A sonometer wire of length 1.5 m is made of steel. The tension in it produces an elastic strain of 1

Q: Why do we use density and area instead of mass?

The problem doesn't provide the total mass or the radius of the wire. By expressing mass per unit length as $\mu = A \cdot d$, we can link it to the stress equation where area $A$ will eventually cancel out.

Q: How does strain relate to tension in this problem?

Strain is the fractional change in length ($\Delta l / l$). Young's modulus relates this strain to the applied stress (Tension / Area). So, $Y = \frac{T/A}{\Delta l/l}$.

Q: Why did the area $A$ disappear from the final equation?

Area appears in the numerator (from the stress term $T/A$) and in the denominator (from the mass per unit length $\mu = A \cdot d$). They perfectly cancel each other out, meaning the frequency is independent of the wire's thickness for a given strain!

Q: What does a strain of 1% mean mathematically?

A strain of 1% means the wire has stretched by 1 part in 100. In equations, this is written as the decimal 0.01 for the term $\Delta l / l$.

Q: Can we use this method for any vibrating string?

Yes, as long as the string is perfectly elastic and obeys Hooke's Law. For very large strains, the material might yield, and Young's modulus would no longer be constant.

Question

A sonometer wire of length $1.5 	ext{ m}$ is made of steel. The tension in it produces an elastic strain of $1\%$. What is the fundamental frequency of steel, if density and elasticity of steel are $7.7 	imes 10^3 	ext{ kg/m}^3$ and $2.2 	imes 10^{11} 	ext{ N/m}^2$, respectively?

Options:

(A) $188.5 	ext{ Hz}$

(B) $178.2 	ext{ Hz}$

(C) $200.5 	ext{ Hz}$

(D) $770 	ext{ Hz}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine a steel wire suspended from a rigid support, holding a weight at its bottom.

This weight creates tension, causing the wire to stretch slightly.

When plucked, the wire vibrates, creating a standing wave.

The fundamental frequency of this standing wave is governed by the formula:
$f = \frac{1}{2l} \sqrt{\frac{T}{\mu}}$

Here, $T$ is the tension and $\mu$ is the mass per unit length.

### The Material Properties

We aren't given the mass directly, but we do have the density of steel.

We can express the mass per unit length, $\mu$, as the cross-sectional area $A$ multiplied by the density $d$:
$\mu = A \cdot d$

Next, how do we find the tension?

The suspended weight causes an elastic strain in the wire.

Using **Young's modulus**, which is defined as stress over strain, we can relate the tension and area to the given strain:
$Y = \frac{T/A}{\Delta l/l}$

Rearranging this gives us an expression for the stress:
$\frac{T}{A} = Y \frac{\Delta l}{l}$

### The Master Equation

Now, let's substitute these relationships back into our frequency formula.

$f = \frac{1}{2l} \sqrt{\frac{T}{A \cdot d}} = \frac{1}{2l} \sqrt{\frac{Y \frac{\Delta l}{l}}{d}}$

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

This means the frequency is independent of the wire's thickness for a given strain.

### Final Calculation

It's time to plug in the given numbers.

The length is $1.5 	ext{ m}$, Young's modulus is $2.2 	imes 10^{11} 	ext{ N/m}^2$, the strain is $1\%$ or $0.01$, and the density is $7.7 	imes 10^3 	ext{ kg/m}^3$.

$f = \frac{1}{2(1.5)} \sqrt{\frac{2.2 	imes 10^{11} 	imes 0.01}{7.7 	imes 10^3}}$

Let's simplify the terms inside the square root.

The powers of $10$ reduce nicely, and $2.2$ over $7.7$ simplifies to $2$ over $7$.

$f = \frac{1}{3} \sqrt{\frac{2.2 	imes 10^9}{7.7 	imes 10^3}} = \frac{1}{3} \sqrt{\frac{2}{7} 	imes 10^6}$

Calculating the final value, we get:
$f = \frac{1000}{3} \sqrt{\frac{2}{7}} \approx 178.2 	ext{ Hz}$

This is the fundamental frequency of our steel wire.

Select Answer:

Visualized Solution

$l = 1.5 m$

$f = \frac{1}{2 l} \frac{T}{μ}$

$μ = A \cdot d$

$Y = \frac{T / A}{Δ l / l}$

$f = \frac{1}{2 l} \frac{Y Δ l / l}{d}$

$Substitute Values$

$Simplify$

$f \approx 178.2 Hz$

$Food for Thought$

The Sigma Insight: Standing Waves in Strings and Organ Pipes

Analyzing the Setup

The Material Properties

The Master Equation

Final Calculation

Similar Questions

A wire of length $L$ and mass per unit length $6.0 \times 1 0^{- 3} kgm^{- 1}$ is put under tension of $540 N$ . Two consecutive frequencies that it resonates at are : $420 Hz$ and $490 Hz$ . Then, $L$ in metres is

A granite rod of 60 cm length is clamped at its middle point and is set into longitudinal vibrations. The density of granite is $2.7 \times 1 0^{3} kg/m^{3}$ and its Young's modulus is $9.27 \times 1 0^{10} Pa$ . What will be the fundamental frequency of the longitudinal vibrations?

A string of length $1 m$ and mass $2 \times 1 0^{- 5} kg$ is under tension $T$ . when the string vibrates, two successive harmonics are found to occur at frequencies $750 Hz$ and $1000 Hz$ . The value of tension $T$ is _______ Newton.

A one metre long (both ends open) organ pipe is kept in a gas that has double the density of air at STP. Assuming the speed of sound in air at STP is $300 m/s$ , the frequency difference between the fundamental and second harmonic of this pipe is ...... Hz. [2020, 8 Jan Shift-I]

Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } l=1.5 m

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f=2l1​μT​​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } μ=A⋅d

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Y=Δl/lT/A​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f=2l1​dYΔl/l​​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substitute Values

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Simplify

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f≈178.2 Hz

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Food for Thought