In Exercises 65–66, an object moves in simple harmonic motion des...

Question

In Exercises 65–66, an object moves in simple harmonic motion described by the given equation, where t is measured in seconds and d in centimeters. In each exercise, find:      a. the maximum displacement     b. the frequency     c. the time required for one cycle.d = 20 cos π/4 t

Accepted Answer

Welcome back everyone. The following equation describes the periodic motion of a particle where the equation is S equals 48 multiplied by the cosine of an eighth of pi multiplied by X for which the unit of S is in inches. And the unit of X is in seconds. We want to find the amplitude, the frequency and the period of our equation, we have four choices where A tells us that the amplitude is inches, the period is 16 seconds and the frequency is 1/16 cycles per second. B tells us that the amplitude is 48 inches, the period is 16 seconds and the frequency is 1/16 cycle per second. C tells us that the amplitude is 24 inches. The period is 1/16 of a second and the frequency is cycles per second. And finally, D tells us that the amplitude is 48 inches, the period is 1/16 of a second and the frequency is 16 cycles per second. So let's first start by defining the amplitude. Let's put that in red, the frequency, let's put that in blue and then the period, let's put that in purple. So let's start with the amplitude. Generally, every equation that represents a trigonometric function can be written in this form where Y is equal to a multiplied by whatever the trick function is. In this case, the cosine R B X plus C plus D. OK. And to help us figure that out, I'm gonna draw a graph that we can follow along. And I'm going to put the cosine function on it, which generally looks something like this. So first, let's start with the amplitude, the amplitude of a trigonometric function is referred to as a in the general formula. And it just tells us how tall our graph is. So in this case, that would be the amplitude and here we'd have our Y axis and our X axis and the amplitude is basically the constant that's multiplying the function. So in this case, since our constant is 48 our amplitude is next or period, let's go to the period first before we talk about the frequency because they're a bit connected. But the period basically tells us how long our wavelength is. So in this case, from our peak here, going back to our peak, this would represent our period. And the formula for the period is equal to two pi divided by B where B is the coefficient of the angle of the trigonometric function. So in this case, because our anger is an eighth of pi multiplied by X, that means our coefficient would be equal to an eighth of pi. So that would be equal to B. So that means our period would be equal to two pi divided by an eighth of pi. And if we rewrite that, that would be two pi multiplied by eight, divided by pi, the pi is cancel and two multiplied by eight is equal to 16. So that would be our period. Finally or frequency generally, let me get the color, the frequency. The frequency just tells us how much times the cycle repeats per second. OK. So in this graph, this frequency, if I were to draw another graph that looked like this, then it would have a higher frequency. It just tells us how much times it repeats. Therefore, the frequency is equal to one divided by the period. OK? And if we were to do that, that would have been one divided by 16. So our amplitude is 48 our period is 16 and our frequency is 1/16 and that means our answer has to be choice. B Thanks for listening. I hope this helped.

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In Exercises 65–66, an object moves in simple harmonic motion described by the given equation, where t is measured in seconds and d in centimeters. In each exercise, find: a. the maximum displacement b. the frequency c. the time required for one cycle.d = 20 cos π/4 t

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