Problem 23 - University Physics with Modern Physics

Question

A mass of \(10.0 \mathrm{~kg}\) is hanging by a steel wire \(1.00 \mathrm{~m}\) long and \(1.00 \mathrm{~mm}\) in diameter. If the mass is pulled down slightly and released, what will be the frequency of the resulting oscillations? Young's modulus for steel is \(2.0 \cdot 10^{11} \mathrm{~N} / \mathrm{m}^{2}\)

Step-by-Step Solution

Answer

Answer: To find the frequency of oscillation of the mass, follow these steps: 1. Find the cross-sectional area of the steel wire: A = \(π(\frac{0.001}{2})^2\) 2. Calculate the tension in the wire: T = mg 3. Calculate the linear mass density of the wire: μ = \(\frac{ρA}{L}\) 4. Calculate the speed of the transverse wave in the wire: v = \(\sqrt{\frac{T}{μ}}\) 5. Find the frequency of oscillation of the mass: f = \(\frac{v}{2L}\) Plug in the given values and solve for the frequency of oscillation.

Step 1: Identify the known variables

We are given: - mass, m = 10.0 kg - length of the steel wire, L = 1.00 m - diameter of the steel wire, d = 1.00 mm - Young's modulus for steel, Y = \(2.0 \cdot 10^{11} \mathrm{~N/m^2}\)

Step 2: Calculate the cross-sectional area of the steel wire

We will first convert the diameter of the steel wire to meters to maintain the consistency of units. Then, we will find the cross-sectional area (A) of the steel wire using the formula A = \(π(\frac{d}{2})^2\) where d is the diameter of the wire. - diameter in meters, d = 1.00 mm = 0.001 m - A = \(π(\frac{0.001}{2})^2\)

Step 3: Calculate the tension in the wire and the force exerted on the mass

We can calculate the tension (T) in the wire due to the hanging mass using gravitational force. The tension in the wire is equal to the force exerted on the mass (mg), where g is the gravitational acceleration: - T = mg

Step 4: Calculate the speed of the transverse wave in the wire

Now we calculate the speed (v) of the transverse wave in the wire using the formula: v = \(\sqrt{\frac{T}{μ}}\), where μ is the linear mass density of the wire (mass per unit length). We can find μ by dividing the mass of the wire by its length, which can be calculated using m = ρV, where ρ is the density of the steel and V is the volume of the wire (V = AL). In this case, we can use the formula μ = \(\frac{ρA}{L}\) - ρ (density of steel) = \(\frac{mass~of~wire}{volume~of~wire} = \frac{m}{AL}\) - μ = \(\frac{ρA}{L}\) - v = \(\sqrt{\frac{T}{μ}}\)

Step 5: Calculate the frequency of oscillation

Finally, we are now able to solve for the frequency (f) of the resulting oscillations using the formula: f = \(\frac{v}{2L}\) - f = \(\frac{v}{2L}\) Now, plug in the values from the previous steps and solve for the frequency of oscillation of the mass.