/*! 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 79 Two waves represented by \(y=a \... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Two waves represented by \(y=a \sin (\omega t-k x)\) and \(y=a \cos (\omega t-k x)\) are superposed. The resultant wave will have an amplitude (a) \(a\) (b) \(\sqrt{2 a}\) (c) \(2 a\) (d) zero

Short Answer

Expert verified
The resultant amplitude is \(\sqrt{2} a\), which is option (b).

Step by step solution

01

Identify the Formulas

We have two wave functions: \(y_1 = a \sin (\omega t - k x)\) and \(y_2 = a \cos (\omega t - k x)\). Our goal is to find the amplitude of the resultant wave when these waves are superposed.
02

Write the Expression for Superposition

When the waves are superposed, the resultant wave is given by \(y = y_1 + y_2 = a \sin (\omega t - k x) + a \cos (\omega t - k x)\).
03

Use Trigonometric Identity

We can use the identity for the sum of sine and cosine: \(y = a [\sin (\omega t - k x) + \cos (\omega t - k x)].\) Applying the identity: \(\sin A + \cos A = \sqrt{2} \sin(A + \pi/4)\).
04

Calculate the Resultant Amplitude

Thus, \(y = a \cdot \sqrt{2} \sin((\omega t - k x) + \pi / 4)\). The amplitude of the resultant wave is \(a \sqrt{2}\).
05

Select the Correct Option

Comparing with the given options, the amplitude of the resultant wave \(a \sqrt{2}\) corresponds to option (b) \(\sqrt{2} a\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Wave Interference
Wave interference occurs when two or more waves overlap in a medium, causing them to combine and produce a new wave pattern. In simple terms, it's like when two musicians play the same note in sync, creating a richer sound. When dealing with two waves, such as the ones given by:
  • \( y_1 = a \sin(\omega t - k x) \)
  • \( y_2 = a \cos(\omega t - k x) \)
we need to add them algebraically to see how they interact.
In constructive interference, the waves are in phase, meaning their peaks and troughs align, resulting in a larger wave. Conversely, in destructive interference, they are out of phase, leading to a smaller wave or even cancellation. Here, the waves are sine and cosine, which relate but don't perfectly overlap. Understanding these interactions helps us find their resultant amplitude and how the wave behaves.
Superposition allows us to see the mix of their individual characteristics. Sprinkling in some math, the resultant wave is simply their addition, showcasing the beautiful math-based harmony of physics.
Trigonometric Identities
To solve problems involving superposition of waves, trigonometric identities become extremely helpful. These are rules that relate trigonometric functions like sine and cosine in ways that make complex equations more manageable.
In our problem, the trigonometric identity used is:
  • \( \sin A + \cos A = \sqrt{2} \sin(A + \frac{\pi}{4}) \)
This identity transforms the separate sine and cosine terms into a single sine term with an adjusted argument.
By applying this identity to the problem, we simplify the expression:
  • \( y = a[\sin(\omega t - k x) + \cos(\omega t - k x)] \)
becomes
  • \( y = a \sqrt{2} \sin((\omega t - k x) + \frac{\pi}{4}) \)
The identity not only provides a straightforward path to solving equations but also highlights the close relationship between different trigonometric functions.
Resultant Amplitude
The resultant amplitude is the final height of the wave formed after interference occurs. It's determined by how the individual waves combine, influenced by both their amplitudes and phase differences.
In this instance, after using the trigonometric identity, we found that the waveform is
  • \( y = a \sqrt{2} \sin((\omega t - k x) + \frac{\pi}{4}) \)
from waves \( y_1 \) and \( y_2 \).
This reveals that the resultant amplitude is \( a \sqrt{2} \), which reflects that the two original amplitudes, both \( a \), have combined to give a larger, single amplitude.
Because \( \sin A \) and \( \cos A \) have the same magnitude and are oriented such that their phases differ exactly by \( \frac{\pi}{2} \), they constructively interfere at 45 degrees. Hence, the maximum size of this combined wave is stretched by a factor of \( \sqrt{2} \).
This highlights how waves with similar frequencies, but differing in phase, can enhance each other when calculated appropriately.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A tuning fork of frequency \(480 \mathrm{~Hz}\) produces 10 beats \({ }^{-1}\) when sounded with a vibrating sonometer string. What must have been the frequency of string if slight increase in tension produces fewer beats \(\mathrm{s}^{-1}\) than before? (a) \(490 \mathrm{~Hz}\) (b) \(470 \mathrm{~Hz}\) (c) \(460 \mathrm{~Hz}\) (d) \(480 \mathrm{~Hz}\)

A closed organ pipe and an open organ pipe of same length produce 2 beats \(\mathrm{s}^{-1}\) when they are set into vibrations together in fundamental mode. The length of open pipe is now halved and that of closed pipe is doubled. The number of beats produced will be (a) 7 (b) 4 (c) 8 (d) 2

A source of sound emitting a tone of frequency \(200 \mathrm{~Hz}\) moves towards an observer with a velocity \(v\) equal to the velocity of sound. If the observer also moves away from the source with the same velocity \(v\), the apparent frequency heard by the observer is (a) \(50 \mathrm{~Hz}\) (b) \(100 \mathrm{~Hz}\) (c) \(150 \mathrm{~Hz}\) (d) \(200 \mathrm{~Hz}\)

An open pipe is suddenly closed at one end with the result that the frequency of third harmonic of the closed pipe is found to be higher at \(100 \mathrm{~Hz}\). The fundamental frequency of the open pipe is (a) \(200 \mathrm{~Hz}\) (b) \(480 \mathrm{~Hz}\) (c) \(240 \mathrm{~Hz}\) (d) \(300 \mathrm{~Hz}\)

Two uniform strings \(A\) and \(B\) made of steel are made to vibrate under the same tension. If the first overtone of \(A\) is equal to the second overtone of \(B\) and if the radius of \(A\) is twice that of \(B\), the ratio of the lengths of the strings is (a) \(2: 1\) (b) \(3: 4\) (c) \(3: 2\) (d) \(1: 3\)

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.