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Chapter 10: Problem 114

\(\Lambda\) wave represented by \(y=2 \cos (4 x-\pi t)\) is supcrposed with another wave to form a stationary wave such that the point \(x=0\) is a node. The equation of other wave is (a) \(2 \sin (4 x+\pi t)\) (b) \(-2 \cos (4 x-\pi t)\) (c) \(-2 \cos (4 x+\pi t)\) (d) \(-2 \sin (4 x-\pi t)\)

Short Answer

Expert verified
The equation of the other wave is \(-2 \cos (4 x + \pi t)\) (option c).

Step by step solution

01

Understand Stationary Waves

A stationary wave forms when two waves of the same amplitude and frequency, but traveling in opposite directions, superpose. At a node, the total displacement is always zero.
02

Analyze Given Wave

The given wave is represented by \(y_1 = 2 \cos (4x - \pi t)\). This wave has amplitude 2, wavenumber 4, and angular frequency \(\pi\). It travels in the positive x-direction.
03

Understand Conditions for a Node

For \(x = 0\) to be a node, the sum of the given wave \(y_1\) and the other wave \(y_2\) at \(x=0\) should always be zero for any value of \(t\).
04

Consider Node Formation at x=0

At \(x=0\), the given wave becomes \(y_1 = 2\cos(-\pi t) = 2\cos(\pi t)\), as cosine is an even function. For this to form a node with another wave, their sum at any time \(t\) must be zero.
05

Identify Potential Wave from Options

The wave \(-2\cos(4x + \pi t)\) will, at \(x=0\), be \(-2\cos(\pi t)\). Adding this to the given wave \(2\cos(\pi t)\), the result is zero, which is necessary for a node. Thus, the correct other wave is option (c).
06

Verify the Correct Wave Equation

Since combining these waves at \(x=0\) results in a zero amplitude for all times \(t\), \(x=0\) is indeed a node, confirming that the other wave is \(-2 \cos (4 x + \pi t)\).

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Key Concepts

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

Node Formation
Nodes are fundamental to the understanding of stationary waves. They are points along a wave where the displacement is consistently zero throughout time. This interesting feature arises from the constructive and destructive interference between two overlapping waves.
When two waves travel in opposite directions with the same frequency and equal amplitude, they perfectly align to create nodes. At these points, the peaks of one wave align with the troughs of the other wave, cancelling each other out completely.
  • If you see a location with constant zero displacement over time, it typically signifies a node is present there.
  • One of the key requirements for node formation is that the waves must have the same frequency.
  • The presence of nodes is a characteristic sign of stationary waves, which are a result of maintained interference patterns.
Understanding where nodes form helps in predicting how and where a stationary wave will exist, as seen in solving for specific conditions like those at point x=0 in a problem.
Wave Superposition
Wave superposition is a principle that explains how waves overlap and interact with each other. When two or more waves traverse the same space, their amplitudes add together.
This superposition can lead to either constructive interference, where waves reinforce each other, or destructive interference, where they cancel each other out.
  • In stationary waves, superposition causes certain points (nodes) to have zero amplitude due to perfectly destructive interference.
  • The sum of two waves with equal and opposite phases will lead to a net displacement of zero, illustrating the superposition principle.
  • The overall pattern of superposition in stationary waves leads to regions alternating between nodes and antinodes (points of maximum amplitude).
The way these waves add up as they superpose entirely shapes the characteristics of the stationary wave, making the understanding of superposition fundamental in physics.
Angular Frequency
Angular frequency is an important concept in wave motion. It鈥檚 expressed in radians per second and provides information about how fast a wave oscillates.
For a sinusoidal wave like those involved in stationary wave formation, the angular frequency is part of how quickly the wave oscillates over time.
  • This value is typically denoted by \( \omega \) and linked to the period and frequency of the wave through the formula: \( \omega = 2\pi f \), where \( f \) is the frequency.
  • In the wave equation, you鈥檒l find angular frequency as part of the term inside the cosine or sine functions, determining how rapidly the wave peaks and troughs alternate.
  • Knowing the angular frequency helps in predicting the wave鈥檚 behavior over time and is crucial in calculating other phenomena like resonance.
In the exercise context, understanding the angular frequency helps to solve problems involving wave superposition and node formation, as it defines the periodicity of the interfering waves.

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Most popular questions from this chapter

When source and detector are stationary and wind blows at speed \(V_{w}=10 \mathrm{~m} / \mathrm{s}\), speed of sound is \(v=330 \mathrm{~m} / \mathrm{s}\), find apparent wavelength of sound in dircction of wind if wavelength of sound is \(33 \mathrm{~m}\). (a) \(33 \mathrm{~m}\) (b) \(1 \mathrm{~m}\) (c) \(34 \mathrm{~m}\) (d) \(\frac{1089}{32} \mathrm{~m}\)

The energy per unit area associated with a progressive sound wave will be doubles if then (a) the amplitude of the wave is doubled (b) the amplitude of the wave is increused by \(50 \%\) (c) the amplitude of the wave is increased by \(43 \%\) (d) the frequency of the wave is increased by \(43 \%\)

A string of length \(L\) is stretched along the \(x\)-axis and is rigidly clamped at its two ends. It undergoes transverse vibrations. If \(n\) is an integer, which of the followings relations may represent the shape of the string at any time \(t ?\) (a) \(y-\Lambda \sin \left(\frac{n \pi x}{L}\right) \cos \omega\) (b) \(y-\Lambda \sin \left(\frac{n \pi x}{L}\right) \sin \alpha\) (c) \(y-A \cos \left(\frac{n \pi x}{L}\right) \cos \omega\) (d) \(y-A \cos \left(\frac{n \pi x}{L}\right) \sin \omega\)

Statement- \(1: \Lambda\) particle moves on \(x\)-axis according to equation \(x=2+6 \sin (\pi t)\). Motion is nol SIIM. Statement-2 : The acceleration is not proportional to displacement from origin.

$$ \begin{aligned} &\text { (a) } \rightarrow(\mathbf{p}),(\mathbf{r}) ; \text { (b) } \rightarrow(\mathbf{q}),(\mathbf{s}) \\ &\text { (c) } \rightarrow \text { (q), }(\mathrm{s}) ; \text { (d) } \rightarrow \text { (p), (r) } \end{aligned} $$ For closed organ pipe $$ L=\left(n+\frac{1}{2}\right) \frac{\lambda}{2} $$ for 2 nd overtone \(n=2\) \(\therefore \quad L=\frac{5}{2} \times \frac{\lambda}{2} \quad \Rightarrow \quad \lambda=\frac{4 L}{5}\) now no, of loop in \(\xi \sim x\) or \(p \sim x\) graph is $$ \left(2+\frac{1}{2}\right)=2 \frac{1}{2} $$ (a) Displacement nodes are at closed end, at \(\frac{\lambda}{2}=\frac{4 L}{2 \times 5}=\frac{2 L}{5}\) from closed cnd and al \(\lambda=\frac{4 L}{5}\) from closed end. (b) Displacement antinode are al open cnd, at \(\frac{\lambda}{2}=\frac{2 L}{5}\) from open end and at \(\boldsymbol{\lambda}=\frac{4 L}{5}\) from open cnd i.e., \(L-\frac{4 L}{5}=\frac{L}{5}\) ?rom closcd end, (c) Pressure nodes are same as displacement antinodes(d) Pressure antinodes are same as displacement nodes
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