/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 105 Two waves get superposed on a st... [FREE SOLUTION] | 91Ó°ÊÓ

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Two waves get superposed on a string \(y_{1}=3 \sin 2 \pi(x-10 t) \quad y_{2}=3 \sin 2 \pi(x+10 t)\) where \(x, y_{1}\) and \(y_{2}\) are in meter and \(t\) in sccond. Then the distance between two adjacent nodcs on the string is (a) \(25 \mathrm{~cm}\) (b) \(50 \mathrm{~cm}\) (c) \(75 \mathrm{~cm}\) (d) \(100 \mathrm{~cm}\)

Short Answer

Expert verified
The distance between two adjacent nodes is 50 cm (option b).

Step by step solution

01

Understanding Superposition of Waves

We have two wave equations \( y_1 = 3 \sin 2\pi(x - 10t) \) and \( y_2 = 3 \sin 2\pi(x + 10t) \). These represent two waves traveling in opposite directions. To find the resultant wave, we will use the principle of superposition, which states that the resultant wave is the sum of the individual waves.
02

Applying the Superposition Principle

The superposition principle gives us the resultant wave as \( y = y_1 + y_2 \). Substituting the provided wave equations, we have:\[ y = 3 \sin 2\pi(x - 10t) + 3 \sin 2\pi(x + 10t) \]. Using the identity \( \sin A + \sin B = 2 \sin\left( \frac{A+B}{2} \right)\cos\left( \frac{A-B}{2} \right) \), we can simplify this expression.
03

Simplifying the Resultant Wave

Let \( A = 2\pi(x - 10t) \) and \( B = 2\pi(x + 10t) \). Then the resultant wave becomes:\[ y = 2 \times 3 \times \sin\left( \pi x \right) \cos\left( -20\pi t \right) = 6 \sin(2\pi x) \cos(20\pi t) \]. The spatial part of the wave is \( 6 \sin(2\pi x) \), which describes the nodes and antinodes.
04

Finding Node Positions

Nodes occur where the amplitude is zero, which happens at \( \sin(2\pi x) = 0 \). This condition is satisfied when \( 2\pi x = n\pi \), where \( n \) is an integer. Solving for \( x \), we find \( x = \frac{n}{2} \), where \( n = 0, \pm 1, \pm 2, \ldots \). The distance between two adjacent nodes is the difference between successive \( x \): \( \frac{n+1}{2} - \frac{n}{2} = \frac{1}{2} \).
05

Calculating the Node Separation

Substitute \( \frac{1}{2} \) meters into centimeters to find the distance in more commonly used units. Thus, \( \frac{1}{2} \) meter equals \( 50 \) cm. This is the distance between two adjacent nodes on the string.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Nodes and Antinodes
In a standing wave pattern, created by the superposition of two identical waves traveling in opposite directions, we encounter unique points known as nodes and antinodes. Nodes are the points along the medium that remain stationary—they have zero amplitude. Antinodes, on the other hand, are where the amplitude of the wave is at its maximum.

The distance between these nodes is crucial to understanding a standing wave. In our exercise, the equation for the resultant wave simplifies the spatial part as: \[ 6 \sin(2\pi x) \]To locate nodes, we set this part equal to zero, leading to the condition \( \sin(2\pi x) = 0 \).This satisfies points at integer multiples of half wavelengths: \( x = \frac{n}{2} \)where \( n \) is an integer. The distance between two adjacent nodes, such as from \( \frac{n}{2} \) to \( \frac{n+1}{2} \), is \( \frac{1}{2} \) meters or 50 centimeters.

Antinodes are halfway between nodes, and these exhibit maximum vibrational amplitude—these points are equally important when analyzing wave patterns. Such regular positioning of nodes and antinodes is typical in many physical systems, such as musical instruments.
Wave Interference
When two or more waves meet, they combine in a process known as interference. In our exercise, wave interference is demonstrated through the superposition of two identical waves traveling in opposite directions along a string. This creates a standing wave pattern. Interference can be constructive or destructive, depending on the phase of the waves involved.

- **Constructive Interference**: When the peaks (crest) and troughs of the waves align, they add together to increase amplitude. - **Destructive Interference**: When peaks align with troughs, they cancel each other out, resulting in reduced or zero amplitude.

The standing wave pattern in this problem arises from such interference. In our specific scenario, the constructive and destructive interference creates the stationary nodes and the oscillating antinodes, as a consequence of the repeating interaction of the waves. This helps in illustrating how wave interference can lead to the distinctive patterns observed in various practical applications such as sound resonance and vibration modes.
Trigonometric Identities
Trigonometric identities serve as the mathematical toolset that enables the simplification and analysis of wave functions. In this exercise, we encounter the identity: \( \sin A + \sin B = 2 \sin\left( \frac{A+B}{2} \right)\cos\left( \frac{A-B}{2} \right) \).This is crucial in breaking down the wave equations: \( y_1 = 3 \sin 2\pi(x - 10t) \) and \( y_2 = 3 \sin 2\pi(x + 10t) \).

By applying this identity, the combined waves can be expressed more simply in terms of one sine and one cosine function, leading to a revelation of the standing wave pattern.The identity helps to reveal the spatial and temporal dependence separately, facilitating easy identification of nodes and antinodes in terms of periodic behavior.

Utilizing such trigonometric identities not only simplifies complex expressions but also provides deeper insight into the behavior of waves in interference. This transformation underpins the theoretical understanding of wave phenomena, allowing for quicker assessment of significant patterns and mathematical features of combined waveforms.

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