Question

What is the wave speed along a brass wire with a radius of \(0.500 \mathrm{~mm}\) stretched at a tension of \(125 \mathrm{~N}\) ? The density of brass is \(8.60 \cdot 10^{3} \mathrm{~kg} / \mathrm{m}^{3}\).

Short answer

Answer: The wave speed along the brass wire is approximately 54.17 m/s.

Step-by-step solution

Calculate the cross-sectional area

Given the radius of the wire \(r = 0.500 \mathrm{~mm}\), we can find the cross-sectional area \(A\) using the formula for the area of a circle: \(A = \pi r^2\). So, $$A = \pi (0.500 \times 10^{-3} \mathrm{~m})^2 \approx 7.854 \times 10^{-7} \mathrm{~m^2}$$

Calculate the volume of one meter of the wire

To find the volume of one meter of the wire, we need to multiply the cross-sectional area by the length of the wire. Since we are considering one meter of the wire, we simply use the area we calculated in Step 1: $$V = A \times 1\mathrm{~m} = 7.854 \times 10^{-7} \mathrm{~m^3}$$

Calculate the mass of one meter of the wire

Now that we have the volume of one meter of the wire, we can find its mass using the density of brass, \(\rho = 8.60 \cdot 10^{3} \mathrm{~kg} / \mathrm{m}^{3}\). The formula for mass is \(m = \rho V\): $$m = (8.60 \cdot 10^{3} \mathrm{~kg} / \mathrm{m}^{3})(7.854 \times 10^{-7} \mathrm{~m^3}) \approx 6.753 \times 10^{-3} \mathrm{~kg}$$

Calculate the linear mass density of the wire

The linear mass density \(\mu\) is the mass per unit length, in this case, the mass of one meter of the wire. We already calculated the mass of one meter of the wire in Step 3, so we can directly use that value: $$\mu = \frac{m}{1 \mathrm{~m}} = 6.753 \times 10^{-3} \mathrm{~kg/m}$$

Find the wave speed along the wire

With the linear mass density \(\mu\) and the given tension \(T = 125 \mathrm{~N}\), we can now use the formula for wave speed on a string to find the wave speed \(v\): $$v = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{125 \mathrm{~N}}{6.753 \times 10^{-3} \mathrm{~kg/m}}} \approx 54.17 \mathrm{~m/s}$$ Thus, the wave speed along the brass wire is approximately \(54.17 \mathrm{~m/s}\).

Key concepts

Brass Wire

Brass is an alloy made primarily of copper and zinc, known for its high durability and excellent tensile strength. These properties make brass an ideal material for manufacturing wires used in a wide range of applications, such as in musical instruments, electrical conductors, and structural reinforcements. When considering a brass wire in physics, we often look at its density, tensile strength, and capability to conduct waves.

The density of brass is a crucial factor for calculations involving wave speed and other dynamic properties. In this exercise, brass has a density of \(8.60 \times 10^3 \, \text{kg/m}^3\), which allows us to understand how a certain mass of the material occupies a space. Knowing the material's density informs us about how it will behave under specific forces, such as tension, and how efficient it is in transmitting waves.

• This alloy's physical traits, such as toughness and ease of fabrication, make it versatile for various physical and mechanical purposes.
• Its aesthetic appeal from a shiny surface also adds value in applications beyond strictly practical ones.

Linear Mass Density

Linear mass density, denoted by \(\mu\), is the measure of mass per unit length of a wire or string, which is critical in calculating wave speed. For our brass wire, it helps determine how the mass distribution affects wave transmission.
In the step-by-step solution, the linear mass density \(\mu\) was calculated as \(6.753 \times 10^{-3} \, \text{kg/m}\). This means that each meter of the brass wire contains about \(6.753 \times 10^{-3}\) kilograms of mass.

An easy way to visualize this is to think of how a denser wire might carry more mass in each segment, impacting how waves travel. Heavier or denser wires generally transmit waves at different speeds compared to lighter wires.

• The formula \(\mu = \frac{m}{L}\) (mass divided by length) gives a quick way to find linear mass density, where \(m\) is mass and \(L\) is length.
• This value is essential when determining the effects of forces, as more mass per length often means more force is needed to achieve similar acceleration or wave transmission effects.
  

Tension

Tension is the force exerted along the wire, pulling it tight. In this exercise, tension is given as \(125 \, \text{N}\) (Newtons). Understanding tension is vital for calculating wave speed, as it directly influences how fast waves move through a wire.
Higher tension increases the force with which the particles in the wire interact when a wave passes through, typically resulting in faster wave speeds. Think of tension as stretching the wire; as it gets tighter, waves can pass more efficiently with fewer disturbances.

In formulas, tension is usually represented by \(T\), and it is a key component along with linear mass density in determining wave speed. The wave speed \(v\) is calculated using the equation \( v = \sqrt{\frac{T}{\mu}} \).

• Greater tension (higher \(T\)) allows for faster wave transmission by maintaining firm interactions between particles in the wire.
• It shows the direct relationship between the forces applied to a material and the resultant physical behavior observed, such as wave movement.

Cross-Sectional Area

Understanding the cross-sectional area of a wire is important to grasp how its physical dimensions affect its properties such as resistance and sound transmission. In the original exercise, the cross-sectional area \( A \) of the brass wire is calculated using the formula for the area of a circle: \( A = \pi r^2 \), where \( r \) is the radius.
Given a radius of \( 0.500 \text{ mm} \), the cross-sectional area was found to be approximately \( 7.854 \times 10^{-7} \text{ m}^2 \). This small yet crucial area determines the volume of the wire per unit length.
When considering one meter of wire, the cross-sectional area becomes vital because it helps calculate the total mass (and hence mass distribution) of the wire. This impacts calculations on wave speed, as variations in cross-sectional areas may alter tension and subsequently wave velocity.

• With a smaller cross-sectional area, the wire's capacity to hold mass is reduced, affecting its density and potential wave carrying capabilities.
• Calculating this area is essential for understanding how different properties of a wire, like size and mass, interact for effective wave transmission.