/*! 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 1 A spring is stretched \(175 \mat... [FREE SOLUTION] | 91影视

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Chapter 22: Problem 1

A spring is stretched \(175 \mathrm{~mm}\) by an \(8-\mathrm{kg}\) block. If the block is displaced \(100 \mathrm{~mm}\) downward from its equilibrium position and given a downward velocity of \(1.50 \mathrm{~m} / \mathrm{s}\), determine the differential equation which describes the motion. Assume that positive displacement is downward. Also, determine the position of the block when \(t=0.22 \mathrm{~s}\).

Short Answer

Expert verified
The differential equation which describes the block's motion is \(\frac{{d^2x}}{{dt^2}} + 56.06x = 0\). The position of the block at \(t=0.22 \, \mathrm{s}\) is approximately \(0.161 \, \mathrm{m}\) downward from the equilibrium position.

Step by step solution

01

Determine spring constant

The force exerted by the block on the spring is given by \(F = mg\), where \(m = 8 \, \mathrm{kg}\) is the mass of the block and \(g = 9.81 \, \mathrm{m/s^2}\) is the acceleration due to gravity. Hence, \(F = 8 \, \mathrm{kg} \cdot 9.81 \, \mathrm{m/s^2} = 78.48 \, \mathrm{N}\). The spring constant \(k\) can be determined using Hooke's law as \(k = -F/x\), where \(x = 175 \, \mathrm{mm} = 0.175 \, \mathrm{m}\). Thus, \(k = -78.48 \, \mathrm{N} / -0.175 \, \mathrm{m} = 448.46 \, \mathrm{N/m}\).
02

Set up the differential equation

The differential equation for the motion of the block is obtained by setting Newton's second law equal to Hooke's law. This gives \(m \frac{{d^2x}}{{dt^2}}=-kx\). But \(m = 8 \, \mathrm{kg}\) and \(k = 448.46 \, \mathrm{N/m}\). Thus, the differential equation becomes \(\frac{{d^2x}}{{dt^2}}+56.06x=0\).
03

Solve the differential equation

The solution to the differential equation \(\frac{d^2x}{dt^2} + 56.06x = 0\) is \(x(t) = A \cos(\sqrt{56.06}t) + B \sin(\sqrt{56.06}t)\). Using the initial conditions \(x(0) = 0.1 \, \mathrm{m}\) and \(v(0) = \frac{dx}{dt}(0) = 1.5 \, \mathrm{m/s}\), we get \(A = 0.1 \, \mathrm{m}\) and \(B = 1.5 / \sqrt{56.06} \approx 0.2\). Thus, the position of the block at any time \(t\) is \(x(t) = 0.1 \cos(\sqrt{56.06}t) + 0.2 \sin(\sqrt{56.06}t)\).
04

Find the position of the block at \(t=0.22 \, \mathrm{s}\)

Putting \(t = 0.22 \, \mathrm{s}\) in the equation \(x(t) = 0.1 \cos(\sqrt{56.06} \cdot 0.22) + 0.2 \sin(\sqrt{56.06} \cdot 0.22)\), we find that the block's position is approximately \(0.161 \, \mathrm{m}\) downward from the equilibrium position.

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Most popular questions from this chapter

A 6-lb weight is suspended from a spring having a stiffness \(k=3 \mathrm{lb} / \mathrm{in} .\) If the weight is given an upward velocity of \(20 \mathrm{ft} / \mathrm{s}\) when it is \(2 \mathrm{in}\). above its equilibrium position, determine the equation which describes the motion and the maximum upward displacement of the weight, measured from the equilibrium position. Assume positive displacement is downward.

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.

The bar has a weight of \(6 \mathrm{lb}\). If the stiffness of the spring is \(k=8 \mathrm{lb} / \mathrm{ft}\) and the dashpot has a damping coefficient \(c=60 \mathrm{lb} \cdot \mathrm{s} / \mathrm{ft}\), determine the differential equation which describes the motion in terms of the angle \(\theta\) of the bar's rotation. Also, what should be the damping coefficient of the dashpot if the bar is to be critically damped?

A 4-lb weight is attached to a spring having a stiffness \(k=10 \mathrm{lb} / \mathrm{ft}\). The weight is drawn downward a distance of 4 in. and released from rest. If the support moves with a vertical displacement \(\delta=(0.5 \sin 4 t)\) in., where \(t\) is in seconds, determine the equation which describes the position of the weight as a function of time.

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.
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