A wire of density 9 x 10^-3 kg-cm^-3 is stretched between two clamps 1 m apart. The resulting strain

Q: Why do we use the fundamental frequency when the question asks for the 'lowest frequency'?

The fundamental frequency corresponds to the simplest mode of vibration (one antinode), which has the longest wavelength ($\lambda = 2L$) and therefore the lowest possible frequency for a given wave speed.

Q: How did the cross-sectional area $A$ disappear from the formula?

Both the tension $T$ (which is $Y \cdot A \cdot \epsilon$) and the linear mass density $\mu$ (which is $\rho \cdot A$) are directly proportional to the cross-sectional area. When you take their ratio $\frac{T}{\mu}$, the area $A$ perfectly cancels out!

Q: Why did we multiply the density by $10^6$?

The density was given in $\text{kg/cm}^3$. Since $1 \text{ m} = 100 \text{ cm}$, $1 \text{ m}^3 = (100 \text{ cm})^3 = 10^6 \text{ cm}^3$. Therefore, to convert $\text{kg/cm}^3$ to the standard SI unit $\text{kg/m}^3$, we must multiply by $10^6$.

Q: Can we use this formula for any type of wave?

No, this specific formula $v = \sqrt{\frac{T}{\mu}}$ is strictly for the speed of *transverse* mechanical waves on a stretched string or wire.

Q: What if the wire was vibrating in its second harmonic?

For the second harmonic, the wire would form two loops, making the wavelength equal to $L$. The frequency would be $f_2 = \frac{v}{L}$, which is exactly twice the fundamental frequency.

Question

A wire of density $9 	imes 10^{-3} 	ext{ kg-cm}^{-3}$ is stretched between two clamps 1 m apart. The resulting strain in the wire is $4.9 	imes 10^{-4}$. The lowest frequency of the transverse vibrations in the wire is (Young's modulus of wire, $Y = 9 	imes 10^{10} 	ext{ Nm}^{-2}$), (to the nearest integer) ……… .

Sigma Percentile · Accepted Answer

### Setting the Stage: The Stretched Wire

Imagine a wire tightly stretched between two rigid clamps, exactly $1 	ext{ m}$ apart. This wire is ready to vibrate, much like a guitar string waiting to be plucked.

When we pluck this wire, it vibrates. The lowest frequency it can produce is known as its **fundamental frequency**.

This fundamental frequency corresponds to the simplest standing wave pattern, forming a single loop between the two clamps.

### Bridging Mechanics and Waves

To find this fundamental frequency, we rely on the classic formula for a stretched string:

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

Here, $L$ is the length of the wire, $T$ is the tension, and $\mu$ is the linear mass density.

But there is a catch! The problem doesn't hand us the tension or the linear mass density on a silver platter. Instead, we are given the material's **Young's modulus ($Y$)**, the **strain ($\epsilon$)**, and the **volume density ($ho$)**.

We need to build a bridge between the mechanics of elasticity and the physics of waves.

Let's recall the definition of Young's modulus:

$$Y = \frac{	ext{Stress}}{	ext{Strain}} = \frac{T/A}{\epsilon}$$

From this, we can express the tension as:

$$T = Y \cdot A \cdot \epsilon$$

Similarly, the linear mass density $\mu$ (mass per unit length) can be written in terms of the volume density $ho$ and the cross-sectional area $A$:

$$\mu = \frac{	ext{Mass}}{	ext{Length}} = \frac{ho \cdot A \cdot L}{L} = ho \cdot A$$

### The Master Equation for Wave Speed

Now, let's substitute these expressions back into our wave speed formula, $v = \sqrt{\frac{T}{\mu}}$.

$$v = \sqrt{\frac{Y \cdot A \cdot \epsilon}{ho \cdot A}}$$

Notice something beautiful? The cross-sectional area $A$ perfectly cancels out!

This leaves us with an elegant equation for the wave speed that depends entirely on the intrinsic properties of the material and its state of strain:

$$v = \sqrt{\frac{Y \epsilon}{ho}}$$

### Navigating the Units and Calculation

Before we rush into calculating the numbers, we must be incredibly careful with our units.

The density is given as $ho = 9 	imes 10^{-3} 	ext{ kg/cm}^3$. To use this in our SI formula, we must convert it to $	ext{kg/m}^3$.

Since $1 	ext{ m} = 100 	ext{ cm}$, a cubic meter is $10^6 	ext{ cm}^3$. Therefore, we multiply by $10^6$:

$$ho = 9 	imes 10^{-3} 	imes 10^6 = 9 	imes 10^3 	ext{ kg/m}^3$$

Now, we can safely substitute our values into the wave speed equation:

$$v = \sqrt{\frac{9 	imes 10^{10} 	imes 4.9 	imes 10^{-4}}{9 	imes 10^3}}$$

Let's simplify the expression inside the square root. The $9$s cancel out immediately.

Combining the powers of $10$, we get $10^{10} 	imes 10^{-4} / 10^3 = 10^3$.

This leaves us with:

$$v = \sqrt{4.9 	imes 10^3} = \sqrt{4900}$$

Taking the square root, we find the wave speed:

$$v = 70 	ext{ m/s}$$

### The Final Crescendo: Fundamental Frequency

With the wave speed in hand, we are ready for the final step.

We substitute $v = 70 	ext{ m/s}$ and $L = 1 	ext{ m}$ back into our fundamental frequency formula:

$$f = \frac{v}{2L} = \frac{70}{2 	imes 1}$$

Calculating this gives us:

$$f = 35 	ext{ Hz}$$

And there we have it! The lowest frequency of the transverse vibrations in the wire is **$35 	ext{ Hz}$**.

This problem beautifully demonstrates how the macroscopic wave behavior of a string is deeply rooted in its microscopic material properties.

Enter Numerical Value:

Visualized Solution

Visualizing the Stretched Wire

The Fundamental Frequency Formula

Linking Elasticity to Wave Mechanics

Deriving the Wave Speed $v$

Unit Conversion and Substitution

Computing the Wave Speed $v$

Calculating the Final Frequency $f$

Exploring Higher Harmonics

The Sigma Insight: Standing Waves in Strings and Organ Pipes

Setting the Stage

Bridging Mechanics and Waves

The Master Equation for Wave Speed

Navigating the Units and Calculation

The Final Crescendo

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