In Exercises 21–28, an object moves in simple harmonic motion des...

Question

In Exercises 21–28, 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, find the following:
a. the maximum displacement
b. the frequency
c. the time required for one cycle.

d = −4 sin 3π/2 t

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 meters. And the unit of T is in minutes we want to solve for the amplitude, the frequency and the period. So the equation that is given to us is H is equal to negative six sine A five pi over four T. So one way to help us identify these three pieces of information, which is the amplitude frequency and period is to look at this equation in a generalized form. Now, we know that this is the equation of a sine function and the generalized form of the equation of a sine function is a sign of B X plus C plus K. In this case here A represents the amplitude B is going to be the coefficient from the X term which we use to help find as the period C represents or B X plus C represents a horizontal shift and K represents a vertical shift of the graph. So let's go ahead and start by finding the amplitude. Now, one thing to note is that the amplitude is the deviation from the center of the particle. And this means that the amplitude can never be negative. So the amplitude is represented by a which is the leading coefficient of the sine function. And in this case here, the leading coefficient is negative six. But we need to take the absolute value of this leading coefficient. So the amplitude is represented by the absolute value of negative six which is going to give us six. That means the amplitude of the particle is going to be a total of six m. Next, let's find the period of the particle. Now, since we are given a sine function, the period is going to be defined as two pi over B B represents the coefficient in front of our T term in this case. And that coefficient is going to be five pi over four. So we can represent the period as two pi over five pi over four. In order to simplify the period, we're going to multiply the denominator by its reciprocal which is 4/5 pi. And we're also gonna multiply the numerator by this reciprocal which is 4/ pi, this simplifies the denominator to just one and it leaves us with a period of two pi multiplied by 4/5 pi. We can represent two pies, a fraction by putting it over one. And we can simplify the pies in both numerator and denominator to just one that leaves us with two times four in the numerator which is 8/1 times five, which is five. And that means that the period of the particle is going to be 8/5 minutes. Now that we have the period of the particle, let's go ahead and find its frequency. The frequency for a sign function is represented by one over the period. Now again, we stated that the period was 8/5. So if we plug this into the frequency, the frequency is represented by 1/8, over five, this is once again a complex fraction. So we multiply the denominator by its reciprocal and multiply the numerator by the reciprocal, the denominator, the denominator simplifies to just one. And what we are left with is one times 5/8, which gives us 5/8. And that means that the frequency is going to be 5/8 cycles per minute. So just to simplify the amplitude is six m, the period is 8/5 minutes and the frequency is 5/8 cycles per minute. With that being said, the answer to this problem is going to be D. 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.

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In Exercises 21–28, 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, find the following: a. the maximum displacement b. the frequency c. the time required for one cycle. d = −4 sin 3π/2 t