Speed of a transverse wave on a straight wire (mass 6.0 g, length 60 cm and area of cross-section 1.

Q: Why do we use the original length $l$ for calculating $\mu$ instead of $l + \Delta l$?

The extension $\Delta l$ is extremely small compared to the original length $l$ (as seen in the final answer, $0.03\text{ mm}$ vs $600\text{ mm}$). Therefore, the change in mass per unit length is negligible, and we can safely use the original length.

Q: How did the length $l$ cancel out in the derivation?

Tension $T$ is proportional to $\frac{1}{l}$ ($T = \frac{YA\Delta l}{l}$), and mass per unit length $\mu$ is proportional to $\frac{1}{l}$ ($\mu = \frac{m}{l}$). When we divide $T$ by $\mu$, the $\frac{1}{l}$ terms cancel each other out perfectly!

Q: What are the common mistakes to avoid in this problem?

The most common mistake is forgetting to convert units to the SI system. Mass must be in kg ($6.0 \times 10^{-3}\text{ kg}$) and area in $\text{m}^2$ ($1.0 \times 10^{-6}\text{ m}^2$).

Q: Does the wave speed depend on the length of the wire?

Interestingly, in this specific setup where tension is provided by stretching the wire itself, the wave speed $v = \sqrt{\frac{YA\Delta l}{m}}$ does not explicitly depend on the original length $l$, but rather on the total mass $m$ and the absolute extension $\Delta l$.

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

No, this specific formula $v = \sqrt{\frac{T}{\mu}}$ applies only to transverse waves on a stretched string or wire. Longitudinal waves (like sound in a solid rod) have a different speed formula ($v = \sqrt{\frac{Y}{\rho}}$).

Question

Speed of a transverse wave on a straight wire (mass $6.0	ext{ g}$, length $60	ext{ cm}$ and area of cross-section $1.0	ext{ mm}^2$) is $90	ext{ ms}^{-1}$. If the Young's modulus of wire is $16 	imes 10^{11}	ext{ Nm}^{-2}$, the extension of wire over its natural length is

Options:

(A) $0.01	ext{ mm}$

(B) $0.04	ext{ mm}$

(C) $0.03	ext{ mm}$

(D) $0.02	ext{ mm}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup
Imagine a straight wire stretched tightly between two supports. When you pluck it, a transverse wave travels along its length. The speed of this wave, $v$, is fundamentally determined by two physical properties: how tight the wire is (the tension $T$) and how heavy it is per unit length (the linear mass density $\mu$).

The formula governing this is beautifully simple:
$$v = \sqrt{\frac{T}{\mu}}$$

### The Master Equations
To solve our problem, we need to break down the variables $T$ and $\mu$ into the quantities given in the question. First, the linear mass density $\mu$ is simply the total mass $m$ divided by the original length $l$:
$$\mu = \frac{m}{l}$$

Next, we must find an expression for the tension $T$. We know that applying a tension force causes the wire to stretch by a small amount, $\Delta l$. This brings Young's modulus $Y$ into the picture. Young's modulus is defined as the ratio of tensile stress to tensile strain:
$$Y = \frac{\frac{T}{A}}{\frac{\Delta l}{l}}$$

By rearranging this definition, we can express the tension $T$ in terms of the extension:
$$T = \frac{Y A \Delta l}{l}$$

### The Elegant Cancellation
Now, let's combine these ideas. We start by squaring our wave speed equation to remove the square root:
$$v^2 = \frac{T}{\mu}$$

Substitute the expressions for $T$ and $\mu$ that we just derived:
$$v^2 = \frac{\left(\frac{Y A \Delta l}{l}ight)}{\left(\frac{m}{l}ight)}$$

Notice something magical here? The original length $l$ appears in the denominator of both the numerator and the denominator. It beautifully cancels out! This leaves us with a much cleaner equation:
$$v^2 = \frac{Y A \Delta l}{m}$$

### Final Calculation
Since our goal is to find the extension of the wire, let's rearrange our equation to make $\Delta l$ the subject:
$$\Delta l = \frac{m v^2}{Y A}$$

It is time to plug in the numbers. But beware—this is where many students fall into a trap! We must ensure all units are strictly in the SI system. We convert the mass to kilograms ($m = 6.0 	imes 10^{-3} 	ext{ kg}$) and the area to square meters ($A = 1.0 	imes 10^{-6} 	ext{ m}^2$).

Substituting these values into our rearranged equation:
$$\Delta l = \frac{(6.0 	imes 10^{-3}) (90)^2}{(16 	imes 10^{11}) (1.0 	imes 10^{-6})}$$

Let's calculate the numerator and denominator separately to avoid silly mistakes. The numerator becomes $6.0 	imes 10^{-3} 	imes 8100 = 48.6$. The denominator simplifies to $16 	imes 10^5$. 
$$\Delta l = \frac{48.6}{16 	imes 10^5}$$

Dividing these gives:
$$\Delta l = 3.0375 	imes 10^{-5} 	ext{ m}$$

Finally, converting this result into millimeters (by multiplying by $10^3$), we get:
$$\Delta l \approx 0.03 	ext{ mm}$$

This tiny extension justifies our assumption that the linear mass density $\mu$ remains practically constant during the stretching process.

Animated Solution for Physics - Waves: Speed of a transverse wave on a straight wire (mass $6.0 g$ , length $60 cm$ and area of cross-section $1.0 mm^{2}$ ) is $90 ms^{- 1}$ . If the Young's modulus of wire is $16 \times 1 0^{11} Nm^{- 2}$ , the extension of wire over its natural length is

Select Answer:

Visualized Solution

$v = \frac{T}{μ}$

$Y = \frac{\frac{T}{A}}{\frac{Δ l}{l}}$

$v^{2} = \frac{Y A Δ l}{m}$

$Δ l = \frac{m v ^{2}}{Y A}$

$Δ l = \frac{( 6.0 \times 1 0 ^{- 3} ) ( 90 ) ^{2}}{( 16 \times 1 0 ^{11} ) ( 1.0 \times 1 0 ^{- 6} )}$

$Δ l \approx 0.03 mm$

$Δ l ≪ l$

The Sigma Insight: Wave Equation and Wave Speed

Analyzing the Setup Imagine a straight wire stretched tightly between two supports

The Master Equations To solve our problem, we need to break down the variables $T$ and $μ$ into the quantities given in the question

The Elegant Cancellation Now, let's combine these ideas

Final Calculation

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Visualized Solution

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } v2=mYAΔl​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Δl=YAmv2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Δl=(16×1011)(1.0×10−6)(6.0×10−3)(90)2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Δl≈0.03 mm

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