A metal wire of linear mass density of 9.8 g/m is stretched with a tension of 10 kg-wt between two r

Q: Why does the wire vibrate in the first place?

When an alternating current flows through the wire, it interacts with the permanent magnetic field. According to Lorentz force law ($F = IL \times B$), the wire experiences a force. Since the current is alternating, the direction of the force alternates, causing the wire to vibrate up and down.

Q: Why do we assume it vibrates in the fundamental mode?

The magnet is placed exactly at the middle of the wire. This means the alternating force is applied at the center. The fundamental mode has its antinode (point of maximum displacement) at the center, making it the most easily excited mode by a force applied there.

Q: What is 'kg-wt' and how do I convert it?

'kg-wt' stands for kilogram-weight, which is the gravitational force exerted on a mass of 1 kg. To convert it to the standard SI unit of force (Newtons), you multiply by the acceleration due to gravity, $g$ (typically $9.8 \text{ m/s}^2$). So, $10 \text{ kg-wt} = 10 \times 9.8 \text{ N} = 98 \text{ N}$.

Q: Why did we multiply the linear mass density by $10^{-3}$?

The linear mass density was given as $9.8 \text{ g/m}$. In physics, to get the final answer in standard SI units (like Hertz for frequency), all inputs must be in SI units. The SI unit for mass is kilograms, so we convert grams to kilograms by multiplying by $10^{-3}$, giving $9.8 \times 10^{-3} \text{ kg/m}$.

Q: What would happen if the AC frequency didn't match the wire's natural frequency?

The wire would still vibrate, but with a very small amplitude. This is called a forced vibration. When the frequencies match, resonance occurs, and the amplitude of vibration increases dramatically.

Question

A metal wire of linear mass density of $9.8 	ext{ g/m}$ is stretched with a tension of $10 	ext{ kg-wt}$ between two rigid supports $1 	ext{ m}$ apart. The wire passes at its middle point between the poles of a permanent magnet and it vibrates in resonance when carrying an alternating current of frequency $n$. The frequency $n$ of the alternating source is

Options:

(A) $50 	ext{ Hz}$

(B) $100 	ext{ Hz}$

(C) $200 	ext{ Hz}$

(D) $25 	ext{ Hz}$

Sigma Percentile · Accepted Answer

### The Symphony of Electromagnetism and Mechanics

Imagine a stretched wire passing between the poles of a permanent magnet. At first glance, it is just a static mechanical setup. But the moment you pass an alternating current through that wire, magic happens. The wire springs to life, vibrating up and down, blurring into a beautiful standing wave.

This problem beautifully marries two seemingly distinct branches of physics: electromagnetism and mechanical waves. To solve it, we must understand the driving force behind the motion and the conditions required for the wire to sing at its loudest—a phenomenon known as resonance.

### The Driving Force: Lorentz Force in Action

Why does the wire move in the first place? When a current $I$ flows through a wire of length $L$ in a magnetic field $B$, it experiences a magnetic force given by the Lorentz force law: $F = I L B \sin	heta$.

Because the current is alternating, its direction reverses periodically. Consequently, the direction of the magnetic force also reverses periodically. This creates a rhythmic push and pull on the wire, acting exactly like a parent pushing a child on a swing. This is a classic example of a **forced vibration**.

### The Phenomenon of Resonance

If you push a swing at random intervals, it won't go very high. But if you push it exactly at its natural rhythm, the amplitude of the swing grows dramatically.

This is **resonance**. The problem states that the wire "vibrates in resonance." This is a massive clue! It tells us that the frequency of the alternating current (the driving frequency) exactly matches the natural frequency of the wire.

### The Master Equation: Natural Frequency of a String

For a string stretched between two rigid supports, the natural fundamental frequency $f$ is given by the master equation:

$$f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$$

Here, $L$ is the length of the wire, $T$ is the tension, and $\mu$ is the linear mass density (mass per unit length). The term $\sqrt{\frac{T}{\mu}}$ represents the speed of the transverse wave traveling along the wire.

### The Trap of Units: Proceed with Caution

This is the phase where many students lose marks. The physics is simple, but the units are a trap. We must convert everything into standard SI units before plugging them into our equation.

First, look at the tension: $T = 10 	ext{ kg-wt}$. 
This is not a mass; it is a force! "Kilogram-weight" is the gravitational force exerted on a $1 	ext{ kg}$ mass. To convert it to Newtons, we multiply by the acceleration due to gravity ($g = 9.8 	ext{ m/s}^2$):

$$T = 10 	imes 9.8 = 98 	ext{ N}$$

Next, look at the linear mass density: $\mu = 9.8 	ext{ g/m}$. 
We must convert grams to kilograms to maintain SI consistency:

$$\mu = 9.8 	imes 10^{-3} 	ext{ kg/m}$$

### The Final Calculation: Bringing It Home

Now that our units are pristine, let's calculate the wave speed $v$ inside the square root:

$$v = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{98}{9.8 	imes 10^{-3}}}$$

Simplifying the fraction, $98 / 9.8 = 10$. Dividing by $10^{-3}$ is equivalent to multiplying by $1000$.

$$v = \sqrt{10 	imes 10^3} = \sqrt{10000} = 100 	ext{ m/s}$$

Finally, we substitute this wave speed back into our fundamental frequency equation. The length $L$ is given as $1 	ext{ m}$:

$$f = \frac{v}{2L} = \frac{100}{2 	imes 1} = 50 	ext{ Hz}$$

For resonance to occur, the frequency of the alternating source must be exactly **$50 	ext{ Hz}$**.

### Beyond the Problem: The Secret of the Magnet's Placement

You might be wondering: *Why did we assume the wire vibrates in its fundamental mode? Could it not vibrate in the second or third harmonic?*

The secret lies in a subtle detail in the question: "The wire passes at its **middle point** between the poles of a permanent magnet."

This means the alternating magnetic force is applied exactly at the center of the wire. The fundamental mode of a vibrating string has its antinode (the point of maximum displacement) right at the center. By applying the force at the center, we are perfectly feeding energy into the fundamental mode. If the magnet had been placed at $L/4$, the force would have been much more effective at exciting the second harmonic instead!

Select Answer:

Visualized Solution

The Physical Setup

Condition for Resonance

Converting Units and Substitution

Calculating Wave Speed $v$

Final Resonant Frequency

Why Fundamental Mode?

The Sigma Insight: Standing Waves in Strings and Organ Pipes

The Symphony of Electromagnetism and Mechanics

The Driving Force

The Phenomenon of Resonance

The Master Equation

The Trap of Units

The Final Calculation

Beyond the Problem

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