A steel wire of length 1 m, mass 0.1 kg and uniform cross-sectional area 10^-6 m^2 is rigidly fixed

Q: Why does lowering the temperature create tension in the wire?

When the temperature of a metal wire is lowered, it naturally attempts to contract. Since the ends of the wire are rigidly clamped, they prevent this contraction by pulling outward on the wire. This outward pulling force creates tension inside the wire, known as thermal tension.

Q: What is Young's Modulus and why is it used here?

Young's Modulus ($Y$) is a measure of a material's stiffness under tension or compression. It relates stress to strain ($Stress = Y \times Strain$). We use it here because the prevented thermal contraction creates a mechanical strain, which translates to stress and tension through Young's Modulus.

Q: How does plucking the string in the middle guarantee the fundamental mode?

Plucking a string at its midpoint forces that point to be a displacement antinode (maximum vibration). Since the clamped ends must be nodes, the simplest standing wave pattern that fits these boundary conditions is a single loop, which is the fundamental mode (first harmonic).

Q: What would happen to the frequency if the wire's cross-sectional area was doubled?

If the cross-sectional area $A$ is doubled, the tension $T = YA\alpha\Delta\theta$ doubles. However, the mass of the wire (and thus $\mu$) also doubles because mass is density times volume ($m = \rho A L$). Since both $T$ and $\mu$ double, their ratio $T/\mu$ remains constant, meaning the wave speed and the fundamental frequency would remain unchanged!

Q: Why is the unit of linear mass density kg/m and how does it affect wave speed?

Linear mass density ($\mu$) represents how heavy the string is per unit length. A heavier string has more inertia, making it harder for the tension force to accelerate the particles of the medium. Therefore, a higher linear mass density decreases the wave speed ($v = \sqrt{T/\mu}$).

Question

A steel wire of length $1	ext{ m}$, mass $0.1	ext{ kg}$ and uniform cross-sectional area $10^{-6}	ext{ m}^2$ is rigidly fixed at both ends. The temperature of the wire is lowered by $20^\circ	ext{C}$. If transverse waves are set-up by plucking the string in the middle, calculate the frequency of the fundamental mode of vibration.

Given: $Y_{	ext{steel}} = 2 	imes 10^{11}	ext{ N/m}^2$, $\alpha_{	ext{steel}} = 1.21 	imes 10^{-5}/^\circ	ext{C}$.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine a heavy steel wire stretched tightly between two rigid, unyielding walls.

Initially, at room temperature, the wire is just taut with negligible tension.

Now, we cool the environment, lowering the temperature of the wire by $20^\circ	ext{C}$.

Naturally, metals contract when cooled.

But because the ends of this wire are clamped to rigid supports, it is physically prevented from shrinking.

This restriction creates a massive internal pulling force—**thermal tension**—within the wire.

Our goal is to find the fundamental frequency of transverse waves when this tensioned wire is plucked in the middle.

### The Physics of Thermal Stress and Tension

If the wire were free to contract, the fractional change in its length (thermal strain) would be:

$$	ext{Strain} = \alpha \Delta	heta$$

where $\alpha$ is the coefficient of linear expansion and $\Delta	heta$ is the temperature drop.

Since the rigid walls prevent any change in length, they exert an equal and opposite stress on the wire.

According to Hooke's Law, this thermal stress is:

$$	ext{Stress} = Y 	imes 	ext{Strain} = Y \alpha \Delta	heta$$

where $Y$ is the Young's modulus of steel.

Since stress is force per unit area, the tension $T$ developed in the wire is:

$$T = 	ext{Stress} 	imes A = Y A \alpha \Delta	heta$$

Let's substitute the given values to find this tension:

$$T = (2 	imes 10^{11}	ext{ N/m}^2) 	imes (10^{-6}	ext{ m}^2) 	imes (1.21 	imes 10^{-5}/^\circ	ext{C}) 	imes 20^\circ	ext{C}$$

$$T = 2 	imes 1.21 	imes 20 	imes 10^{11 - 6 - 5}$$

$$T = 48.4	ext{ N}$$

### Wave Speed on the String

The speed $v$ of a transverse wave on a stretched string depends on the tension $T$ and the linear mass density $\mu$ (mass per unit length):

$$v = \sqrt{\frac{T}{\mu}}$$

First, let's calculate the linear mass density $\mu$ of the wire:

$$\mu = \frac{	ext{mass}}{	ext{length}} = \frac{0.1	ext{ kg}}{1	ext{ m}} = 0.1	ext{ kg/m}$$

Now, substituting $T = 48.4	ext{ N}$ and $\mu = 0.1	ext{ kg/m}$ into the velocity equation:

$$v = \sqrt{\frac{48.4}{0.1}} = \sqrt{484} = 22	ext{ m/s}$$

### Finding the Fundamental Frequency

When the string is plucked in the middle, it vibrates in its **fundamental mode** (the first harmonic).

This mode consists of a single loop with nodes at both clamped ends and a single antinode in the middle.

The length $L$ of the wire corresponds to half of a wavelength:

$$L = \frac{\lambda}{2} \implies \lambda = 2L = 2 	imes 1	ext{ m} = 2	ext{ m}$$

Finally, the fundamental frequency $f_0$ is:

$$f_0 = \frac{v}{\lambda} = \frac{22	ext{ m/s}}{2	ext{ m}} = 11	ext{ Hz}$$

Thus, the fundamental frequency of vibration of the wire is **$11	ext{ Hz}$**.

Enter Numerical Value:

Visualized Solution

Visualizing the Setup

Understanding Thermal Stress

Relating Stress to Tension

Substituting the Given Values for Tension

Calculating the Tension Force

Wave Speed on a Stretched String

Calculating Linear Mass Density

Computing the Wave Velocity

Fundamental Mode of Vibration

Calculating the Fundamental Frequency

The Way Forward

The Sigma Insight: Standing Waves in Strings and Organ Pipes

Analyzing the Setup

The Physics of Thermal Stress and Tension

Wave Speed on the String

Finding the Fundamental Frequency

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