*x(t) = 0 \), which has the solution \( x_h(t) = A \cos(\sqrt{60}*t) + B \sin(\sqrt{60}*t) \), where A and B are constants. The particular solution \( x_p(t) \) can be guessed as \( x_p(t) = C \sin(8*t) + D \cos(8*t) \). Upon substituting in the given equation and comparing coefficients, we get \( C = -0.044 \) and \( D = 0 \). So the complete solution becomes \( x(t) = x_h(t) + x_p(t) \).
04

Find Constants Using Initial Conditions

After substituting \( t = 0 \) in \( x(t) = x_h(t) + x_p(t) \) we get \( A = -0.1 - D \) or \( A = -0.1 \). Similarly, differentiating \( x(t) \) once, and substitifying \( t = 0 \), we obtain \( B = 0 \). Substituting A and B in the expression of \( x(t) \), we get the final position as function of time \( x(t) = -0.1 \cos(\sqrt{60}*t) - 0.044 \sin(8*t) \).

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Key Concepts

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

Spring-Mass System
Understanding the spring-mass system is crucial when studying physics or engineering as it forms the foundation of vibratory and harmonic systems. This system consists of a mass attached to a spring which can move freely. When an external force is applied or the mass is displaced from its equilibrium position, it oscillates around that position.
The behavior of a spring-mass system is commonly described by Hooke’s Law, which states that the force exerted by a spring is proportional to its displacement from the equilibrium. Mathematically, this can be written as \( F = -kx \), where \( k \) is the spring constant, and \( x \) is the displacement. In the context of the differential equation \( m*x''(t) + k*x(t) = F(t) \), the terms represent the mass's inertial resistance to acceleration (\( m*x''(t) \)), the restoring force of the spring (\( k*x(t) \)), and the external force (\( F(t) \)).
This setup results in a periodic motion, which can be simple harmonic motion if the system is undamped, or it can include more complex dynamics depending on additional forces.
Harmonic Motion
Harmonic motion is a type of periodic motion where the restoring force is directly proportional to the displacement and acts in the direction opposite to that of displacement. In practical terms, it's the back-and-forth vibratory motion of the spring-mass system we discussed earlier.
In the absence of damping and external forces, the mass would undergo simple harmonic motion, described by \( x(t) = A \times \text{cos}(\omega t) + B \times \text{sin}(\omega t) \), where \( A \) and \( B \) are constants determined by initial conditions, and \( \omega \) is the angular frequency. The angular frequency itself is connected to the physical characteristics of the system, specifically mass \( m \) and spring constant \( k \), through the relation \( \omega = \text{sqrt}(\frac{k}{m}) \). The presence of an external force, like in our exercise, leads to the superposition of an additional type of motion on top of this harmonic motion, resulting in a complex oscillatory pattern.
Initial Conditions
In the realm of differential equations, initial conditions are essential as they allow us to find the specific solution that fits the physical scenario being analyzed. These conditions describe the state of the system at a certain initial time—usually, \( t = 0 \).
For our spring-mass system, the initial conditions were that the block was released from rest after being pulled down 100 mm from the equilibrium position. Mathematically, this translates to \( x(0) = -0.1m \) and \( x'(0) = 0 \). These initial conditions are pivotal for defining the constants in the homogeneous solution, which represent the amplitude and phase of the resulting motion. Without these, we would not be able to provide the precise description of motion our system would exhibit after being disturbed.
Homogeneous and Particular Solutions
Solving differential equations typically involves finding two types of solutions: the homogeneous and the particular solution. The homogeneous solution \( x_h(t) \) addresses the characteristic response of the system without the external force and consists of components that make up the complementary function.
In contrast, the particular solution \( x_p(t) \) is found by considering the form of the external force acting on our system and represents the non-homogeneous aspect of the differential equation, which in this case includes the \( 7 \text{sin}(8t) \) driving force.
To combine these solutions, we superimpose them, resulting in the general solution of the differential equation that models the full motion of the mass in our spring-mass system. Subsequently, the initial conditions are used to solve for the unknown constants. As seen in steps 3 and 4 above, for our specific exercise, the homogeneous solution gave us the constants \( A \) and \( B \) upon applying the initial conditions, yielding the final equation of motion that captures all aspects of the block's movement.

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Most popular questions from this chapter

A block of mass \(m\) is suspended from two springs having a stiffness of \(k_{1}\) and \(k_{2}\), arranged a) parallel to each other, and \(\mathrm{b}\) ) as a series. Determine the equivalent stiffness of a single spring with the same oscillation characteristics and the period of oscillation for each case.

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 machine has a mass \(m\) and is uniformly supported by four springs, each having a stiffness \(k\). Determine the natural period of vertical vibration.

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.

The block, having a weight of \(12 \mathrm{lb}\), is immersed in a liquid such that the damping force acting on the block has a magnitude of \(F=(0.7|v|) \mathrm{lb}\), where \(v\) is in \(\mathrm{ft} / \mathrm{s}\). If the block is pulled down \(0.62 \mathrm{ft}\) and released from rest, determine the position of the block as a function of time. The spring has a stiffness of \(k=53 \mathrm{lb} / \mathrm{ft}\). Assume that positive displacement is downward.

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