Question

A metal wire having linear mass density \(10 \mathrm{~g} / \mathrm{m}\) is passed over two supports separated by a distance of \(1 \mathrm{~m}\). The wire is kept in tension by suspending a \(10 \mathrm{~kg}\) mass. The mid point of the wire passes through a magnetic field provided by magnets and an a. c. supply having frequency \(\mathrm{n}\) is passed through the wire. If the wire starts vibrating with its resonant frequency, what is the frequency of a. c. supply? (A) \(50 \mathrm{~Hz}\) (B) \(100 \mathrm{~Hz}\) (C) \(200 \mathrm{~Hz}\) (D) \(25 \mathrm{~Hz}\)

Short answer

The frequency of the AC supply that causes the wire to vibrate with its resonant frequency is 50 Hz. Therefore, the correct answer is (A) \(50 \mathrm{~Hz}\).

Step-by-step solution

Calculate the tension in the wire

In order to find the frequency of the AC supply, we need to calculate the tension in the wire. The tension is created by the weight of the 10 kg mass that is suspended from the wire. Using the formula for tension: Tension = Mass × acceleration due to gravity T = 10 kg × 9.81 m/s² = 98.1 N

Calculate the linear mass density of the wire

The linear mass density of the wire (μ) is given as 10 g/m. We need to convert it to kg/m, as our units for mass must be consistent. μ = 10 g/m × (1 kg/1000 g) = 0.01 kg/m

Calculate the fundamental resonant frequency

Now we will calculate the fundamental resonant frequency (f) of the wire by using the formula: \[ f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \] Where L is the length of the wire between the supports, which is given as 1 m in the problem. Plugging in the values we found in step 1 and step 2: \[ f = \frac{1}{2(1~\text{m})} \sqrt{\frac{98.1~\text{N}}{0.01~\text{kg/m}}} \]

Solve for the frequency

Simplify and solve the equation to find the fundamental resonant frequency: \[ f = \frac{1}{2} \sqrt{\frac{98.1}{0.01}} \] \[ f = \frac{1}{2} \sqrt{9810} \] f ≈ 50 Hz

Choose the correct answer

The frequency of the AC supply that causes the wire to vibrate with its resonant frequency is 50 Hz. Therefore, the correct answer is: (A) \(50 \mathrm{~Hz}\)

Key concepts

Linear Mass Density

Linear mass density is a measure of mass per unit length of an object. When we talk about a wire, this concept helps us understand how much the wire weighs along its length. It's an important factor in calculating how a wire will behave when under tension, as it directly influences the wave speed along the wire.
For instance, in the exercise, the wire has a linear mass density of 10 g/m. For calculations involving equations in physics, we often need this in kilograms per meter, so we convert it by dividing by 1000. As a result, the linear mass density of the wire is 0.01 kg/m.
This conversion allows us to use uniform units for further calculations such as wave speed and resonant frequency.

Tension in Wire

Tension is the force exerted along the length of a wire, usually because of weights hanging from it or forces pulling it from each side. In the exercise, a 10 kg mass is hanging from the wire, creating tension.
To calculate the tension, we use the formula: \[ T = ext{mass} \times ext{acceleration due to gravity} \]The acceleration due to gravity is generally about 9.81 m/s² on the surface of the Earth. So, for a 10 kg mass:

• Tension = 10 kg × 9.81 m/s² = 98.1 N.

Understanding tension is crucial because it affects the wave speed and the resonant frequency of the wire.

Fundamental Frequency

The fundamental frequency is the lowest frequency at which a system naturally vibrates. For a wire or string under tension, this frequency is determined by its length, tension, and linear mass density.
The equation to calculate the fundamental frequency of a wire is:\[ f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \]where \(L\) is the length of the wire, \(T\) is the tension, and \(\mu\) is the linear mass density.
This equation shows that the frequency depends on the square root of tension divided by linear mass density, indicating that both have a significant impact on how a string vibrates. A longer string or higher mass density results in a lower fundamental frequency, while a higher tension increases it.

Wave Equation on String

The wave equation on a string describes how waves travel along it. In simple terms, it ties together velocity, tension, and linear mass density. Wave speed on a string can be calculated from:\[ v = \sqrt{\frac{T}{\mu}} \]where \(v\) is the wave speed, \(T\) is the tension in the wire, and \(\mu\) is the linear mass density.
This equation tells us that wave speed is directly affected by the tension and inversely affected by linear mass density. A wire with more tension and lower mass density allows waves to travel faster.
Understanding how these factors interplay is crucial for solving problems about vibrations and resonance in systems like musical instruments or engineering applications.