The mass per unit length of a uniform wire is 0.135 g/cm. A transverse wave of the form y = -0.21sin

Q: Why does the wave equation have a positive sign inside the sine function? Doesn't that mean it's moving backwards?

Yes, the positive sign in $(kx + \omega t)$ indicates that the wave is traveling in the negative x-direction. However, the direction of propagation does not affect the magnitude of the wave speed $v$, which is what we need to calculate the tension.

Q: How exactly do we convert g/cm to kg/m without getting confused?

Think of it step-by-step. To convert grams to kilograms, you multiply by $10^{-3}$. To convert centimeters to meters in the denominator, you divide by $10^{-2}$. Combining these, $\frac{10^{-3}}{10^{-2}} = 10^{-1}$. So, you simply multiply the given value by $10^{-1}$ (or divide by 10).

Q: Why did we completely ignore the amplitude (-0.21) in our calculations?

The amplitude of a wave tells us how high the string oscillates, but it has absolutely no effect on how fast the wave travels along the string. Wave speed is determined solely by the medium's properties: tension and mass density.

Q: What would happen if I forgot to convert the units of $\mu$?

If you used $\mu = 0.135$ directly with $v = 30$, you would be mixing CGS and SI units. Your resulting tension would be numerically meaningless in Newtons, leading to a completely incorrect integer answer for $x$.

Q: Is the tension $T$ constant everywhere in the wire?

Yes, in this idealized problem, we assume the wire is uniform (constant $\mu$) and horizontal, meaning the tension $T$ is uniform throughout its length. If the wire were hanging vertically, tension would vary due to gravity.

Question

The mass per unit length of a uniform wire is $0.135 	ext{ g/cm}$. A transverse wave of the form $y = -0.21\sin(x + 30t)$ is produced in it, where $x$ is in metre and $t$ is in second. Then, the expected value of tension in the wire is $x 	imes 10^{-2} 	ext{ N}$. Value of $x$ is ……… (Round-off to the nearest integer)

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine a uniform wire stretched tight, acting as a medium for a transverse wave. We are provided with the mathematical equation governing this wave:

$$y = -0.21 \sin(x + 30t)$$

Alongside this, we are given the physical property of the wire, its mass per unit length (linear mass density), $\mu = 0.135 	ext{ g/cm}$. Our ultimate objective is to determine the tension $T$ that is keeping this wire taut.

### Extracting the Wave Parameters

To unlock the physics hidden within the wave equation, we must compare it to the standard form of a traveling wave:

$$y = A \sin(kx + \omega t)$$

By aligning our given equation with this standard template, we can directly extract two critical parameters. The coefficient of $t$ gives us the angular frequency, $\omega = 30 	ext{ rad/s}$. The coefficient of $x$ gives us the wave number, $k = 1 	ext{ m}^{-1}$.

Why are these parameters so important? Because they are the keys to finding the speed at which the wave propagates through the wire. The wave speed $v$ is defined as the ratio of the angular frequency to the wave number:

$$v = \frac{\omega}{k} = \frac{30}{1} = 30 	ext{ m/s}$$

### The Trap of Units

Before we rush into calculating the tension, we must address a common pitfall: **units**. The linear mass density $\mu$ is given in CGS units ($	ext{g/cm}$). If we use this directly with our SI wave speed, our final answer will be disastrously wrong. We must convert $\mu$ into standard SI units ($	ext{kg/m}$).

$$\mu = \frac{0.135 	imes 10^{-3} 	ext{ kg}}{10^{-2} 	ext{ m}} = 0.0135 	ext{ kg/m}$$

### The Master Equation

Now that we have the wave speed $v$ and the linear mass density $\mu$ in compatible units, we can bridge the gap to the tension $T$. The fundamental relationship for the speed of a transverse wave on a stretched string is:

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

To isolate the tension, we simply square both sides of the equation:

$$T = v^2 \mu$$

### Final Calculation

Let's substitute our carefully prepared values into the master equation. We know $v = 30 	ext{ m/s}$, so $v^2 = 900$. Multiplying this by our converted mass density:

$$T = (30)^2 	imes 0.0135 = 900 	imes 0.0135 = 12.15 	ext{ N}$$

The tension in the wire is $12.15 	ext{ N}$. However, the question demands the answer in a very specific format: $x 	imes 10^{-2} 	ext{ N}$. To match this, we rewrite our result:

$$12.15 	ext{ N} = 1215 	imes 10^{-2} 	ext{ N}$$

By direct comparison, we find that the integer value we seek is **$x = 1215$**.

Enter Numerical Value:

Visualized Solution

$Visualizing the Wave$

$Extracting Wave Parameters$

$Calculating Wave Speed$

$Unit Conversion for μ$

$Relating Speed and Tension$

$Computing the Tension$

$Final Formatting$

The Sigma Insight: Wave Equation and Wave Speed

Analyzing the Setup

Extracting the Wave Parameters

The Trap of Units

The Master Equation

Final Calculation

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Enter Numerical Value:

Visualized Solution

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

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

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

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

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

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

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