In Exercises 37–40, an object moves in simple harmonic motion ...

Question

In Exercises 37–40, an object moves in simple harmonic motion described by the given equation, where t is measured in seconds and d in inches. In each exercise, graph one period of the equation. Then find the following: a. the maximum displacement b. the frequency c. the time required for one cycle d. the phase shift of the motion. d = 3 cos(πt + π/2)

Accepted Answer

Hello, today we're going to be solving the following problem. So we are told that the following equation describes the periodic motion of a particle for which the unit of H is in feet. And the unit of X is in ours, we want to solve for the function's amplitude, the frequency, the period and the phase shift. So the function that is given to us is X is H is equal to five cosine of two pi X plus the quantity pi over four. Now let's go ahead and first solve for the phase shift. Now the phase shift of a cosine graph is represented by some horizontal shift in the graph. And a horizontal shift of any graph is given to us as the quantity X minus H. The argument of cosine is given to us as two pi X plus pi over four. So what we want to go ahead and do is we wanna try to rewrite this quantity to be as close as possible to X minus H. Now notice that X has a coefficient of one and X minus H. So we want to get X to have a coefficient of one in our given argument in order to do that, we can factor out two pi from the entire quantity doing so is going to leave us with two pi multiplied by the quantity X plus pi over four divided by two pi. The next thing we wanna do is we want to simplify this quantity. So we have pi over four divided by two pi, we can represent two pi as a fraction by putting the quantity over one. And now what we have is a complex fraction. In order to simplify the complex fraction, we're gonna multiply the denominator by its reciprocal. And we're going to multiply the numerator by the reciprocal of the denominator. So the denominator is gonna simplify to just give us one. And what we are left with is pi over four, multiplied by 1/2 pi pi is going to simplify to just give us one. And what we are left with is one times one in the numerator which gives us 1/4 times two in the denominator which gives us eight. So the argument here is going to simplify to give us two pi times the quantity X plus 1/8. Now what we wanna do is you want to pay attention to the quantity X plus 1/8 as it closely resembles the tra the traditional form of X minus H. So X plus 1/8 is not in the standard form of X minus H, but this quantity can be rewritten as X minus negative 1/8. This is where H is going to equal to negative 1/8. And that means that the phase shift is going to be negative 1/8 radiance for the entirety of this cosine graph. Now that we have the phase shift, let's go ahead and find the amplitude of the graph. Now the amplitude represents the deviation that the particle is going to be traveling from the center of the graph. So in order to get the amplitude, we want to take the absolute value of the leading coefficient of the trig graph. And in this case here, the leading coefficient is going to be five. So the amplitude is going to be the absolute value of five which is just going to give us five. And that means that the amplitude of the particle is going to be five ft. Next, let's solve for the period of the particle. Now the period of the particle for a cosine graph is defined as two pi over B B represents the coefficient in front of our X term, which in this case is going to be two pi so we can represent the period as two pi over two pi which is going to simplify to one. And that means that the period of the particle is going to be one hour. Finally let's solve for the frequency. Now the frequency is defined as one over the period for, for a cosine graph. Now recall that we stated that the period was one. So we can represent the frequency as 1/1, which is going to simplify to just one. And that means that the frequency of the particle is going to be one cycle per hour. So just to summarize, we found that the phase shift of the graph is going to be negative 1/8 radiant. We found that the amplitude is five ft, the period is going to be one hour and the frequency is going to be one cycle per hour. With that being said, the answer to this problem is going to be a. So I hope this video helps you in understanding how to approach this problem. And I'll go ahead and see you all in the next video.

Key Concepts

Simple Harmonic Motion (SHM)

Amplitude, Frequency, and Period

Phase Shift in Trigonometric Functions

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