Question

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

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

Step-by-step solution

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 = \(4.906 \mathrm{cm^2}\)

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/s²). 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 = \(\frac{196.2 \mathrm{N}}{3.33 \times 10^{-4} \mathrm{m}} = 5.88 \times 10^5 \mathrm{\frac{N}{m}}\)

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.

Key concepts

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.