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Chapter 10: Problem 1456

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

Expert verified
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

01

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
02

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
03

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: \[ f = \frac{1}{2(1~\text{m})} \sqrt{\frac{98.1~\text{N}}{0.01~\text{kg/m}}} \]
04

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
05

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) \(50 \mathrm{~Hz}\)

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