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Chapter 17: Problem 70

A steel wire has density \(7800 \mathrm{~kg} / \mathrm{m}^{3}\) and mass \(2.50 \mathrm{~g}\). It is stretched between two rigid supports separated by \(0.400 \mathrm{~m}\). (a) When the temperature of the wire is \(20.0^{\circ} \mathrm{C}\), the frequency of the fundamental standing wave for the wire is \(440 \mathrm{~Hz}\). What is the tension in the wire? (b) What is the temperature of the wire if its fundamental standing wave has frequency \(460 \mathrm{~Hz}\) ? For steel the coefficient of linear expansion is \(1.2 \times 10^{-5} \mathrm{~K}^{-1}\) and Young's modulus is \(20 \times 10^{10} \mathrm{~Pa}\)

Short Answer

Expert verified
The tension in the wire is 780 N. Assuming the wire is in thermal equilibrium with the surrounding and the initial temperature is 20掳C, the final temperature is calculated in Celsius.

Step by step solution

01

Calculate the linear mass density \(\mu\)

From the given mass and length, the linear mass density \(\mu\) of the string can be calculated as \(\mu = \frac{mass}{length}= \frac{2.5 g}{0.4 m} = 6.25 \times 10^{-3} \mathrm{~kg/m}\)
02

Calculate the tension in the wire

We can rearrange the wave speed equation to solve for Tension as \(T = \mu v^{2}\). The speed of wave \(v\) is given as \(v = f \lambda\). The frequency \(f\) is given as 440 Hz. For a string fixed at both ends, the wavelength of the fundamental frequency is twice the length of the string, i.e. \(\lambda = 2L = 2 \times 0.400 m = 0.800 m\). So, \(v = 440 Hz \times 0.800 m = 352 m/s\). Substituting the values, we get \(T = 6.25 \times 10^{-3} \mathrm{~kg/m} \times (352 m/s)^{2} = 780 N\) for the tension.
03

Calculate the change in the length of the wire

Using Young鈥檚 modulus \(Y = \frac{\sigma}{\epsilon}\), we can calculate the strain \(\epsilon = \frac{\sigma}{Y}\). The stress \(\sigma\) is the force per unit area. Assuming the cross section area A is uniform along the wire's length, the force on it is just the tension T. So, \(\sigma = \frac{F}{A} = \frac{T}{A}\). Substituting the values, \(\epsilon = \frac{780 N}{\pi r^{2}} / 20 \times 10^{10} Pa\). This gives the fractional change in length.
04

Calculate the temperature change

The change in length from original length is equal to the fractional length change \( \Delta L/ L = \alpha \Delta T\). The coefficient of linear expansion is given as \( \alpha = 1.2 \times 10^{-5} K^{-1}\). Solving for the change in temperature \( \Delta T = \Delta L / L \alpha\). The new length of the wire \(L_{new} = L + \Delta L = L (1 + \epsilon) = 0.400 m (1 + \epsilon)\). From step 3, we can find the value of \(\epsilon\). The tension does not change when the temperature changes, so the string's fundamental frequency with new length is given by \(f_{new} = \frac{v}{\lambda_{new}} = \frac{v}{2L_{new}}\). We can solve for \(L_{new}\) and thereafter \(\Delta L/L\). Hence, we can solve for \( \Delta T = \Delta L / L \alpha \). The final temperature is \(T_{final} = T_{initial} + \Delta T = 20.0 掳C + \Delta T\). The absolute temperature is calculated in Celsius so the final answer should be transformed back into Celsius.

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Key Concepts

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

Density of Material
Density is a key concept in physics that describes how much mass is contained in a given volume of a substance. It is a measure of how tightly the material's particles are packed together.
In mathematical terms, density (\(\rho\)) is defined as \(\rho = \frac{m}{V}\), where \(m\) is mass and \(V\) is volume.
For the problem mentioned, the density of the steel wire is given as \(7800 \, \text{kg/m}^3\). This value tells us that each cubic meter of the steel wire contains \(7800\) kilograms of mass.
Understanding density helps in calculating other properties like linear mass density, which is essential in determining characteristics such as tension and wave speed in a wire.
Coefficient of Linear Expansion
Materials expand or contract with temperature changes, and the coefficient of linear expansion helps quantify this behavior. It indicates how much a material will expand per degree of temperature increase.
For steel, the coefficient of linear expansion (\(\alpha\)) is given as \(1.2 \times 10^{-5} \, \text{K}^{-1}\). This coefficient represents the fractional change in length per temperature unit change.
In practical terms, if a steel wire is heated or cooled, the total change in its length (\(\Delta L\)) can be calculated using the formula \(\Delta L = \alpha \times L \times \Delta T\), where \(\Delta T\) is the change in temperature and \(L\) is the original length.
This property is vital in scenarios where precision in length is essential, such as wire tension applications, as it directly affects properties like wave frequency.
Young's Modulus
Young's Modulus (\(E\)) is a measure of a material's ability to withstand changes in length when subjected to tension or compression forces. It is a fundamental concept that describes the stiffness of a material.
The formula for Young's Modulus is given as:\[E = \frac{\sigma}{\epsilon}\]where \(\sigma\) is the stress (force per unit area) and \(\epsilon\) is the strain (the fractional change in length).
In the context of a steel wire, Young's Modulus can be utilized to determine how the wire will deform under a given load (tension). For steel, it's typically large, indicating that steel is relatively stiff and doesn't stretch much under tension.
Calculating Young's Modulus allows us to predict how materials will behave when subject to various forces, which is crucial when dealing with materials under tension.
Fundamental Frequency of Standing Waves
Standing waves occur when a wave reflects back upon itself, creating areas of constructive and destructive interference. For a wire fixed at both ends, its fundamental frequency is the lowest frequency at which it vibrates.
The fundamental frequency (\(f\)) is given by the formula:\[f = \frac{v}{2L}\]where \(v\) is the wave speed and \(L\) is the length of the wire.
For the steel wire problem, the fundamental frequency was given as \(440 \, \text{Hz}\) at \(20^{\circ} \text{C}\).
Variables such as tension, length, and linear mass density help determine this frequency. A change in any of these aspects alters the frequency, as demonstrated by the temperature-induced frequency shift to \(460 \, \text{Hz}\) in the problem.
Understanding the fundamentals of standing waves is crucial for applications like musical instruments and other technologies that rely on specific wave frequencies.

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Most popular questions from this chapter

A laboratory technician drops a \(0.0850 \mathrm{~kg}\) sample of unknown solid material, at \(100.0^{\circ} \mathrm{C}\), into a calorimeter. The calorimeter can, initially at \(19.0^{\circ} \mathrm{C},\) is made of \(0.150 \mathrm{~kg}\) of copper and contains \(0.200 \mathrm{~kg}\) of water. The final temperature of the calorimeter can and contents is \(26.1^{\circ} \mathrm{C}\). Compute the specific heat of the sample.

A \(\mathrm 500.0 \mathrm{~g}\) chunk of an unknown metal, which has been in boiling water for several minutes, is quickly dropped into an insulating Styrofoam beaker containing \(1.00 \mathrm{~kg}\) of water at room temperature \(\left(20.0^{\circ} \mathrm{C}\right) .\) After waiting and gently stirring for 5.00 minutes, you observe that the water's temperature has reached a constant value of \(22.0^{\circ} \mathrm{C}\). (a) Assuming that the Styrofoam absorbs a negligibly small amount of heat and that no heat was lost to the surroundings, what is the specific heat of the metal? (b) Which is more useful for storing thermal energy: this metal or an equal weight of water? Explain. (c) If the heat absorbed by the Styrofoam actually is not negligible, how would the specific heat you calculated in part (a) be in error? Would it be too large, too small, or still correct? Explain.

\(Camels require very little }}\) water because they are able to tolerate relatively large changes in their body temperature. While humans keep their body temperatures constant to within one or two Celsius degrees, a dehydrated camel permits its body temperature to drop to \(34.0^{\circ} \mathrm{C}\) overnight and rise to \(40.0^{\circ} \mathrm{C}\) during the day. To see how effective this mechanism is for saving water, calculate how many liters of water a \(400 \mathrm{~kg}\) camel would have to drink if it attempted to keep its body temperature at a constant \(34.0^{\circ} \mathrm{C}\) by evaporation of sweat during the day ( 12 hours) instead of letting it rise to \(40.0^{\circ} \mathrm{C}\). (Note: The specific heat of a camel or other mammal is about the same as that of a typical human, \(3480 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\). The heat of vaporization of water at \(34^{\circ} \mathrm{C}\) is \(2.42 \times 10^{6} \mathrm{~J} / \mathrm{kg} .\) )

On-Demand Water Heaters. Conventional hot-water heaters consist of a tank of water maintained at a fixed temperature. The hot water is to be used when needed. The drawbacks are that energy is wasted because the tank loses heat when it is not in use and that you can run out of hot water if you use too much. Some utility companies are encouraging the use of on-demand water heaters (also known as flash heaters), which consist of heating units to heat the water as you use it. No water tank is involved, so no heat is wasted. A typical household shower flow rate is \(2.5 \mathrm{gal} / \mathrm{min}(9.46 \mathrm{~L} / \mathrm{min})\) with the tap water being heated from \(50^{\circ} \mathrm{F}\left(10^{\circ} \mathrm{C}\right)\) to \(120^{\circ} \mathrm{F}\left(49^{\circ} \mathrm{C}\right)\) by the on-demand heater. What rate of heat input (either electrical or from gas) is required to operate such a unit, assuming that all the heat goes into the water?

An aluminum tea kettle with mass \(1.10 \mathrm{~kg}\) and containing \(1.80 \mathrm{~kg}\) of water is placed on a stove. If no heat is lost to the surroundings, how much heat must be added to raise the temperature from \(20.0^{\circ} \mathrm{C}\) to \(85.0^{\circ} \mathrm{C} ?\)
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