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Chapter 4: Problem 47

Find a model for simple harmonic motion satisfying the specified conditions. Displacement \((t=0)\) $$0$$ Amplitude 4 centimeters Period 2 seconds

Short Answer

Expert verified
The model for simple harmonic motion satisfying the given conditions is \(x(t) = 4 sin(\pi t)\).

Step by step solution

01

Identify given values

From the problem, we can identify these values: Amplitude \(A = 4\) cm, Period \(T = 2\) s.
02

Calculate the angular frequency

The angular frequency \(\omega\) can be found from the period \(T\). Use the relation \(\omega = \frac{{2\pi}}{{T}}\). Substitute \(T = 2\) s into the formula, yielding \(\omega = \frac{{2\pi}}{{2}} = \pi\) rad/s.
03

Write the equation for the simple harmonic motion

The equation for the simple harmonic motion based on the given parameters is \(x(t) = 4 sin(\pi t)\). This equation represents the motion of an object with amplitude 4 cm, period 2 s and initial displacement 0.

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Key Concepts

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

Amplitude
In simple harmonic motion, the **amplitude** is the maximum extent of a vibration or oscillation, measured from the position of equilibrium. It is a crucial factor in describing how far the object moves away from its rest position. Given the amplitude value of 4 centimeters in our exercise, it means the object will move 4 cm away from the center position at its peak.

Amplitude is always positive, as it represents distance. The actual displacement can vary from + 4 cm to - 4 cm as the object moves. You can think about it this way to visualize: If you stretch a spring 4 cm from its equilibrium position, 4 cm is our amplitude, regardless of direction.
  • The amplitude is a measure of energy in a wave. Larger amplitudes mean more energy.
  • The amplitude affects how 'loud' or 'intense' a wave is 鈥� for instance, in sound waves.
  • In this exercise, an amplitude of 4 cm signifies the wave reaches a peak of 4 cm above and below the equilibrium position.
Angular Frequency
Angular frequency, often symbolized by \(\omega\)is related to how fast the object is oscillating back and forth, within a simple harmonic motion. It translates to the rate of change of the phase angle regarding time. Here in our context, it shows how quickly the object moves through the cycle of motion.

From the original exercise, we calculated the angular frequency using the relation \(\omega = \frac{{2\pi}}{T}\)where \(T\)is the period. In this exercise, the period is given as 2 seconds, leading to an angular frequency of \(\pi\)radians per second (\(\omega = \pi\)).
  • Angular frequency \(\omega\)is in radians per second.
  • Higher angular frequency means more cycles in a given time.
  • It connects with frequency \(f\) by the equation \(\omega = 2\pi f\).
  • In our case: \(\omega = \pi\) rad/s reflects a simple harmonic motion with one complete cycle every 2 seconds.
Period
The period of simple harmonic motion relates to how long it takes to complete one full cycle of motion. It is the reciprocal of frequency and is an essential measure to understand the behavior of oscillatory systems.

In the context of our exercise, the period is given as 2 seconds. This means that once the motion starts, it takes 2 seconds for the object to return to the starting position and complete one full oscillation.
  • The period \(T\)is measured in seconds (s).
  • A shorter period means the object oscillates more quickly.
  • It links with frequency (\(f\)in hertz) by the relation \(T = \frac{1}{f}\).
  • For our simple harmonic motion, a single oscillation completed in 2 seconds implies a cycle per 2 seconds.

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Most popular questions from this chapter

A ball that is bobbing up and down on the end of a spring has a maximum displacement of 3 inches. Its motion (in ideal conditions) is modeled by \(y=\frac{1}{4} \cos 16 t, t>0,\) where \(y\) is measured in feet and \(t\) is the time in seconds. (a) Graph the function. (b) What is the period of the oscillations? (c) Determine the first time the weight passes the point of equilibrium \((y=0)\)

A sprinkler on a golf green sprays water over a distance of 15 meters and rotates through an angle of \(140^{\circ} .\) Draw a diagram that shows the region that the sprinkler can irrigate. Find the area of the region.

The normal monthly high temperatures \(H\) (in degrees Fahrenheit) in Erie, Pennsylvania, are approximated by $$H(t)=56.94-20.86 \cos \left(\frac{\pi t}{6}\right)-11.58 \sin \left(\frac{\pi t}{6}\right)$$ and the normal monthly low temperatures \(L\) are approximated by $$L(t)=41.80-17.13 \cos \left(\frac{\pi t}{6}\right)-13.39 \sin \left(\frac{\pi t}{6}\right)$$ where \(t\) is the time (in months), with \(t=1\) corresponding to January (see figure). (Source: National Climatic Data Center (GRAPH CANNOT COPY). (a) What is the period of each function? (b) During what part of the year is the difference between the normal high and normal low temperatures greatest? When is it smallest? (c) The sun is northernmost in the sky around June \(21,\) but the graph shows the warmest temperatures at a later date. Approximate the lag time of the temperatures relative to the position of the sun.

For the simple harmonic motion described by the trigonometric function, find (a) the maximum displacement, (b) the frequency, (c) the value of \(d\) when \(t=5,\) and (d) the least positive value of \(t\) for which \(d=0 .\) Use a graphing utility to verify your results. $$d=9 \cos \frac{6 \pi}{5} t$$

Fill in the blank. If not possible, state the reason. $$\text { As } x \rightarrow \infty, \text { the value of } \arctan x \rightarrow\text { _____ } .$$
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