Problem 64 - University Physics with Modern Physics

Question

One type of steel has a density of \(7.8 \times 10^{3} \mathrm{kg} / \mathrm{m}^{3}\) and a breaking stress of \(7.0 \times 10^{3} \mathrm{N} / \mathrm{m}^{2}\) . A cylindrical guitar string is to be made of 4.00 \(\mathrm{g}\) of this steel. (a) What are the length and radius of the longest and thinnest string that can be placed under a tension of 900 \(\mathrm{N}\) without breaking? (b) What is the highest fundamental frequency that this string could have?

Step-by-Step Solution

Answer

The string can be up to 12.6 m long with a radius of 6.39 mm, and its highest fundamental frequency is about 42.5 Hz.

Step 1: Convert Mass to Kilograms

The mass of the steel is given as 4.00 g. First, we need to convert this mass into kilograms (kg) since the density is given in kg/m³.Recall that 1 g = 0.001 kg.Convert the mass: \[4.00 \, \text{g} = 0.004 \, \text{kg} \]

Step 2: Calculate the Volume of the Steel

Using the density formula, \( \text{density} = \frac{\text{mass}}{\text{volume}} \), we can solve for the volume:\[ \text{Volume} = \frac{\text{mass}}{\text{density}} \]Substitute the known values:\[ \text{Volume} = \frac{0.004 \, \text{kg}}{7.8 \times 10^3 \, \text{kg/m}^3} \approx 5.13 \times 10^{-7} \, \text{m}^3 \]

Step 3: Express the Volume in Terms of Length and Radius

For a cylindrical string, the volume is given by the formula:\[ \text{Volume} = \pi r^2 L \]where \( r \) is the radius and \( L \) is the length of the string. Therefore:\[ 5.13 \times 10^{-7} = \pi r^2 L \]

Step 4: Determine Maximum Radius and Corresponding Length for Tension

The breaking stress gives the maximum force per unit area before breaking.\[ \text{Breaking Stress} = \frac{\text{Tension}}{\pi r^2} \]Set the tension under which the string can be placed without breaking:\[ 7.0 \times 10^3 = \frac{900}{\pi r^2} \]Solve for \( r^2 \):\[ r^2 = \frac{900}{7.0 \times 10^3 \pi} \approx 4.09 \times 10^{-5} \]Solve for \( r \):\[ r \approx 6.39 \times 10^{-3} \, \text{m} \]Substitute back to find \( L \):\[ L = \frac{5.13 \times 10^{-7}}{\pi (6.39 \times 10^{-3})^2} \approx 12.6 \, \text{m} \]

Step 5: Calculate the Highest Fundamental Frequency

The fundamental frequency \( f \) of a string is given by:\[ f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \]where \( T \) is the tension (900 N) and \( \mu \) is the linear density (mass per unit length).First, find \( \mu \):\[ \mu = \frac{0.004}{12.6} \approx 3.17 \times 10^{-4} \, \text{kg/m} \]Now calculate \( f \):\[ f = \frac{1}{2 \times 12.6} \sqrt{\frac{900}{3.17 \times 10^{-4}}} \approx 42.5 \, \text{Hz} \]

Key Concepts

Density Calculations

Understanding density is crucial in solving various physics problems, including those pertaining to materials and their properties. Density is defined as the mass per unit volume of a substance and is typically expressed in kilograms per cubic meter \(\left( \text{kg/m}^3 \right)\). In our given problem, we focus on a particular type of steel with a density of \(7.8 \times 10^3 \, \text{kg/m}^3\). To utilize this information, we convert the mass of the steel from grams to kilograms. This allows us to employ the formula: \\[ \text{Volume} = \frac{\text{mass}}{\text{density}} \] \to find the volume of the steel. In this case, a mass of 0.004 kg results in a volume of approximately \(5.13 \times 10^{-7} \, \text{m}^3\). This volume, once expressed in terms of string geometry—such as its radius and length—becomes the foundation for subsequent calculations.

Breaking Stress

Breaking stress is a measure of the maximum stress that a material can withstand before failure or fracture. It's an essential factor in determining the limits of structural components. The unit of breaking stress is typically Newtons per square meter \(\left( \text{N/m}^2 \right)\). In our exercise, the steel has a breaking stress of \(7.0 \times 10^3 \, \text{N/m}^2\). Under a given tension of 900 N, we need to ensure that the stress remains within the material's safe limits.To solve for the maximum radius under these conditions, use the relationship:\[ \text{Breaking Stress} = \frac{\text{Tension}}{\pi r^2} \]This relationship helps us find the radius \(r\) that ensures the stress does not exceed breaking stress. Solving the formula, we obtain a radius \( r \approx 6.39 \times 10^{-3} \, \text{m} \).This allows us to set the dimensions of the string, taking care not to exceed its breaking stress limits while maintaining desired performance.

Tension and Strings

Tension plays a critical role in the behavior of strings used in various applications, such as musical instruments or structural elements. For an object like a guitar string, maintaining the proper tension ensures that the instrument produces the desired sound.In physics, tension refers to the force transmitted through a string, cable, or similar object. Concerning our steel string, the tension applied is 900 N. This value factors into determining the string's radius using the equation for breaking stress:\[ \text{Breaking Stress} = \frac{\text{Tension}}{\pi r^2} \]and subsequently informs its potential length. Given the mass, density, and specific tension, we also calculate the string's maximum length \(L\) while adhering to geometric and stress constraints. In our example, this calculates to an approximate length of 12.6 meters.

Fundamental Frequency

The fundamental frequency, sometimes called the first harmonic, is the lowest frequency at which a system oscillates. For strings, like those on a guitar, fundamental frequency is vital in determining the pitch of the note produced. The frequency can be calculated using the formula:\[ f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \]where \(L\) is the string length, \(T\) the tension, and \(\mu\) the linear density (mass per unit length). In our problem, the linear density is calculated as \(\mu = \frac{0.004 \, \text{kg}}{12.6 \, \text{m}} \approx 3.17 \times 10^{-4} \, \text{kg/m}\). Inserting these values into our frequency equation gives us the highest possible fundamental frequency, calculated to be approximately 42.5 Hz, which determines the highest pitch the string can produce under these conditions.