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Two particles oscillate in simple harmonic motion along a common straight-line segment of length \(A\). Each particle has a period of \(1.5 \mathrm{~s}\), but they differ in phase by \(\pi / 6 \mathrm{rad}\). (a) How far apart are they (in terms of \(A\) ) \(0.50 \mathrm{~s}\) after the lagging particle leaves one end of the path? (b) Are they then moving in the same direction, toward each other, or away from each other?

Short Answer

Expert verified
(a) Distance is \(\frac{A\sqrt{3}}{2}\). (b) They are moving towards each other.

Step by step solution

01

Determine the Equation of Motion

The position of a particle in simple harmonic motion can be described by the equation \(x(t) = A \cos\left(\omega t + \phi\right)\), where \(\omega\) is the angular frequency and \(\phi\) is the phase. For both particles, the angular frequency \(\omega\) can be found from the period \(T = 1.5\text{ s}\) using \(\omega = \frac{2\pi}{T}\). Substituting the given period, \(\omega = \frac{2\pi}{1.5}\).
02

Calculate the Positions at t = 0.5 s

For the leading particle, the equation is \(x_1(t) = A \cos\left(\omega t\right)\). For the lagging particle, the equation is \(x_2(t) = A \cos\left(\omega t + \frac{\pi}{6}\right)\). Substituting \(t = 0.5\text{ s}\) and \(\omega = \frac{2\pi}{1.5}\) into both equations, calculate the positions: \(x_1(0.5) = A \cos\left(\frac{2\pi}{1.5} \times 0.5\right)\) and \(x_2(0.5) = A \cos\left(\frac{2\pi}{1.5} \times 0.5 + \frac{\pi}{6}\right)\).
03

Find the Distance Between Particles

Subtract the position of the lagging particle from the position of the leading particle: \(d(t) = |x_1(t) - x_2(t)|\), for \(t = 0.5\text{ s}\). Solve the trigonometric expressions to find the distance \(d(0.5)\) in terms of \(A\).
04

Determine Direction of Movement

Consider the derivatives of the position functions, \(v_1(t) = -A \omega \sin\left(\omega t\right)\) and \(v_2(t) = -A \omega \sin\left(\omega t + \frac{\pi}{6}\right)\), which give the velocity. Evaluate these at \(t = 0.5\text{ s}\) to determine their signs. A positive sign indicates movement in one direction (say, to the right), and a negative indicates the opposite direction. Compare the signs to determine if they are moving towards each other or in the same direction.

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Key Concepts

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

Angular Frequency
Angular frequency is a crucial concept in understanding simple harmonic motion. It is denoted by the symbol \( \omega \) and is defined as the rate at which an object oscillates in radians per second. To compute the angular frequency of a particle undergoing simple harmonic motion, you can use the formula \( \omega = \frac{2\pi}{T} \), where \( T \) is the period of the motion.
In our exercise, the period \( T \) is given as 1.5 seconds. By substituting this value into our formula, we find that \( \omega = \frac{2\pi}{1.5} \). This value of \( \omega \) helps us determine how rapidly the particles oscillate along their path. Understanding angular frequency ties together the concepts of time and position and allows us to precisely predict the positions of the particles at any given moment.
Phase Difference
Phase difference refers to the amount by which one oscillation leads or lags behind another. In the context of simple harmonic motion, it describes when two waves, such as those of our particles, start or differ in their oscillation cycles.
For two particles in this scenario, they differ in phase by \( \frac{\pi}{6} \) radians. This phase difference means that while one particle reaches a maximum or minimum point, the other does so later or earlier, depending on whether it is leading or lagging.
  • A phase difference of zero would mean that the particles move in perfect synchrony.
  • Phase differences allow us to analyze how the oscillations compare at a given moment in time.
Understanding the phase difference is essential for calculating the exact positions of particles relative to each other at any time, such as at \( t = 0.5 \) seconds in our exercise.
Equation of Motion
The equation of motion in simple harmonic oscillation allows us to predict the position of an object at any time. In mathematical terms, it is expressed as \( x(t) = A \cos(\omega t + \phi) \), where \( x(t) \) is the position of the particle, \( A \) is the amplitude, \( \omega \) is the angular frequency, and \( \phi \) is the phase angle.
For the two particles in our problem, we set their equations of motion as:
  • Leading particle: \( x_1(t) = A \cos(\omega t) \)
  • Lagging particle: \( x_2(t) = A \cos(\omega t + \frac{\pi}{6}) \)
By substituting \( t = 0.5 \) seconds and the computed \( \omega \) value, we can find the positions of each particle along their path at that specific time. This calculation allows us to understand not only their motion but also their relative position and distance from one another.
Trigonometric Expressions
Trigonometric expressions come into play through the cosine functions in simple harmonic motion. In this scenario, we use the expressions \( \cos(\omega t) \) and \( \cos(\omega t + \frac{\pi}{6}) \) to describe the oscillations of our particles.
These expressions are crucial as they help us determine the specific positions of our particles at any moment. When solving problems like these, it often involves simplifying complex trigonometric equations to find clear, concise answers.
Calculating the distance between the particles at a specific time involves:
  • Substituting the given values of \( t \), \( \omega \), and phase \( \phi \)
  • Evaluating the cosine expressions
  • Subtracting the positions \( |x_1 - x_2| \) for the desired outcome.
Understanding these trigonometric expressions and the method to work with them will give insights into how the particles are moving relative to one another.

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Most popular questions from this chapter

A massless spring with spring constant \(19 \mathrm{~N} / \mathrm{m}\) hangs vertically. A body of mass \(0.20 \mathrm{~kg}\) is attached to its free end and then released. Assume that the spring was unstretched before the body was released. Find (a) how far below the initial position the body descends, and the (b) frequency and (c) amplitude of the resulting SHM.

A grandfather clock has a pendulum that consists of a thin brass disk of radius \(r=15.00 \mathrm{~cm}\) and mass \(1.000\) kg that is attached to a long thin rod of negligible mass. The pendulum swings freely about an axis perpendicular to the rod and through the end of the rod opposite the disk, as shown in Fig. \(15.56 .\) If the pendulum is to have a period of \(2.000 \mathrm{~s}\) for small oscillations at a place where \(g=9.800 \mathrm{~m} / \mathrm{s}^{2}\), what must be the rod length \(L\) to the nearest tenth of a millimeter?

A \(10 \mathrm{~g}\) particle undergoes SHM with an amplitude of \(2.0 \mathrm{~mm}\), a maximum acceleration of magnitude \(8.0 \times 10^{3} \mathrm{~m} / \mathrm{s}^{2}\), and an unknown phase constant \(\phi\). What are (a) the period of the motion, (b) the maximum speed of the particle, and (c) the total mechanical energy of the oscillator? What is the magnitude of the force on the particle when the particle is at (d) its maximum displacement and (e) half its maximum displacement?

A \(2.0 \mathrm{~kg}\) block is attached to the end of a spring with a spring constant of \(350 \mathrm{~N} / \mathrm{m}\) and forced to oscillate by an applied force \(F=\) \((15 \mathrm{~N}) \sin \left(\omega_{d} t\right)\), where \(\omega_{d}=35 \mathrm{rad} / \mathrm{s}\). The damping constant is \(b=\) \(15 \mathrm{~kg} / \mathrm{s}\). At \(t=0\), the block is at rest with the spring at its rest length. (a) Use numerical integration to plot the displacement of the block for the first \(1.0 \mathrm{~s}\). Use the motion near the end of the \(1.0 \mathrm{~s}\) interval to estimate the amplitude, period, and angular frequency. Repeat the calculation for (b) \(\omega_{d}=\sqrt{k / m}\) and (c) \(\omega_{d}=20 \mathrm{rad} / \mathrm{s}\)

A pendulum is formed by pivoting a long thin rod about a point on the rod. In a series of experiments, the period is measured as a function of the distance \(x\) between the pivot point and the rod's center. (a) If the rod's length is \(L=2.20 \mathrm{~m}\) and its mass is \(m=22.1 \mathrm{~g}\), what is the minimum period? (b) If \(x\) is cho- sen to minimize the period and then \(L\) is increased, does the period increase, decrease, or remain the same? (c) If, instead, \(m\) is increased without \(L\) increasing, does the period increase, decrease, or remain the same?

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