/*! 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 56 Three simple harmonic motions of... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Three simple harmonic motions of equal amplitudes \(A\) and equal time periods in the same direction combine. The phase of the second motion is \(60^{\circ}\) ahead of the first and the phase of the third motion is \(60^{\circ}\) ahead of the second. Find the amplitude of the resultant motion.

Short Answer

Expert verified
The amplitude of the resultant motion is \(2A\).

Step by step solution

01

Express the Harmonic Motions

To solve the problem, begin by expressing each of the three simple harmonic motions in terms of their respective equations. Each motion can be represented as:- First motion: \( x_1 = A\cos(\omega t) \) - Second motion: \( x_2 = A\cos(\omega t + 60^{\circ}) \)- Third motion: \( x_3 = A\cos(\omega t + 120^{\circ}) \).Here, \( A \) is the amplitude, and \( \omega \) is the angular frequency.
02

Use Trigonometric Identities

Knowing the equations of each motion, use trigonometric identities to simplify them:1. \( x_2 = A[\cos(\omega t)\cos(60^{\circ}) - \sin(\omega t)\sin(60^{\circ})] \).2. \( x_3 = A[\cos(\omega t)\cos(120^{\circ}) - \sin(\omega t)\sin(120^{\circ})] \).Recall: \( \cos(60^{\circ}) = \frac{1}{2} \), \( \sin(60^{\circ}) = \frac{\sqrt{3}}{2} \), \( \cos(120^{\circ}) = -\frac{1}{2} \), and \( \sin(120^{\circ}) = \frac{\sqrt{3}}{2} \).
03

Combine the Harmonic Motions

Combine the three motions to find the resultant motion, \( x_r \):\[\begin{align*}x_r &= x_1 + x_2 + x_3 \&= A\cos(\omega t) + A\left(\frac{1}{2}\cos(\omega t) - \frac{\sqrt{3}}{2}\sin(\omega t)\right) + A\left(-\frac{1}{2}\cos(\omega t) - \frac{\sqrt{3}}{2}\sin(\omega t)\right) \&= A\left(1 + \frac{1}{2} - \frac{1}{2}\right)\cos(\omega t) + A\left(-\frac{\sqrt{3}}{2} - \frac{\sqrt{3}}{2}\right)\sin(\omega t) \&= A\cos(\omega t) - A\sqrt{3}\sin(\omega t).\end{align*}\]
04

Calculate Amplitude of the Resultant Motion

The resultant equation represents another simple harmonic motion:\[ x_r = A[\cos(\omega t) - \sqrt{3}\sin(\omega t)] \].To find the resultant amplitude, recall that the formula for the amplitude of a resultant motion is given by:\[ R = \sqrt{(\text{Coefficient of } \cos(\omega t))^2 + (\text{Coefficient of } \sin(\omega t))^2} \].Here, \[ R = \sqrt{(A)^2 + (-A\sqrt{3})^2} \].
05

Simplify the Expression

Simplify the amplitude expression:\[R = \sqrt{A^2 + (-A\sqrt{3})^2} = \sqrt{A^2 + 3A^2} = \sqrt{4A^2} = 2A.\]The amplitude of the resultant motion is \(2A\).

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.

Amplitude
Amplitude is a fundamental concept in simple harmonic motion (SHM) that represents the maximum displacement from the equilibrium position. Think of it as how far the object moves away from its center point. When multiple harmonic motions interact, as in the given problem, their amplitudes contribute to the overall magnitude of the resultant motion.
In this case, we have three equal amplitude harmonic motions, denoted by \( A \). Each has the same time period, indicating they oscillate at the same rate but get displaced by different initial phase angles. We aim to find the amplitude of their combined effect, the resultant motion.
To combine these motions, each with amplitude \( A \), we employ trigonometric identities to determine how they reinforce or cancel each other out. The process shows the real power of the amplitude concept in describing and predicting motion patterns easily.
Phase Difference
Understanding phase difference is key in analyzing waves and SHM. It indicates the relative timing of two periodic functions, often measured in degrees or radians. In our problem, the phase difference is given explicitly: the second motion is \( 60^{\circ} \) ahead of the first, and the third is another \( 60^{\circ} \) ahead of the second.
When combining waves or motions, phase differences determine how these displacements add up at any point in time. They influence whether the waves constructively add to produce a larger amplitude, or destructively interfere, possibly reducing the overall amplitude.
  • If the phase difference is \( 0^{\circ} \), waves are in phase and add constructively.
  • A \( 180^{\circ} \) phase difference causes destructive interference, cancelling the amplitudes.
  • Any other angle, like \( 60^{\circ} \), leads to partial interference where magnitudes vary based on the specific configuration and timing.
Recognizing these patterns helps predict the behavior of the resultant wave as reflected in the problem solution.
Trigonometric Identities
Trigonometric identities are crucial tools in simplifying the equations of harmonic motions. They help express complex wave interactions in more manageable terms, often revealing underlying patterns and simplifying calculations.
In our solution, trigonometric identities are employed to handle phase differences easily. For example, the second motion in the problem is expressed using the identity:
  • \( \cos(a + b) = \cos(a)\cos(b) - \sin(a)\sin(b) \)
This results in simplifications such as:
  • \( x_2 = A[\cos(\omega t)\cos(60^{\circ}) - \sin(\omega t)\sin(60^{\circ})] \)
  • \( x_3 = A[\cos(\omega t)\cos(120^{\circ}) - \sin(\omega t)\sin(120^{\circ})] \)
These transformations allow us to understand how each phase-shifted motion contributes to the resultant motion. Using identities correctly aligns our calculations and reveals the clear path to the final amplitude solution, showing that the resultant amplitude is indeed \( 2A \).

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 small block of mass \(m\) is kept on a bigger block of mass \(M\) which is attached to a vertical spring of spring constant \(k\) as shown in the figure. The system oscillates vertically. (a) Find the resultant force on the smaller block when it is displaced through a distance \(x\) above its equilibrium position. (b) Find the normal force on the smaller block at this position. When is this force smallest in magnitude ? (c) What can be the maximum amplitude with which the two blocks may oscillate together?

The position, velocity and acceleration of a particle executing simple harmonic motion are found to have magnitudes \(2 \mathrm{~cm}, 1 \mathrm{~m} \mathrm{~s}^{-1}\) and \(10 \mathrm{~m} \mathrm{~s}^{-2}\) at a certain instant. Find the amplitude and the time period of the motion.

A simple pendulum is constructed by hanging a heavy ball by a \(5-0 \mathrm{~m}\) long string. It undergoes small oscillations. (a) How many oscillations does it make per second? (b) What will be the frequency if the system is taken on the moon where acceleration due to gravitation of the moon is \(1.67 \mathrm{~m} \mathrm{~s}^{-2}\) ?

A block suspended from a vertical spring is in equilibrium. Show that the extension of the spring equals the length of an equivalent simple pendulum, i.e., a pendulum having frequency same as that of the block.

A simple pendulum of length \(l\) is suspended from the ceiling of a car moving with a speed \(v\) on a circular horizontal road of radius \(r\). (a) Find the tension in the string when it is at rest with respect to the car. (b) Find the time period of small oscillation.

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.