/*! 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 21 Consider two objects, A and B, b... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Consider two objects, A and B, both undergoing SHM, but with different frequencies, as described by the equations \(x_{\mathrm{A}}=(2.0 \mathrm{~m}) \sin (2.0 t)\) and \(x_{\mathrm{B}}=(5.0 \mathrm{~m}) \sin (3.0 t),\) where \(t\) is in seconds. After \(t=0,\) find the next three times \(t\) at which both objects simultaneously pass through the origin.

Short Answer

Expert verified
The next three times are \( \frac{3\pi}{2}, 3\pi, \frac{9\pi}{2} \) seconds.

Step by step solution

01

Understanding Simple Harmonic Motion (SHM)

The equation for SHM is given by \( x(t) = A \sin(\omega t + \phi) \), where \( A \) is the amplitude, \( \omega \) is the angular frequency, and \( \phi \) is the phase constant. For the problem, we have two equations: \( x_A = (2.0 \, \text{m}) \sin(2.0t) \) and \( x_B = (5.0 \, \text{m}) \sin(3.0t) \). These equations describe the motion of objects A and B, respectively. They pass through the origin when \( x(t) = 0 \), which occurs when the sine function equals zero.
02

Setting the Criteria for Passing Through Origin

An object passes through the origin when \( \sin(\omega t) = 0 \). This occurs at \( \omega t = n\pi \), where \( n \) is any integer. For object A, this becomes \( 2.0t = n\pi \), and for object B, \( 3.0t = m\pi \). Here, \( n \) and \( m \) are integers.
03

Equating and Solving for Time 't'

To find when both objects pass through the origin simultaneously, set the equations equal to each other: \( 2.0t = n\pi \) and \( 3.0t = m\pi \). Equating them gives us \( \frac{n}{2} = \frac{m}{3} \). Solving for common values, we set \( n = 3k \) and \( m = 2k \) where \( k \) is an integer. Thus, \( t = \frac{3k\pi}{2} \) or \( t = \frac{2k\pi}{3} \).
04

Finding the First Three Times Both Pass Through the Origin

Substitute \( k = 1, 2, 3 \) into either \( t = \frac{3k\pi}{2} \) or \( t = \frac{2k\pi}{3} \). We shall use \( t = \frac{3k\pi}{2} \):- For \( k = 1, \, t = \frac{3\pi}{2} \)- For \( k = 2, \, t = 3\pi \)- For \( k = 3, \, t = \frac{9\pi}{2} \).
05

Summarizing the Result

The first three positive times \( t \) when both objects A and B pass through the origin simultaneously are \( \frac{3\pi}{2} \), \( 3\pi \), and \( \frac{9\pi}{2} \). These values indicate the instances when both sine waves are at zero crossings at the same time.

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
In Simple Harmonic Motion (SHM), the amplitude is a crucial concept. It represents the maximum displacement from the equilibrium position. In simpler words, it's how far the object swings from its central position during SHM.
For objects A and B, the amplitudes are given in the motion equations as 2.0 meters and 5.0 meters respectively. This means that object A moves up to 2.0 meters away from its center point, while object B moves up to 5.0 meters.
Having a larger amplitude means the object travels a greater distance from its central position. However, amplitude doesn't affect the period of the motion. It simply shows how 'big' or 'loud' the motion is.
Angular Frequency
Angular frequency, denoted by \( \omega \), tells us how fast the object undergoes its periodic motion. It is measured in radians per second and involves rotations or oscillations within a particular time frame.
For the SHM equations provided, object A has an angular frequency of 2.0, and object B has 3.0. The higher the angular frequency, the faster the object completes a cycle.
Angular frequency is directly related to the period of the motion, which is the time taken to complete one full cycle. It is calculated using the formula \( \omega = \frac{2\pi}{T} \), where \( T \) is the period. This means a higher angular frequency results in a shorter period, signifying quicker oscillations.
Sine Function
The sine function in SHM describes how the displacement varies with time. It's an essential mathematical function in trigonometry, often dealing with waveforms and oscillations.
In our exercise, both objects A and B have their motion described by the sine function, involving time \( t \). The sine function gives a cyclic nature to the motion, meaning the objects move back and forth in a regular repeating pattern.
The function \( \sin(\omega t) \) equals zero whenever \( \omega t = n\pi \) where \( n \) is an integer. It represents the moments when both objects pass through the equilibrium point. The periodic nature of the sine function makes it ideal for describing SHM.
Phase Constant
The phase constant, denoted by \( \phi \), is an important aspect in SHM. It determines where the object starts along its cycle. In simple terms, it affects how 'shifted' or 'offset' the wave is from the usual position.
Although the phase constant is not explicitly shown in this particular exercise, it plays a significant role when comparing SHM graphs of different starting points.
If two SHM objects start their cycles at different positions, the phase constant will help denote these initial differences. It shifts the entire sine wave along the time axis. Hence, understanding the phase constant is crucial when multiple oscillatory motions overlap or synchronize.

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

(II) A physical pendulum consists of an 85 -cm-long, 240 -g-mass, uniform wooden rod hung from a nail near one end (Fig, 38 ). The motion is damped because of friction in the pivot; the damping force is approximately proportional to \(d \theta / d t .\) The rod is set in oscillation by displacing it \(15^{\circ}\) from its equilibrium position and releasing it. After 8.0 \(\mathrm{s}\) the amplitude of the oscillation has been reduced to \(5.5^{\circ} .\) If the angular displacement can be written as \(\theta=A e^{-\gamma t} \cos \omega^{\prime} t,\) find \((a) \gamma,(b)\) the approximate period of the motion, and \((c)\) how long it takes for the amplitude to be reduced to \(\frac{1}{2}\) of its original value.

(II) Estimate the stiffness of the spring in a child's pogo stick if the child has a mass of 35 \(\mathrm{kg}\) and bounces once every 2.0 seconds.

A vertical spring of spring constant \(115 \mathrm{~N} / \mathrm{m}\) supports a mass of \(75 \mathrm{~g}\). The mass oscillates in a tube of liquid. If the mass is initially given an amplitude of \(5.0 \mathrm{~cm},\) the mass is observed to have an amplitude of \(2.0 \mathrm{~cm}\) after \(3.5 \mathrm{~s}\). Estimate the damping constant \(b\). Neglect buoyant forces.

The human leg can be compared to a physical pendulum, with a "natural" swinging period at which walking is easiest. Consider the leg as two rods joined rigidly together at the knee; the axis for the leg is the hip joint. The length of each rod is about the same, \(55 \mathrm{~cm} .\) The upper rod has a mass of \(7.0 \mathrm{~kg}\) and the lower rod has a mass of \(4.0 \mathrm{~kg} .\) ( \(a\) ) Calculate the natural swinging period of the system. ( \(b\) ) Check your answer by standing on a chair and measuring the time for one or more complete back-andforth swings. The effect of a shorter leg is a shorter swinging period, enabling a faster "natural" stride.

An \(1150 \mathrm{~kg}\) automobile has springs with \(k=16,000 \mathrm{~N} / \mathrm{m}\). One of the tires is not properly balanced; it has a little extra mass on one side compared to the other, causing the car to shake at certain speeds. If the tire radius is \(42 \mathrm{~cm},\) at what speed will the wheel shake most?

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.