/*! 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 84 If two orthogonal SHM of same fr... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

If two orthogonal SHM of same frequency having initial phase difference of \(\pi / 2\) act simultaneously on particle free to move. The particle will move in (a) Straight line (b) Parabola (c) Cirele (d) Ellipse

Short Answer

Expert verified
The particle moves in a circle.

Step by step solution

01

Understanding Orthogonal SHM

When two simple harmonic motions (SHM) are orthogonal, it means their directions are perpendicular to each other. If they have the same frequency, their motions can be represented as x-motion and y-motion. The frequencies matching ensure synchronization by time, and an initial phase difference of \( \frac{\pi}{2} \) indicates that the two SHMs are out of phase by 90 degrees.
02

Equation Setup for SHM

Let's represent the two SHM as follows: \( x(t) = A \sin(\omega t) \) and \( y(t) = A \sin(\omega t + \frac{\pi}{2}) \). By using the trigonometric identity, \( \sin(\omega t + \frac{\pi}{2}) = \cos(\omega t) \), the y-motion can be rewritten as \( y(t) = A \cos(\omega t) \). Hence, we have \( x(t) = A \sin(\omega t) \) and \( y(t) = A \cos(\omega t) \).
03

Forming the Relationship

Considering the parametric equations for \( x(t) \) and \( y(t) \): \( x(t) = A \sin(\omega t) \) and \( y(t) = A \cos(\omega t) \). These can be rearranged to represent a trigonometric identity of a circle: \( x^2(t) + y^2(t) = A^2 \sin^2(\omega t) + A^2 \cos^2(\omega t) = A^2 \). This equation \( x^2 + y^2 = A^2 \) is the equation of a circle.
04

Recognizing the Shape

Since the equation \( x^2 + y^2 = A^2 \) describes a circle with radius \( A \) centered at the origin (0, 0), analysis of these SHM components confirms that the particle traces out a circular path as it moves.

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.

Orthogonal SHM
When discussing simple harmonic motion, or SHM, the term "orthogonal" may seem a bit complex. Essentially, when two SHMs are orthogonal, it means they occur in perpendicular directions. Imagine two different motions; one moving along the x-axis and the other along the y-axis. In such cases, the x-motion and y-motion are at right angles to each other.
To further understand orthogonal SHM, consider how these motions occur at the same frequency. This frequency alignment ensures that both motions are in sync as time progresses. Additionally, an initial phase difference of \( \frac{\pi}{2} \) or 90 degrees means the two motions do not start at the same point. Instead, one lags behind the other by a quarter cycle. This makes one motion reach its maximum while the other is at zero, maintaining their orthogonality effectively.
  • Directions perpendicular to each other (x and y).
  • Same frequency ensures synchronized timing.
  • \( \frac{\pi}{2} \) phase difference represents a 90-degree phase lag.
Phase Difference
Phase difference is crucial in explaining how two oscillating systems interact over time. It is the difference in the phase angle between two oscillating quantities. In the context of orthogonal SHM, the phase difference is \( \frac{\pi}{2} \), or 90 degrees. This specific value results in a quarter cycle phase shift between the two motions.
The significance of this phase difference is that it causes one SHM to be at its maximum positive or negative amplitude when the other SHM is passing through its equilibrium position. This leads to interesting and predictable patterns in the resultant path of a particle.
  • Describes how much one wave lags or leads another.
  • \( \frac{\pi}{2} \) phase difference results in right angles between sine and cosine components.
  • Ensures maximum and minimum alignments for orthogonal motions.
Equation of a Circle
In the context of SHM, understanding the equation of a circle helps determine the path traced by a particle subjected to orthogonal SHM. Consider the two motions given by the equations: \( x(t) = A \sin(\omega t) \) and \( y(t) = A \cos(\omega t) \).
By applying trigonometric identities, we find that these two equations can be rewritten as a single expression that combines both x and y motions. By squaring and adding both equations, we derive \( x^2(t) + y^2(t) = A^2 \).
This equation, \( x^2 + y^2 = A^2 \), represents a circle with a radius equal to \( A \), centered at the origin. Hence, when a particle moves under the influence of orthogonal SHM with appropriate phase difference, it traces a perfect circle.
  • Combines x and y SHM into a single geometric entity.
  • \( x^2 + y^2 = A^2 \) is the standard equation of a circle.
  • Circle's center is at the origin (0,0) with radius \( A \).

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 tuning forks \(\Lambda\) and \(B\) when sounded togcther produce 4 beats \(/ \mathrm{s}\). When \(B\) is loaded with wax, the beat frequency remains same. If frequency of \(A\) is 212 Ilz then frequency of \(B\) before loading is (a) \(208 \mathrm{IIz}\) (b) \(212 \mathrm{IIz}\) (c) \(216 \mathrm{IIz}\) (d) Irrelevant

Two corks are \(10 \mathrm{~m}\) aparl in a lake. Each goes up and down with period \(5 \mathrm{~s}\). It is observed that when one is at its highest point, oucr one is at lowest point. The possible speed of wave is (a) \(2.5 \mathrm{~m} / \mathrm{s}\) (b) \(5 \mathrm{~m} / \mathrm{s}\) (c) \(40 \mathrm{~m} / \mathrm{s}\) (d) \(4 \mathrm{~m} / \mathrm{s}\)

A plane sound wave is travelling in a medium. In reference to a frame \(A\), its equation is: \(y=a \cos (\omega l-k x)\) In reference to a frame \(B\), moving with a constant velocity \(v\) in the direction of propagation of the wave, equation of the wave cannot be (a) \(y=a \cos [(\omega+k v) l-k x \mid\) (b) \(y=-a \cos [(\omega-k v) t-k x]\) (c) \(y=a \cos [(\omega-k v) l-k x \mid\) (d) \(y=a \cos |(\omega+k v) t+k x|\)

Standing wave \(\xi=a \sin k x \cos \omega t\) is maintained in a homogeneous rod with cross-sectional area \(S\) and density \(\rho\). Find the total mechanical cnergy confined between the sections corresponding to the adjacent displacement nodes.

Statement-1: When a sound wave propagates through air, the displacements take place in the direction of propagation. Statement-2 : Sound waves can be polarized as in the case of light rays.

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.