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A \(30 \mathrm{~cm}\) aluminum rod possessing a circular cross section of \(1.25 \mathrm{~cm}\) radius is inserted into a testing machine where it is fixed at one end and attached to a load cell at the other end. At some point during a tensile test the clamp at the load cell slips, releasing that end of the rod. If the \(20 \mathrm{~kg}\) clamp remains attached to the end of the rod, determine the frequency of the oscillations of the rod-clamp system?

Short Answer

Expert verified
Answer: The frequency of the oscillations of the rod-clamp system is 11.9 Hz.

Step by step solution

01

Calculate the area of the rod's cross-section

We will first need to find the area of the rod's circular cross-section to use it in further calculations. The formula for the area of a circle is given by \(A = \pi r^2\). A = \(\pi (1.25 \mathrm{cm})^2\) A = \(4.906 \mathrm{cm^2}\)
02

Find the spring constant of the rod

We can use Hooke's Law to find the spring constant of the rod, which states that \(F = k \cdot x\). We can rearrange the formula for the spring constant: \(k = \frac{F}{x}\). We know the force F applied to the rod when the clamp slips is equal to the weight of the clamp, which can be calculated using the formula \(F = m \cdot g\), where m is the mass and g is the gravitational acceleration (9.81 m/sĀ²). F = (20 \mathrm{kg}) (9.81 \mathrm{\frac{m}{s^2}}) = 196.2 \mathrm{N} To find the elongation x, we'll use the formula \(x = \frac{F \cdot L}{A \cdot E}\), where L is the length of the rod, A is the cross-sectional area, and E is the Young's modulus for aluminum. The Young's modulus for aluminum is approximately \(6.9 \times 10^{10} \mathrm{Pa}\). x = \(\frac{(196.2 \mathrm{N})(0.30 \mathrm{m})}{(4.906 \times 10^{-4} \mathrm{m^2})(6.9\times10^{10} \mathrm{Pa})} = 3.33 \times 10^{-4} \mathrm{m}\) Now, we can find the spring constant k. k = \(\frac{196.2 \mathrm{N}}{3.33 \times 10^{-4} \mathrm{m}} = 5.88 \times 10^5 \mathrm{\frac{N}{m}}\)
03

Determine the frequency of oscillations

We can use the formula for the frequency of oscillations of a mass-spring system to find the frequency: \(f = \frac{1}{2\pi}\sqrt{\frac{k}{m}}\). f = \(\frac{1}{2\pi}\sqrt{\frac{5.88 \times 10^5 \mathrm{\frac{N}{m}}}{20 \mathrm{kg}}}\) f = 11.9 Hz So, the frequency of the oscillations of the rod-clamp system is 11.9 Hz.

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Key Concepts

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

Hooke's Law
Understanding Hooke's Law is crucial in the study of engineering vibrations. It describes the behavior of springs and other elastic materials, stating that the force (\f\( F \f\)) needed to extend or compress a spring by some distance (\f\( x \f\)) is directly proportional to that distance, as long as the material's elastic limit is not exceeded. The formula is expressed as \f\( F = k \times x \f\), where \f\( k \f\) is known as the spring constant, which measures the stiffness of the spring.

When engineers design machines and structures, they must ensure that the materials used can withstand the forces acting upon them without deforming permanently. By applying Hooke's Law, they can predict how the material will behave under load and design the system to maintain safety and functionality. In the case of the aluminum rod mentioned in the exercise, the force applied was due to the weight of the clamp, and we calculated how much the rod stretched before it reached the spring constant.
Young's Modulus
Young's modulus, often represented by the symbol \f\( E \f\), is an essential property of materials in the field of mechanical engineering. It quantifies the relationship between tensile stress (force per unit area) and tensile strain (proportional deformation) in a material. It is a measure of the stiffness of a given material, indicating how much it will deform under a certain load.

To put it simply, the higher the Young's modulus, the stiffer the material is, and hence, the less it will stretch or compress under tension or compression. Young's modulus is particularly useful for predicting how materials will behave when forces are applied along their length, as with the rod in the textbook exercise. Knowing aluminum's Young's modulus allowed us to determine the elongation of the rod when a specific force was applied, which, as per our exercise, directly relates to the spring constant and the subsequent vibrational frequency of the system.
Mass-Spring System
The mass-spring system is a fundamental model used in the analysis of mechanical vibrations. An ideal system consists of a mass (\f\( m \f\)) that is attached to a spring with a spring constant (\f\( k \f\)). As the mass moves, the spring either stretches or compresses, storing potential energy. If the mass is set in motion, it will oscillate about an equilibrium position due to the restoring force of the spring that follows Hooke's Law.

The frequency of these oscillations, as indicated in the solution of the exercise, can be described by the formula \f\( f = \frac{1}{2\fp} \tc \frac{k}{m} \f\), where \f\( f \f\) is the frequency, \f\( k \f\) is the spring constant, and \f\( m \f\) is the mass of the object. Essentially, this formula encapsulates how oscillation frequency is influenced by the mass of the object and the stiffness of the spring. In our example, after calculating the spring constant using Hooke's Law and Young's modulus, we used the mass of the attached clamp to find the system's frequency of oscillation.

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Most popular questions from this chapter

A single degree of freedom system is represented as a \(2 \mathrm{~kg}\) mass attached to a spring possessing a stiffness of \(4 \mathrm{~N} / \mathrm{m}\) and a viscous damper whose coefficient is \(2 \mathrm{~N}-\mathrm{sec} / \mathrm{m}\). (a) Determine the response of the horizontally configured system if the mass is displaced 1 meter to the right and released from rest. Plot and label the response history of the system. (b) Determine the response and plot its history if the damping coefficient is \(8 \mathrm{~N}-\mathrm{sec} / \mathrm{m}\).

A screen door of mass \(m\), height \(L\) and width \(\ell\) is attached to a door frame as indicated. A torsional spring of stiffness \(k_{T}\) is attached as a closer at the top of the door as indicated, and a damper is to be installed near the bottom of the door to keep the door from slamming. Determine the limiting value of the damping coefficient so that the door closes gently, if the damper is to be attached a distance a from the hinge.

The system shown consists of a rigid rod, a flywheel of radius \(R\) and mass \(m\), and an elastic belt of stiffness \(k\). Determine the natural frequency of the system. The belt is unstretched when \(\theta=0\).

A \(12 \mathrm{~kg}\) spool that is \(1 \mathrm{~m}\) in radius is pinned to a viscoelastic rod of negligible mass with effective properties \(k=10 \mathrm{~N} / \mathrm{m}\) and \(c=8 \mathrm{~N}-\mathrm{sec} / \mathrm{m}\). The end of the rod is attached to a rigid support as shown. Determine the natural frequency of the system if the spool rolls without slipping.

A single degree of freedom system is represented as a \(4 \mathrm{~kg}\) mass attached to a spring possessing a stiffness of \(6 \mathrm{~N} / \mathrm{m}\). Determine the response of the horizontally configured system if the mass is displaced 2 meters to the right and released with a velocity of \(4 \mathrm{~m} / \mathrm{sec}\). What is the amplitude, period and phase lag for the motion? Sketch and label the response history of the system.

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