/*! 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 112 Two waves of frequencies \(20 \m... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Two waves of frequencies \(20 \mathrm{~Hz}\) and \(30 \mathrm{~Hz}\) travel out from a common point. How will they differ in phase at the end of \(0.75\) second? (a) \(15 \pi\) (b) \(\pi\) (c) \(7 \pi\) (d) \(2 \pi\)

Short Answer

Expert verified
The waves will differ in phase by \(15 \pi\).

Step by step solution

01

Identify the Problem

We need to find the phase difference between two waves with different frequencies (\(20 \mathrm{~Hz}\) and \(30 \mathrm{~Hz}\)) after \(0.75\) seconds.
02

Determine the Formula for Phase Difference

The phase difference between two waves is given by \(\Delta \phi = 2 \pi \Delta f \cdot t\), where \(\Delta f\) is the frequency difference and \(t\) is the time duration.
03

Calculate Frequency Difference

The frequency difference \(\Delta f\) between the two waves is \(\Delta f = 30 \mathrm{~Hz} - 20 \mathrm{~Hz} = 10 \mathrm{~Hz}\).
04

Compute Phase Difference

Using the formula \(\Delta \phi = 2 \pi \cdot 10 \times 0.75\), compute the phase difference. This simplfies to \(\Delta \phi = 2 \pi \cdot 7.5 = 15 \pi\).
05

Choose the Correct Option

Compare the computed phase difference \(15 \pi\) with the options provided: \((a) 15 \pi\), \((b) \pi\), \((c) 7 \pi\), \((d) 2 \pi\). The correct answer is \(15 \pi\).

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.

Understanding Frequency Difference
Frequency difference is an important concept in wave physics, especially when analyzing the interaction between two or more waves. When two waves have different frequencies, they oscillate at different rates. This difference in oscillation speed is what we refer to as the frequency difference. For instance, in our original exercise, two waves have frequencies of 20 Hz and 30 Hz. The frequency difference is simply calculated as the subtraction of these two frequencies, resulting in a frequency difference of 10 Hz. It is a straightforward calculation but crucial for determining how waves interact over time.
Understanding frequency difference helps in predicting phenomena such as beat frequency, where two waves with slightly different frequencies interfere to produce a new pattern. In various applications, such as sound engineering or signal processing, acknowledging frequency differences allows engineers to design systems that correctly interpret or utilize these wave characteristics.
Exploring Wave Phase
The phase of a wave refers to a specific point in the wave's cycle, which is usually measured in radians. When discussing phase difference, we are looking at the offset between the phases of two waves at a particular point in time.
In this context, the wave phase difference is influenced by both the frequency difference and the time elapsed. The formula to calculate phase difference, \[ \Delta \phi = 2 \pi \Delta f \cdot t \]is used to determine how out of sync two waves are after a given period. Here, \( \Delta f \) represents the frequency difference, and \( t \) is the time over which the waves are observed.
For the exercise given, the calculated phase difference at the end of 0.75 seconds was \( 15 \pi \). It tells us how much one wave leads or lags behind the other in its oscillation cycle. This concept is vital in many fields, such as telecommunications and acoustics, where controlling and predicting wave interactions is needed.
NEET Physics Problems and Their Importance
The NEET exam (National Eligibility cum Entrance Test) includes a variety of physics problems, like the one discussed, which tests a student's understanding of wave properties. These problems often require knowledge of basic physics principles, problem-solving skills, and the ability to apply formulas accurately.
Preparation for NEET physics problems generally involves:
  • Understanding fundamental concepts such as waves, phase, and frequency.
  • Practicing problem-solving techniques to quickly and accurately solve questions under timed conditions.
  • Learning to choose the correct formula and apply it to the given data.
These exams are designed to ensure students have a solid foundational understanding of physics necessary for future studies in medicine and other sciences. By tackling NEET-relevant problems, students enhance their analytical skills and gain confidence in handling complex physics questions, paving the way for success in academic and professional pursuits.

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

Two identical particles each of mass \(m\) are interconnected by a light spring of stiffness \(k\), the time period for small oscillation is equal to (a) \(2 \pi \sqrt{\frac{m}{k}}\) (b) \(\pi \sqrt{\frac{m}{k}}\) (c) \(2 \pi \sqrt{\frac{m}{2 k}}\) (d) \(\pi \sqrt{\frac{2 m}{k}}\)

A wave is represented by the equation \(y=7 \sin \left(7 \pi t-0.04 x+\frac{\pi}{3}\right)\) where, \(x\) is in metres and \(t\) in seconds. The speed of the waves is (a) \((175 \pi) \mathrm{m} / \mathrm{s}\) (b) \((49 \pi) \mathrm{m} / \mathrm{s}\) (c) \((49 / \pi) \mathrm{m} / \mathrm{s}\) (d) \((0.28 \pi) \mathrm{m} / \mathrm{s}\)

Two SHM's are respectively represented by \(y=a \sin\) \((\omega t-k x)\) and \(y=b \cos (\omega t-k x)\). The phase difference between the two is (a) \(\pi / 2\) (b) \(\pi / 4\) (c) \(\pi / 6\) (d) \(3 \pi / 4\)

A simple pendulum of length \(L\) has an energy \(E\) and amplitude \(A .\) The energies of the simple pendulum (i) when the length is doubled but with same amplitude and (ii) when the amplitude is doubled but with the same length, are respectively (a) \(2 E, 2 E\) (b) \(\frac{E}{2}, \frac{E}{2}\) (c) \(\frac{E}{2}, 2 E\) (d) \(\frac{E}{2}, 4 E\)

The amplitude of a wave disturbance propagating in the positive \(Y\) -direction is given by \(y=\frac{1}{1+x^{2}}\) at \(t=0\) and \(y=\frac{1}{\left[1+(x-1)^{2}\right]}\) at \(t=2 \mathrm{~s}\) where, \(x\) and \(y\) are in \(\mathrm{m}\). If the shape of the wave disturbance does not change during the propagation, what is the velocity of the wave? (a) \(1 \mathrm{~m} / \mathrm{s}\) (b) \(1.5 \mathrm{~m} / \mathrm{s}\) (c) \(0.5 \mathrm{~m} / \mathrm{s}\) (d) \(2 \mathrm{~m} / \mathrm{s}\)

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.