Equation of travelling wave on a stretched string of linear density 5 g/m is y = 0.03 sin(450t - 9x)

Q: Why do we use the formula $v = \omega/k$ to find the speed?

The speed of a wave is the distance it travels per unit time. A full wave cycle covers a distance of one wavelength $\lambda$ in one time period $T$. So, $v = \lambda/T$. Since $\omega = 2\pi/T$ and $k = 2\pi/\lambda$, substituting these gives $v = (2\pi/k) / (2\pi/\omega) = \omega/k$.

Q: Why did we have to convert the linear mass density from g/m to kg/m?

In physics, to get a result in standard SI units (like Newtons for force), all input variables must also be in standard SI units. The standard unit for mass is the kilogram (kg), not the gram (g). If you leave it in grams, your answer will be off by a factor of 1000.

Q: What does the negative sign in $(450t - 9x)$ indicate?

The negative sign indicates the direction of wave propagation. An argument of the form $(\omega t - kx)$ or $(kx - \omega t)$ means the wave is travelling in the positive x-direction. If it were $(\omega t + kx)$, the wave would be travelling in the negative x-direction.

Q: Does the amplitude $0.03$ affect the wave speed or the tension?

No, it does not. For ideal strings, the wave speed is determined entirely by the medium's properties (tension and mass density), not by the wave's amplitude or frequency. The amplitude only tells us how high the string is displaced from equilibrium.

Q: What if the given equation was in terms of cosine instead of sine?

It wouldn't change our process at all! Both sine and cosine represent valid harmonic travelling waves, just shifted in phase by $90^\circ$. The coefficients of $t$ and $x$ would still represent $\omega$ and $k$, respectively.

Question

Equation of travelling wave on a stretched string of linear density $5	ext{ g/m}$ is $y = 0.03 \sin(450t - 9x)$, where distance and time are measured in SI units. The tension in the string is

Options:

(A) $5	ext{ N}$

(B) $12.5	ext{ N}$

(C) $7.5	ext{ N}$

(D) $10	ext{ N}$

Sigma Percentile · Accepted Answer

Imagine you are holding one end of a long, taut string while a friend holds the other. If you flick your wrist, a wave pulse travels down the string. The speed at which this pulse travels isn't random; it is strictly governed by the physical properties of the string—specifically, how tightly it is pulled (the tension) and how heavy it is (the linear mass density).

In this problem, we are given the mathematical blueprint of such a wave: $y = 0.03 \sin(450t - 9x)$. Our mission is to reverse-engineer this equation to uncover the hidden physical force—the tension—that allows this wave to exist.

### Decoding the Wave Equation

Every travelling wave can be described by a standard mathematical template:

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

Here, $A$ is the amplitude (how high the wave gets), $\omega$ is the angular frequency (how fast the wave oscillates in time), and $k$ is the wave number (how densely the wave is packed in space).

By placing our given equation side-by-side with the standard template, we can directly extract the vital statistics of our wave. We see that the angular frequency $\omega = 450 	ext{ rad/s}$ and the wave number $k = 9 	ext{ m}^{-1}$.

### The Speed of Propagation

How fast is this wave zipping along the string? The wave speed $v$ is beautifully simple to find once you have $\omega$ and $k$. It is just the ratio of the two:

$$v = \frac{\omega}{k}$$

Substituting our extracted values, we get:

$$v = \frac{450}{9} = 50 	ext{ m/s}$$

So, our wave is travelling at a brisk $50 	ext{ m/s}$.

### The Physics of the Stretched String

Now, we bridge the gap between the abstract math of the wave equation and the tangible physics of the string. For any transverse wave on a stretched string, the speed $v$ is dictated by Newton's laws, resulting in the classic formula:

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

where $T$ is the tension in the string and $\mu$ is the linear mass density (mass per unit length).

We want to find the tension $T$, so let's isolate it by squaring both sides:

$$v^2 = \frac{T}{\mu} \implies T = \mu v^2$$

### The Final Calculation

Before we plug in the numbers, we must navigate a classic physics trap: **units**. The linear mass density is given as $\mu = 5 	ext{ g/m}$. To ensure our final tension is in standard SI units (Newtons), we must convert grams to kilograms.

$$\mu = 5 	imes 10^{-3} 	ext{ kg/m}$$

Now, we substitute our values into the rearranged tension formula:

$$T = (5 	imes 10^{-3} 	ext{ kg/m}) 	imes (50 	ext{ m/s})^2$$

$$T = 5 	imes 10^{-3} 	imes 2500$$

$$T = 12.5 	ext{ N}$$

The tension required to sustain this specific wave on this specific string is exactly $12.5 	ext{ N}$. The math perfectly mirrors the physical reality!

Animated Solution for Physics - Waves: Equation of travelling wave on a stretched string of linear density $5 g/m$ is $y = 0.03 sin (450 t - 9 x)$ , where distance and time are measured in SI units. The tension in the string is

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Visualized Solution

$G i v e n D a t a$

$S t an d a r d W a v e E q u a t i o n$

$W a v e S p ee d$

$S p ee d o n a S t r e t c h e d S t r in g$

$C a l c u l a t in g T e n s i o n$

$F ina l A n s w er$

The Sigma Insight: Wave Equation and Wave Speed

Decoding the Wave Equation

The Speed of Propagation

The Physics of the Stretched String

The Final Calculation

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

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Speed on a Stretched String

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

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