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Chapter 15: Problem 9

\(=9\) The position function \(x=\) \((6.0 \mathrm{~m}) \cos [(3 \pi \mathrm{rad} / \mathrm{s}) t+\pi / 3 \mathrm{rad}]\) gives the simple harmonic motion of a body. At \(t=2.0 \mathrm{~s}\), what are the (a) displacement, (b) velocity, (c) acceleration, and (d) phase of the motion? Also, what are the (e) frequency and (f) period of the motion?

Short Answer

Expert verified
(a) 3.0 m, (b) -49.1 m/s, (c) -267.9 m/s虏, (d) \(\frac{19\pi}{3}\) rad, (e) 1.5 Hz, (f) 0.67 s.

Step by step solution

01

Displacement at t = 2.0 s

The displacement equation is given by \( x = (6.0 \, \text{m}) \cos[(3\pi \, \text{rad/s})\, t + \pi/3 \, \text{rad}] \). Substitute \( t = 2.0 \, \text{s} \) into the equation:\[x = (6.0) \cos[(3\pi)\cdot(2) + \frac{\pi}{3}]\]Calculate the angle: \[x = (6.0) \cos[6\pi + \frac{\pi}{3}]\]Since \( 6\pi \) is a multiple of \( 2\pi \), it does not change the cosine value, thus:\[x = (6.0) \cos[\frac{\pi}{3}] = (6.0) \cdot \frac{1}{2} = 3.0 \, \text{m}\]
02

Velocity at t = 2.0 s

Velocity is the derivative of displacement with respect to time: \[v(t) = \frac{d}{dt}[(6.0) \cos((3\pi) t + \frac{\pi}{3})]\]Using chain rule, we have:\[v(t) = -6.0 \cdot (3\pi) \sin((3\pi) t + \frac{\pi}{3})\]Substitute \( t = 2.0 \, \text{s} \):\[v = -18\pi \sin(6\pi + \frac{\pi}{3}) = -18\pi \sin(\frac{\pi}{3}) = -18\pi \cdot \frac{\sqrt{3}}{2} \approx -49.1 \, \text{m/s} \]
03

Acceleration at t = 2.0 s

Acceleration is the derivative of velocity with respect to time:\[a(t) = \frac{d}{dt}[-18\pi \sin((3\pi)t + \frac{\pi}{3})]\]Using the chain rule again, we get:\[a(t) = -18\pi \cdot (3\pi) \cos((3\pi)t + \frac{\pi}{3})\]Substitute \( t = 2.0 \, \text{s} \):\[a = -54\pi^2 \cos(6\pi + \frac{\pi}{3}) = -54\pi^2 \cdot \frac{1}{2} = -27\pi^2 \approx -267.9 \, \text{m/s}^2\]
04

Phase at t = 2.0 s

The phase at any time \( t \) is given by the expression inside the cosine function:\[\theta(t) = (3\pi)t+\frac{\pi}{3}\]Substitute \( t = 2.0 \) and evaluate:\[\theta = (3\pi)(2) + \frac{\pi}{3} = 6\pi + \frac{\pi}{3} = \frac{19\pi}{3} \, \text{rad}\]
05

Frequency of the motion

The frequency \( f \) of the motion in simple harmonic motion is given by the angular frequency \( \omega = 3\pi \, \text{rad/s}\):\[\omega = 2\pi f \f = \frac{3\pi}{2\pi} = \frac{3}{2} = 1.5 \, \text{Hz}\]
06

Period of the motion

The period \( T \) of the motion is the reciprocal of the frequency:\[T = \frac{1}{f} = \frac{1}{1.5} \approx 0.67 \, \text{s}\]

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Key Concepts

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

Displacement Calculation
Displacement in simple harmonic motion refers to the position of the oscillating object at a given time. It is described using a cosine or sine function. In this problem, the displacement function is given by:
  • \( x(t) = (6.0 \, \text{m}) \cos[(3\pi \, \text{rad/s})\, t + \pi/3 \, \text{rad}] \)
To find the displacement at a specific time, substitute the given time into the equation. For example, at \(t = 2.0 \, \text{s}\), the angle inside the cosine function becomes \(6\pi + \pi/3\). Since \(6\pi\) equals a full rotation, it simplifies, and the displacement for \(t = 2.0 \, \text{s}\) can be calculated as:
  • \( x = 3.0 \, \text{m} \)
This value indicates that at 2 seconds, the object is 3 meters away from its equilibrium position.
Velocity Calculation
Velocity in simple harmonic motion is the rate of change of displacement over time. It is obtained by taking the derivative of the displacement function with respect to time. To calculate the velocity function from the displacement equation \(x(t)\), apply the chain rule:
  • \( v(t) = \frac{d}{dt}[(6.0) \cos((3\pi) t + \frac{\pi}{3})] = -18\pi \sin((3\pi) t + \frac{\pi}{3}) \)
To find the velocity at \(t = 2.0 \, \text{s}\), substitute the value of \(t\) into the velocity equation:
  • \( v = -18\pi \cdot \sin(6\pi + \frac{\pi}{3}) \approx -49.1 \, \text{m/s} \)
This result indicates the object is moving with a velocity of approximately \(-49.1 \, \text{m/s}\) at 2 seconds, in the direction contrary to the positive direction.
Acceleration Calculation
Acceleration in simple harmonic motion is the derivative of the velocity function with respect to time. This represents how quickly the velocity of the object is changing. Starting from the velocity equation:
  • \( v(t) = -18\pi \sin((3\pi) t + \frac{\pi}{3}) \)
The acceleration function is then:
  • \( a(t) = \frac{d}{dt}[-18\pi \sin((3\pi)t + \frac{\pi}{3})] = -54\pi^2 \cos((3\pi)t + \frac{\pi}{3}) \)
At \(t = 2.0 \, \text{s}\), the phase remains \(6\pi + \pi/3\), and substituting gives:
  • \( a = -27\pi^2 \approx -267.9 \, \text{m/s}^2 \)
This means the object is experiencing a significant deceleration at \(2.0 \, \text{s}\), trying to restore the object back to its equilibrium position.
Phase Angle
The phase angle in simple harmonic motion helps establish the position of the wave initially at \(t=0\) and as it progresses over time. It comprises both the angular frequency term and the initial phase offset in this question:
  • \( \theta(t) = (3\pi)t + \frac{\pi}{3} \)
By substituting \(t = 2.0 \, \text{s}\), we calculate:
  • \( \theta = 6\pi + \frac{\pi}{3} = \frac{19\pi}{3} \, \text{rad} \)
This phase angle provides a snapshot of the motion's progression through its cycle, helping clarify where in its oscillation the object currently resides.
Frequency and Period
Frequency and period are fundamental attributes of simple harmonic motion. Frequency indicates how many oscillations occur per second, while the period is the time taken for one complete cycle.To determine frequency \( f \), use the angular frequency \( \omega \) of the system, which is established as \(3\pi \, \text{rad/s}\):
  • \(\omega = 2\pi f \Rightarrow f = \frac{3\pi}{2\pi} = 1.5 \, \text{Hz}\)
The period \( T \) is the reciprocal of frequency:
  • \(T = \frac{1}{f} = \frac{1}{1.5} \approx 0.67 \, \text{s}\)
In summary, the system oscillates 1.5 times per second, and each full cycle takes approximately 0.67 seconds.

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Most popular questions from this chapter

A \(4.00 \mathrm{~kg}\) block hangs from a spring, extending it \(16.0 \mathrm{~cm}\) from its unstretched position. (a) What is the spring constant? (b) The block is removed, and a \(0.500 \mathrm{~kg}\) body is hung from the same spring. If the spring is then stretched and released, what is its period of oscillation?

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?

If the phase angle for a block-spring system in SHM is \(\pi / 6\) rad and the block's position is given by \(x=x_{m} \cos (\omega t+\phi)\), what is the ratio of the kinetic energy to the potential energy at time \(t=0 ?\)

A block rides on a piston (a squat cylindrical piece) that is moving vertically with simple harmonic motion. (a) If the SHM has period \(1.0 \mathrm{~s}\), at what amplitude of motion will the block and piston separate? (b) If the piston has an amplitude of \(5.0 \mathrm{~cm}\), what is the maximum frequency for which the block and piston will be in contact continuously?

A \(2.0 \mathrm{~kg}\) block executes SHM while attached to a horizontal spring of spring constant \(200 \mathrm{~N} / \mathrm{m}\). The maximum speed of the block as it slides on a horizontal frictionless surface is \(3.0 \mathrm{~m} / \mathrm{s}\). What are (a) the amplitude of the block's motion, (b) the magnitude of its maximum acceleration, and (c) the magnitude of its minimum acceleration? (d) How long does the block take to complete \(7.0 \mathrm{cy}\) cles of its motion?
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