/*! 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 93 The damped motion of a vibrating... [FREE SOLUTION] | 91影视

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Chapter 11: Problem 93

The damped motion of a vibrating particle is defined by the position vector \(\mathbf{r}=x_{1}[1-1 /(t+1)] \mathbf{i}+\left(y_{1} e^{-\mathrm{n} t / 2} \cos 2 \pi t\right) \mathbf{j}\), where \(t\) is expressed in seconds. For \(x_{1}=30 \mathrm{mm}\) and \(y_{1}=20 \mathrm{mm},\) determine the position, the velocity, and the acceleration of the particle when \((a) t=0,(b) t=1.5 \mathrm{s}\)

Short Answer

Expert verified
At \(t = 0\), position is \(0 \mathbf{i} + 20 \mathbf{j}\) mm, velocity \(30 \mathbf{i} - 10n \mathbf{j}\) mm/s, acceleration \(-60 \mathbf{i} + 20( \frac{n^2}{4} + 4\pi^2 ) \mathbf{j}\) mm/s虏. At \(t = 1.5\) s, position is \(18.75 \mathbf{i} - 20 e^{-\frac{3n}{2}} \mathbf{j}\) mm, velocity \(4.8 \mathbf{i} + 10n e^{-\frac{3n}{2}} \mathbf{j}\) mm/s, acceleration \(-3.84 \mathbf{i} + 20( \frac{n^2}{4} + 4\pi^2 ) e^{-\frac{3n}{2}} \mathbf{j}\) mm/s虏.

Step by step solution

01

Calculate Position at t=0

Plug in the values of \(x_1 = 30\) mm and \(y_1 = 20\) mm into the position vector formula for \(t=0\):\[ \mathbf{r} = 30 \left[1 - \frac{1}{0+1}\right] \mathbf{i} + \left(20 e^{-\frac{n(0)}{2}} \cos(2 \pi \times 0)\right) \mathbf{j} \]This simplifies to:\[ \mathbf{r} = 30 \times 0 \mathbf{i} + (20 \times 1 \times 1) \mathbf{j} = 0 \mathbf{i} + 20 \mathbf{j} \] mm.
02

Calculate Position at t=1.5s

For \(t=1.5\), substitute the values into the position vector:\[ \mathbf{r} = 30 \left[1 - \frac{1}{1.5+1}\right] \mathbf{i} + \left(20 e^{-\frac{n(1.5)}{2}} \cos(3 \pi)\right) \mathbf{j} \]This becomes:\[ \mathbf{r} = 30 \times \frac{5}{8} \mathbf{i} + \left(20 e^{-\frac{3n}{2}} \times (-1)\right) \mathbf{j} \]Hence, the position vector is \( \mathbf{r} = 18.75 \mathbf{i} - 20 e^{-\frac{3n}{2}} \mathbf{j} \) mm.
03

Find the Velocity Expression

To find the velocity, differentiate the position vector with respect to time \(t\):\[ \mathbf{v}(t) = \frac{d}{dt}[x_1[1-1/(t+1)] \, \mathbf{i} + y_1 e^{-nt/2} \cos(2\pi t) \, \mathbf{j}] \]After differentiating:\[ \mathbf{v}(t) = \frac{x_1}{(t+1)^2} \, \mathbf{i} + y_1 \left(-\frac{n}{2}e^{-nt/2}\cos(2\pi t) - 2\pi e^{-nt/2}\sin(2\pi t)\right) \, \mathbf{j} \]
04

Calculate Velocity at t=0

Substitute \(t=0\) into the velocity expression:\[ \mathbf{v}(0) = \frac{30}{1^2} \mathbf{i} + 20 \left(-\frac{n}{2}\cos(0) - 2\pi\sin(0)\right) \mathbf{j} \]This simplifies to:\[ \mathbf{v}(0) = 30 \mathbf{i} - 10n \mathbf{j} \] mm/s.
05

Calculate Velocity at t=1.5s

Substitute \(t=1.5\) into the velocity expression:\[ \mathbf{v}(1.5) = \frac{30}{(2.5)^2} \mathbf{i} + 20 \left(-\frac{n}{2} e^{-\frac{3n}{2}} \cos(3\pi) - 2\pi e^{-\frac{3n}{2}}\sin(3\pi)\right) \mathbf{j} \]This simplifies to:\[ \mathbf{v}(1.5) = \frac{30}{6.25} \mathbf{i} + 10n e^{-\frac{3n}{2}} \mathbf{j} \] mm/s.
06

Find the Acceleration Expression

To determine acceleration, differentiate the velocity vector with respect to time \(t\):\[ \mathbf{a}(t) = \frac{d}{dt}(\mathbf{v}(t)) \]Resulting in:\[ \mathbf{a}(t) = -\frac{2x_1}{(t+1)^3} \mathbf{i} + y_1 \left(\frac{n^2}{4} e^{-nt/2} \cos(2\pi t) + n\pi e^{-nt/2} \sin(2\pi t) + 4\pi^2 e^{-nt/2} \cos(2\pi t)\right) \mathbf{j} \]
07

Calculate Acceleration at t=0

Substitute \(t=0\) into the acceleration expression:\[ \mathbf{a}(0) = -60 \mathbf{i} + 20 \left( \frac{n^2}{4} + 4\pi^2 \right) \mathbf{j} \] mm/s虏.
08

Calculate Acceleration at t=1.5s

Substitute \(t=1.5\) into the acceleration expression:\[ \mathbf{a}(1.5) = -\frac{60}{15.625} \mathbf{i} + 20 \left( \frac{n^2}{4} e^{-\frac{3n}{2}} + 4\pi^2 e^{-\frac{3n}{2}} \right) \mathbf{j} \] mm/s虏.

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Key Concepts

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

Position Vector Calculation
The position vector is crucial in describing a particle's location in space at any given time. Imagine it as the coordinates telling us exactly where a particle is situated along an x- and y-axis. In this damped vibration problem, the position vector is given as:
\[ \mathbf{r}=x_{1}\left[1-1 /(t+1)\right] \mathbf{i}+\left(y_{1} e^{-nt/2} \cos 2 \pi t\right) \mathbf{j} \]
In this formula:
  • \(x_1\) represents the initial x-offset and affects how far along the x-axis the particle starts.
  • \(y_1\) is the starting factor for the y-axis displacement.
  • \(t\) is the time variable, showing that the position changes over time.
  • The expression \(1-\frac{1}{t+1}\) helps dictate how fast the x-direction changes.
  • The exponential \(e^{-nt/2}\) captures the damping effect, which looks like reducing oscillations over time.
  • The \(\cos 2\pi t\) oscillation describes the particle's to-and-fro movement.
Understanding and calculating this vector involves plugging in time values as we've done for \(t=0\) and \(t=1.5s\) to explore how a particle's position shifts over time.
Velocity Differentiation
Velocity shows how fast a particle's position is changing over time. It is derived by differentiating the position vector with respect to time. In this exercise, the calculated velocity expression is:
\[ \mathbf{v}(t) = \frac{x_1}{(t+1)^2} \, \mathbf{i} + y_1 \left(-\frac{n}{2}e^{-nt/2}\cos(2\pi t) - 2\pi e^{-nt/2}\sin(2\pi t)\right) \, \mathbf{j} \]
The differentiation process involves:
  • The term \(\frac{x_1}{(t+1)^2}\) indicates how the x component decelerates over time since the denominator grows.
  • The \(j\)-component is a more complex mix of exponential and trigonometric functions representing vertical velocity.
  • The \(-\frac{n}{2}e^{-nt/2}\) factor from \(y_1\) captures the damping aspect in the velocity.
  • Incorporating trigonometric derivatives shows oscillatory effects, which include \(-2\pi e^{-nt/2}\sin(2\pi t)\), accounting for tangential changes.
By evaluating the velocity expression at specific times like \(t=0\) and \(t=1.5s\), we can see how rapidly the particle moves or changes direction.
Acceleration Differentiation
Acceleration is a measure of how quickly the velocity of a particle changes. It is obtained by differentiating the velocity with respect to time again. The derived acceleration expression from the exercise is:
\[ \mathbf{a}(t) = -\frac{2x_1}{(t+1)^3} \mathbf{i} + y_1 \left(\frac{n^2}{4} e^{-nt/2} \cos(2\pi t) + n\pi e^{-nt/2} \sin(2\pi t) + 4\pi^2 e^{-nt/2} \cos(2\pi t)\right) \mathbf{j} \]
Here's what's happening:
  • The \(-\frac{2x_1}{(t+1)^3}\) component further complicates this deceleration in the x-direction with time.
  • The \(i\)-component has additional layers of damping, indicating the rapid decrease with time squared terms.
  • The \(j\)-component factors like \(\frac{n^2}{4}\) and \(4\pi^2\) indicate more intense trigonometric terms and damping effects, leading to refined changes in velocity.
  • The term \(n\pi e^{-nt/2}\sin(2\pi t)\) adds a conducive oscillating effect, signifying how acceleration also spins in rhythm with vibrations.
Substituting in values like \(t=0\) and \(t=1.5s\) helps clarify how these complicated interactions come together to define a vibrating particle's movement.
Vibrating Particle Dynamics
Understanding vibrating particle dynamics is vital when dealing with systems that exhibit oscillatory motion like our damped particle. The dynamics tell us how forces and motions interplay to mold the behavior of the particle over time.
Several key insights from the dynamics in this context include:
  • Damping Mechanism: The damping term \(e^{-nt/2}\) represents energy loss over time, resulting in a reduction of overall oscillation amplitude.
  • Harmonic Motion: The presence of trigonometric terms like \(\cos(2\pi t)\) connects directly to harmonic, cyclic motion typical in vibration analysis.
  • Time Influence: Time variable \(t\) in the vector components continually affects the amplitude and frequency of particle vibrations. As seen, larger time values bring diminishing oscillations.
  • Complex Interaction: Combining exponential damping terms with sinusoidal functions creates a rich dynamic that dictates how the particle will move, slow down, and eventually rest.
Grasping these dynamics provides a more holistic view of how damping shapes the motion, making our calculations not just steps to complete, but a part of understanding motion physics in a vibrating system.

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Most popular questions from this chapter

The path of a particle \(P\) is a limafon. The motion of the particle is defined by the relations \(r=b(2+\cos \pi t)\) and \(\theta=\pi t\) where \(t\) and \(\theta\) are expressed in seconds and radians, respectively. Determine (a) the velocity and the acceleration of the particle when \(t=2 \mathrm{s},(b)\) the value of \(\theta\) for which the magnitude of the velocity is maximum.

The acceleration of a particle is defined by the relation \(a=3 e^{-0.2 t},\) where \(a\) and \(t\) are expressed in \(\mathrm{ft} / \mathrm{s}^{2}\) and seconds, respectively. Knowing that \(x=0\) and \(v=0\) at \(t=0,\) determine the velocity and position of the particle when \(t=0.5 \mathrm{s}\).

The velocity of a particle is \(v=v_{0}[1-\sin (\pi t / T)] .\) Knowing that the particle starts from the origin with an initial velocity \(v_{0}\), determine \((a)\) its position and its acceleration at \(t=3 T,(b)\) its average velocity during the interval \(t=0\) to \(t=T\)

Instruments in an airplane which is in level flight indicate that the velocity relative to the air (airspeed) is \(120 \mathrm{km} / \mathrm{h}\) and the direction of the relative velocity vector (heading) is \(70^{\circ}\) east of north. Instruments on the ground indicate that the velocity of the airplane (ground speed) is \(110 \mathrm{km} / \mathrm{h}\) and the direction of flight (course) is \(60^{\circ}\) east of north. Determine the wind speed and direction.

Determine the speed of a satellite relative to the indicated planet if the satellite is to travel indefinitely in a circular orbit \(100 \mathrm{mi}\) above the surface of the planet. (See information given in Probs. \(11.153-11.154 .)\) $$ \begin{array}{l}{11.155 \text { Venus: } g=29.20 \mathrm{ft} / \mathrm{s}^{2}, R=3761 \mathrm{mi}} \\ {11.156 \mathrm{Mars}: g=12.17 \mathrm{ft} / \mathrm{s}^{2}, R=2102 \mathrm{mi}} \\ {11.157 \text { Jupiter: } g=75.35 \mathrm{ft} / \mathrm{s}^{2}, R=44,432 \mathrm{mi}}\end{array} $$
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