Question

A mass of \(10.0 \mathrm{~kg}\) is hanging by a steel wire \(1.00 \mathrm{~m}\) long and \(1.00 \mathrm{~mm}\) in diameter. If the mass is pulled down slightly and released, what will be the frequency of the resulting oscillations? Young's modulus for steel is \(2.0 \cdot 10^{11} \mathrm{~N} / \mathrm{m}^{2}\)

Short answer

Answer: The frequency of the oscillations for the mass hanging on the steel wire is approximately 6.31 Hz.

Step-by-step solution

Calculate cross-sectional area of the wire

Using diameter \(d\), we can find the cross-sectional area (\(A\)) by applying the formula for the area of a circle: \(A = \pi (d/2)^2\). Convert \(d\) to meters before making calculations. $$A = \pi\left(\frac{1.00 \times 10^{-3}}{2}\right)^2$$ $$A \approx 7.854 \times 10^{-7} \mathrm{~m}^{2}$$

Find the effective spring constant

The steel wire acts like a spring storing potential energy when stretched or compressed. The effective spring constant (\(k\)) for the wire can be determined using Young's modulus (\(Y\)), length (\(l\)), and cross-sectional area (\(A\)) with the following formula: \(k = \frac{YA}{l}\). $$k = \frac{2.0 \cdot 10^{11} \mathrm{~N} / \mathrm{m}^{2} \times 7.854 \times 10^{-7} \mathrm{~m}^{2}}{1.00 \mathrm{~m}}$$ $$k \approx 1.571 \times 10^{5} \mathrm{~N} / \mathrm{m}$$

Determine the angular frequency

We will now find the angular frequency (\(\omega\)) using the formula: \(\omega = \sqrt{\frac{k}{m}}\). $$\omega = \sqrt{\frac{1.571 \times 10^5 \mathrm{~N} / \mathrm{m}}{10.0 \mathrm{~kg}}}$$ $$\omega \approx 39.64 ~\mathrm{rad} / \mathrm{s}$$

Calculate the frequency of oscillations

Finally, we can determine the frequency (\(f\)) of oscillations by dividing the angular frequency by \(2\pi\), as \(f = \frac{\omega}{2\pi}\). $$f = \frac{39.64 ~\mathrm{rad} / \mathrm{s}}{2\pi}$$ $$f \approx 6.31 ~\mathrm{Hz}$$ The frequency of the resulting oscillations for the mass hanging on the steel wire is approximately 6.31 Hz.

Key concepts

Understanding Young's Modulus

Young's modulus is a fundamental material property that describes the relationship between stress and strain in elastic materials, such as steel. It's a measure of the stiffness of a material and is denoted by the symbol \( Y \). In the context of oscillations, we consider the behavior of a wire acting like a spring: when the wire is stretched by an external force, it exerts an equal and opposite force, similar to a spring.

For the steel wire in our problem, Young's modulus (\(2.0 \cdot 10^{11} \mathrm{~N/m}^{2}\)) represents the ability of the steel to resist changes in shape under tension. The higher the Young's modulus, the stiffer the material, and the more force it requires to create a given amount of deformation. This property is significant when calculating the wire's effective spring constant, as a stiffer material will result in a higher spring constant and affect the oscillation frequency when the wire is displaced and released.

Real-World Application of Young's Modulus

Engineers use Young's Modulus to choose materials for buildings and bridges, ensuring they can handle the stresses they'll face without deforming excessively.

Effective Spring Constant

The concept of the effective spring constant is crucial when dealing with objects that aren't traditional springs but exhibit spring-like behavior. In our exercise, a steel wire is being used, and Young's modulus helps us deduce its spring-like properties. The effective spring constant, symbolized by \( k \), is calculated using the formula \( k = \frac{YA}{l} \), where \( Y \) is Young's modulus, \( A \) is the cross-sectional area of the wire, and \( l \) is the length of the wire.

The spring constant essentially determines the force required to stretch or compress the 'spring' (in this case, the wire) by a certain amount. A higher spring constant means a stiffer spring, needing more force for the same displacement. In the step-by-step solution, once we compute the cross-sectional area, the effective spring constant is found, leading us to the study of the system's oscillatory motion - specifically its frequency, based upon this intrinsic 'springy' property of the wire.

Spring Constant in Everyday Use

From vehicle suspension systems to the keys on your keyboard, the concept of the effective spring constant is utilized to design a wide array of functional and comfortable products.

Angular Frequency and its Relation to Oscillations

Angular frequency, represented by \( \omega \), is a measure of how fast an object is rotating or, in the case of oscillations, how quickly the system completes a full cycle. It is related to the traditional frequency, which counts how many cycles occur per second, with the central relationship being \( f = \frac{\omega}{2\pi} \). To calculate the angular frequency, we use the effective spring constant \( k \) and the mass \( m \) using the formula \( \omega = \sqrt{\frac{k}{m}} \).

This expression for angular frequency encapsulates how the mass and the stiffness of the spring (effective spring constant) affect the oscillation's rapidity. A notable outcome is that the heavier the mass or the softer the spring, the lower the angular frequency, and consequently, the slower the oscillation cycle. When we plug in our calculated spring constant and the mass of the hanging object, we find the system's specific angular frequency, which we then convert to the more familiar frequency measure in Hertz, revealing the answer to our original problem - the frequency of the oscillations - which is approximately 6.31 Hz.

Importance of Angular Frequency

From the precision timing in watches to the music we enjoy, angular frequency plays an integral role in technology and nature, defining the rhythms and cycles that are essential to various phenomena.