A copper wire is held at the two ends by rigid supports. At 30°C, the wire is just taut, with neglig

Q: Why does the cross-sectional area of the wire not affect the wave speed?

While a thicker wire experiences more tension ($T \propto A$), it also has a proportionally larger mass per unit length ($\mu \propto A$). Since wave speed depends on the ratio of tension to mass per unit length ($v = \sqrt{T/\mu}$), the area $A$ cancels out completely, making the speed independent of the wire's thickness.

Q: What would happen if the supports were not completely rigid?

If the supports were flexible, they would yield slightly as the wire contracted. This would reduce the thermal strain and consequently lower the tension developed in the wire, resulting in a slower wave speed than the calculated $70.1\text{ m/s}$.

Q: Why is the temperature difference $\Delta \theta$ positive when the temperature decreased?

We are calculating the magnitude of the tension force, which must be positive. The temperature drop causes contraction, which creates tension. Thus, we use the absolute temperature change $\Delta \theta = 30^\circ\text{C} - 10^\circ\text{C} = 20^\circ\text{C}$.

Q: Does the length of the wire affect the wave speed?

No, the length of the wire does not affect the wave speed. Both thermal strain ($\Delta L / L$) and mass per unit length ($\mu = M/L$) are independent of the total length of the wire when clamped rigidly at both ends.

Q: Is Hooke's law always valid for thermal stress calculations?

Hooke's law is valid as long as the thermal stress developed does not exceed the elastic limit of the material. For copper, a temperature drop of $20^\circ\text{C}$ produces stresses well within its elastic range.

Question

A copper wire is held at the two ends by rigid supports. At $30^\circ	ext{C}$, the wire is just taut, with negligible tension. Find the speed of transverse waves in this wire at $10^\circ	ext{C}$.

Given:
Young's modulus of copper, $Y = 1.3 	imes 10^{11}	ext{ N/m}^2$
Coefficient of linear expansion of copper, $\alpha = 1.7 	imes 10^{-5}	ext{ }^\circ	ext{C}^{-1}$
Density of copper, $ho = 9 	imes 10^3	ext{ kg/m}^3$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine a **copper wire** tightly clamped between two rigid, unyielding supports.

At $30^\circ	ext{C}$, the wire is perfectly straight and just taut, meaning there is absolutely no tension pulling on the walls.

But what happens when the temperature drops?

As the environment cools down to $10^\circ	ext{C}$, the copper wire naturally wants to contract and shrink.

However, because the rigid supports hold its ends firmly in place, it is prevented from shortening.

This restriction creates a powerful pulling force, generating **thermal tension** within the wire.

### The Master Equation

To quantify this effect, we must first look at the **thermal strain**.

When a material undergoes a temperature change, its fractional change in length—known as thermal strain—is directly proportional to the temperature difference:

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

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

According to Hooke's Law, within the elastic limit, stress is directly proportional to strain:

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

Since stress is defined as force divided by area, we can easily find the tension, which is the pulling force:

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

Now, let us bring in the physics of wave propagation.

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

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

We can express the mass per unit length $\mu$ in terms of density $ho$ and cross-sectional area $A$:

$$\mu = ho A$$

Substituting $T$ and $\mu$ back into the wave speed formula:

$$v = \sqrt{\frac{Y A \alpha \Delta 	heta}{ho A}}$$

Notice how the cross-sectional area $A$ cancels out completely!

$$v = \sqrt{\frac{Y \alpha \Delta 	heta}{ho}}$$

This is a beautiful result because it means the wave speed is completely independent of the wire's thickness!

### Final Calculation

With our elegant formula ready, we can now substitute the given physical values:

$$v = \sqrt{\frac{1.3 	imes 10^{11} 	imes 1.7 	imes 10^{-5} 	imes 20}{9 	imes 10^3}}$$

Let's calculate the numerator first:

$$	ext{Numerator} = 1.3 	imes 1.7 	imes 20 	imes 10^6 = 44.2 	imes 10^6$$

Now, divide by the density in the denominator:

$$\frac{44.2 	imes 10^6}{9 	imes 10^3} \approx 4.911 	imes 10^3 = 4911.11$$

Finally, taking the square root:

$$v = \sqrt{4911.11} \approx 70.1	ext{ m/s}$$

Thus, the speed of the transverse wave in the wire at $10^\circ	ext{C}$ is **$70.1	ext{ m/s}$**.

Enter Numerical Value:

Visualized Solution

Visualizing the Physical Setup

Understanding Thermal Strain

Relating Strain to Thermal Stress

Calculating Tension in the Wire

The Wave Speed Formula

Expressing Mass per Unit Length

Simplifying the Wave Speed Equation

Substituting the Given Values

Calculating the Numerator

Dividing by Density

Taking the Square Root

The Way Forward & Engineering Insights

The Sigma Insight: Wave Equation and Wave Speed

Analyzing the Setup

The Master Equation

Final Calculation

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