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A \(100 \mathrm{~kg}\) mass suspended from a wire whose unstretched length is \(4.00 \mathrm{~m}\) is found to stretch the wire by \(6.0 \mathrm{~mm}\). The wire has a uniform crosssectional area of \(0.10 \mathrm{~cm}^{2}\). (a) If the load is pulled down a small additional distance and released, find the frequency at which it vibrates. (b) Compute Young's modulus for the wire.

Short Answer

Expert verified
(a) Frequency: 6.43 Hz (b) Young's Modulus: 6.53 脳 10鹿鈦 N/m虏

Step by step solution

01

Identify the Given Values and Formulas

The problem provides the following values:- Mass (\(m\)) = 100 kg- Original length of wire (\(L\)) = 4.00 m- Extension (\(\Delta L\)) = 6.0 mm = 0.006 m- Cross-sectional area (\(A\)) = 0.10 cm虏 = 0.00001 m虏The following formulas are relevant:- Young's modulus (\(Y\)) = \(\frac{F \cdot L}{A \cdot \Delta L}\)- Frequency (\(f\)) = \(\frac{1}{2\pi}\sqrt{\frac{k}{m}}\), where \(k\) is the spring constant of the wire.
02

Calculate the Force due to the Mass

The force due to the mass (\(F\)) acting on the wire is given by the weight of the mass, calculated as:\[ F = m \cdot g \]where \(g\) is the acceleration due to gravity (\(9.8 \frac{m}{s^2}\)).Substitute the given values:\[ F = 100 \times 9.8 = 980 \, \text{N} \]
03

Calculate the Spring Constant

The spring constant (\(k\)) is calculated using Hooke's Law, which states:\[ F = k \cdot \Delta L \]Rearrange to find \(k\):\[ k = \frac{F}{\Delta L} = \frac{980}{0.006} = 163333.33 \, \text{N/m} \]
04

Calculate the Frequency of Vibration

The frequency (\(f\)) of vibration is given by:\[ f = \frac{1}{2\pi} \sqrt{\frac{k}{m}} \]Substitute the values:\[ f = \frac{1}{2\pi} \sqrt{\frac{163333.33}{100}} \]\[ f \approx \frac{1}{2\pi} \times 40.42 \approx 6.43 \, \text{Hz} \]
05

Calculate Young's Modulus

Young's modulus (\(Y\)) for the wire is calculated using:\[ Y = \frac{F \cdot L}{A \cdot \Delta L} \]Substitute the values:\[ Y = \frac{980 \times 4}{0.00001 \times 0.006} \]\[ Y = \frac{3920}{0.00006} \approx 65333333333.33 \, \text{N/m}^2 \]Therefore, Young's modulus is approximately \(6.53 \times 10^{10} \, \text{N/m}^2 \).

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Key Concepts

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

Elasticity
Elasticity is a property that enables materials to resume their original shape after being deformed by an external force. This concept is crucial when analyzing materials such as wires or springs. The degree of elasticity is determined by the material's ability to stretch and return to its initial form. Thin wires, like the one described in the exercise, demonstrate elasticity when they extend under a load and retract when the load is removed. This behavior is evaluated quantitatively using Young's modulus, which is a measure of the stiffness of a solid material.
The stronger a material's elasticity, the higher its Young's modulus value. In this exercise, the calculated Young's modulus gives us insight into the wire's elasticity, showing us how it responds to forces applied lengthwise.
Harmonic Motion
Harmonic motion refers to a type of periodic motion or vibration, where the force acting on an object is proportional to its displacement and acts in the opposite direction. The wire, when stretched and released, will vibrate back and forth due to its elasticity, exemplifying harmonic motion. This movement is typical in oscillating systems like springs, pendulums, and musical instrument strings.
The simple harmonic motion is predictable and often relies on factors such as mass, spring constant, and damping. The wire in the problem experiences harmonic motion when the mass attached is pulled and let go, causing it to oscillate at a certain frequency.
Spring Constant
The spring constant, denoted as **k**, is an essential component in understanding the behavior of systems performing harmonic motion. It measures the stiffness or resistance to deformation of a spring or elastic material. The spring constant is defined by Hooke's Law, which states that the force needed to extend or compress a spring by some distance is proportional to that distance.
This exercise involves calculating the spring constant for the wire using the formula:
  • \[ k = \frac{F}{\Delta L}\]
By understanding the spring constant, one gains insights into how easily the wire can be stretched and how much force is necessary to cause a certain amount of deformation. A high spring constant indicates a stiffer wire that resists deformation more strongly.
Vibration Frequency
Vibration frequency is an important parameter in determining how quickly an oscillating or vibrating system completes its cycles. It is expressed in hertz (Hz), where one Hz equals one cycle per second. In this problem, the frequency of vibration of the wire is determined by the mass hanging at the end and the spring constant.
The formula to find the frequency is:
  • \[ f = \frac{1}{2\pi} \sqrt{\frac{k}{m}}\]
This equation shows that frequency is dependent on both the spring constant and the mass. Higher frequencies indicate quicker vibrations. Understanding vibration frequency helps in predicting the dynamic behavior of an oscillating system and is crucial in designing structures or systems that can withstand specific vibration levels.

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Most popular questions from this chapter

(a) If a vibrating system has total energy \(E_{0},\) what will its total energy be (in terms of \(E_{0}\) ) if you double the amplitude of vibration? (b) If you want to triple the total energy of a vibrating system with amplitude \(A_{0}\), what should its new amplitude be (in terms of \(A_{0}\) )?

A science museum has asked you to design a simple pendulum that will make 25.0 complete swings in \(85.0 \mathrm{~s}\). What length should you specify for this pendulum?

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A steel wire has the following properties: $$ \begin{array}{l} \text { Length }=5.00 \mathrm{~m} \\ \text { Cross-sectional area }=0.040 \mathrm{~cm}^{2} \end{array} $$ Young's modulus \(=2.0 \times 10^{11} \mathrm{~Pa}\) Shear modulus \(=0.84 \times 10^{11} \mathrm{~Pa}\) Proportional limit \(=3.60 \times 10^{8} \mathrm{~Pa}\) Breaking stress \(=11.0 \times 10^{8} \mathrm{~Pa}\) The proportional limit is the maximum stress for which the wire still obeys Hooke's law (see point \(\mathrm{B}\) in Figure 11.12 ). The wire is fastened at its upper end and hangs vertically. (a) How great a weight can be hung from the wire without exceeding the proportional limit? (b) How much does the wire stretch under this load? (c) What is the maximum weight that can be supported?

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