/*! 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 9 A \(3-\mathrm{kg}\) block is sus... [FREE SOLUTION] | 91影视

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Chapter 22: Problem 9

A \(3-\mathrm{kg}\) block is suspended from a spring having a stiffness of \(k=200 \mathrm{~N} / \mathrm{m} .\) If the block is pushed \(50 \mathrm{~mm}\) upward from its equilibrium position and then released from rest, determine the equation that describes the motion. What are the amplitude and the natural frequency of the vibration? Assume that positive displacement is downward.

Short Answer

Expert verified
The equation that describes the motion is \( x(t) = 0.05 \cos (8.166 t + \pi) \) m. The amplitude of the vibration is 0.05 m and the natural frequency is 8.166 rad/s.

Step by step solution

01

- Converting displacement to SI units

Firstly, change the displacement \( x \) from mm to meters, because the spring constant k is in N/m. Therefore, \( x = 50mm = 50 \times 10^{-3} m = 0.05m \).
02

- Determine the natural frequency

Use the formula \( \omega = \sqrt{\frac{k}{m}} \) to find the natural frequency. Therefore, the natural frequency \( \omega = \sqrt{\frac{200 N/m}{3 kg}} = \sqrt{66.67 s^{-2}} = 8.166 s^{-1} \). Thus, the natural frequency of vibration is 8.166 rad/s.
03

- Write the equation of motion

The equation of motion of a simple harmonic motion (SHM) is given by \( x(t) = A \cos (\omega t + \phi) \), where x(t) is the displacement at time t, A is the amplitude, \(\omega\) is the natural frequency and \(\phi\) is the phase angle. Here, final displacement is positive downward, thus to match with final condition, we need to keep our phase angle \(\phi = 180掳 = \pi\) radians in cosine function. So the equation of motion is: \( x(t) = 0.05 \cos (8.166 t + \pi) \).
04

- Identify the amplitude and natural frequency of the vibration

From our equation of motion, we can see that the amplitude A is equal to 0.05 m (same as the initial displacement), and the natural frequency \(\omega\) is 8.166 rad/s, same as earlier calculated.

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Most popular questions from this chapter

If the block-and-spring model is subjected to the periodic force \(F=F_{0} \cos \omega t\), show that the differential equation of motion is \(\ddot{x}+(k / m) x=\left(F_{0} / m\right) \cos \omega t\), where \(x\) is measured from the equilibrium position of the block. What is the general solution of this equation?

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.

A flywheel of mass \(m\), which has a radius of gyration about its center of mass of \(k_{o}\), is suspended from a circular shaft that has a torsional resistance of \(M=C \theta .\) If the flywheel is given a small angular displacement of \(\theta\) and released, determine the natural period of oscillation.

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.

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.
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