Question

A wire stretched between two rigid supports vibrates with a frequency of \(45 \mathrm{~Hz}\). If the mass of the wire is \(3.5 \times 10^{-2} \mathrm{~kg}\) and its linear mass density is \(4.0 \times 10^{-2} \mathrm{~kg} / \mathrm{m}\), what will be the tension in the wire? (A) \(212 \mathrm{~N}\) (B) \(236 \mathrm{~N}\) (C) \(248 \mathrm{~N}\) (D) \(254 \mathrm{~N}\)

Short answer

The tension in the wire is approximately \(248 N\). The correct answer is (C) \(248 \mathrm{~N}\).

Step-by-step solution

Write the formula for the fundamental frequency of a vibrating wire.

The formula for the fundamental frequency of a vibrating wire is given by: \[f = \frac{1}{2L}\sqrt{\frac{T}{\mu}}\] where: - \(f\) is the frequency of the wire (\(45 Hz\)), - \(L\) is the length of the wire, which we need to find, - \(T\) is the tension in the wire, which is what we want to determine, - and \(\mu\) is the linear mass density of the wire (\(4.0 \times 10^{-2} kg/m\)).

Find the length of the wire using the mass and linear mass density.

We can find the length of the wire using the given mass and linear mass density. The linear mass density is defined as the mass per unit length of the wire: \[\mu = \frac{m}{L}\] where: - \(\mu\) is the linear mass density (\(4.0 \times 10^{-2} kg/m\)), - \(m\) is the mass of the wire (\(3.5 \times 10^{-2} kg\)), - and \(L\) is the length of the wire. We can rearrange this equation to solve for the length: \[L = \frac{m}{\mu} = \frac{3.5 \times 10^{-2} kg}{4.0 \times 10^{-2} kg/m} = 0.875 m\]

Rearrange the formula of the fundamental frequency to solve for the tension.

We can rearrange the formula of the fundamental frequency to solve for the tension: \[T = \mu (2Lf)^2\]

Plug in the values and calculate the tension in the wire.

Now, we can plug in the values for the linear mass density, length, and frequency into the equation and calculate the tension in the wire: \[T = (4.0 \times 10^{-2} kg/m)(2 \times 0.875 m \times 45 Hz)^2\] \[T = (4.0 \times 10^{-2} kg/m)(157.5 Hz)^2\] \[T \approx 248 N\] The tension in the wire is approximately \(248 N\). The correct answer is (C) \(248 \mathrm{~N}\).

Key concepts

Frequency of a Vibrating Wire

The frequency of a vibrating wire is an important aspect in understanding how the wire behaves when it is under tension. Think of frequency as the number of times the wire vibrates back and forth in one second. This is measured in Hertz (Hz). The main factors affecting the frequency are the tension in the wire, the length of the wire, and the wire's mass distribution, also known as linear mass density. When you hear a guitar string plucked, it's essentially the frequency at play. More tension usually means a higher frequency, resulting in a higher pitch sound, while less tension results in a lower frequency and lower pitch. To grasp this effectively, we can refer to the formula for fundamental frequency, which mathematically connects these variables and helps predict how the wire will vibrate under different settings.

Linear Mass Density

Linear mass density, denoted by the symbol \( \mu \), is a measure of mass per unit length of a wire. It is measured in kilograms per meter (kg/m). Understanding this concept is crucial because it influences how the wire vibrates. The lower the linear mass density, the easier it is for the wire to vibrate at higher frequencies with the same tension. If you think of two wires with different masses but with the same tension and length, the lighter one (with lower linear mass density) will vibrate faster, producing a higher sound frequency. In practice, to compute the linear mass density, we divide the total mass of the wire by its length. For example, if you know the mass of the wire is \(3.5 \times 10^{-2} \text{ kg}\) and its linear mass density is \(4.0 \times 10^{-2} \text{ kg/m}\), you can easily find the length by rearranging the equation: \[ L = \frac{m}{\mu} \].

Fundamental Frequency Equation

To find the connection between the physical properties of a wire and its vibration, the fundamental frequency equation is used. This is given as:\[ f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \]where:

• \( f \) is the frequency of the vibrating wire.
• \( L \) is the length of the wire.
• \( T \) is the tension in the wire.
• \( \mu \) is the linear mass density.

This equation tells us that the fundamental frequency is directly related to the square root of the tension divided by linear mass density. Therefore, by increasing the tension or decreasing the linear mass density, the frequency will increase.
For practical applications, if you know the values of the other three variables, you can rearrange this formula to solve for any of them, such as tension. For instance, once you compute the wire's length and insert the known values of \( f \) and \( \mu \) into this equation, you can solve for \( T \), determining the tension necessary for the wire to vibrate at a given frequency. This makes the equation a powerful tool for designing musical instruments or engineering systems that require specific vibration frequencies.