Question

Two steel wires are stretched under the same tension. The first wire has a diameter of \(0.500 \mathrm{~mm}\), and the second wire has a diameter of \(1.00 \mathrm{~mm}\). If the speed of waves traveling along the first wire is \(50.0 \mathrm{~m} / \mathrm{s}\), what is the speed of waves traveling along the second wire?

Short answer

In this problem, we have two steel wires with different diameters, both under the same tension. The first wire has a diameter of 0.5 mm and a wave speed of 50 m/s, while the second wire has a diameter of 1 mm. We determined the linear mass densities of both wires using their diameters, and set up a proportion between their wave speeds and linear mass densities. After solving for the wave speed of the second wire, we found it to be 100 m/s.

Step-by-step solution

Identify the relevant equation

We'll use the equation for the wave speed of a string: \(v = \sqrt{\frac{T}{\mu}}\) Where \(v\) is the wave speed, \(T\) is the tension, and \(\mu\) is the linear mass density. Since both wires are under the same tension, we can set up a proportion between their wave speeds and linear mass densities.

Calculate the linear mass density for the first wire

We need to find the linear mass densities of both wires to determine the proportion between their wave speeds. The linear mass density (\(\mu\)) can be calculated using the following formula: \(\mu = \frac{m}{L}\) In our case, we don't have the mass or length of the wires, but we do have their diameters and we know they're made of steel. We can use the volume formula for a cylinder and the density of steel to find the linear mass density. Let the length of both wires be represented as \(L\). The volume of the first wire can be calculated using the formula for the volume of a cylinder: \(V = \pi r^2 L\) For the first wire, the radius is 0.250 mm or 0.00025 m. Therefore, its volume is: \(V_{1} = \pi (0.00025)^2 L\) Now, using the density of steel (\(\rho = 7850 kg/m^3\)), we can find the mass of the first wire: \(m_{1} = \rho V_{1} = 7850 \pi (0.00025)^2 L\) Then, the linear mass density of the first wire is: \(\mu_{1} = \frac{m_{1}}{L} = 7850 \pi (0.00025)^2\)

Calculate the linear mass density for the second wire

Using the same process as in Step 2, we can find the linear mass density for the second wire. The diameter of the second wire is 1 mm or 0.001 m, so its radius is 0.0005 m. The volume of the second wire is: \(V_{2} = \pi (0.0005)^2 L\) The mass of the second wire is: \(m_{2} = \rho V_{2} = 7850 \pi (0.0005)^2 L\) And the linear mass density of the second wire is: \(\mu_{2} = \frac{m_{2}}{L} = 7850 \pi (0.0005)^2\)

Set up a proportion between wave speeds and linear mass densities

Since both wires are under the same tension, we can set up a proportion between their wave speeds and linear mass densities: \(\frac{v_{2}}{v_{1}} = \frac{\sqrt{\frac{T}{\mu_{2}}}}{\sqrt{\frac{T}{\mu_{1}}}}\) We're given that the wave speed of the first wire (\(v_{1}\)) is 50 m/s. Plugging this value and the linear mass densities we found earlier into the equation, we have: \(\frac{v_{2}}{50} = \frac{\sqrt{\frac{T}{7850 \pi (0.0005)^2}}}{\sqrt{\frac{T}{7850 \pi (0.00025)^2}}}\)

Solve for the wave speed of the second wire

Now we can solve for the wave speed of the second wire, \(v_{2}\): \(v_{2} = 50\frac{\sqrt{7850 \pi (0.0005)^2}}{\sqrt{7850 \pi (0.00025)^2}}\) \(v_{2} = 50 \sqrt{\frac{(0.0005)^2}{(0.00025)^2}}\) \(v_{2} = 50\cdot 2\) \(v_{2} = 100\, m/s\) The wave speed of the second wire is 100 m/s.