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 displacementb. the frequencyc. the time required for one cycle.d = 1/3 sin 2t

Accepted Answer

Hello, today we're going to be solving the following problem. So what we are given is 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 the amplitude, the frequency and the period. So let's take a look at the equation that is given to us, we are given H is equal to 1/6 sign of three T. So let's go ahead and start by solving for the amplitude of the sine function. Now the amplitude represents the deviation of the particle from the center of the equation. And what this means is that the amplitude is given as a measurement. So a measurement can never be negative, which means that we want to take the absolute value of the amplitude. The amplitude is given to us by the absolute value of the leading coefficient of the sine function. In this case here, the leading coefficient is 1/6. So the amplitude is going to be defined as the absolute value of 1/6, which is going to give us 1/6. This means that the amplitude of the particle is going to be 1/ meters long. Now that we have the amplitude, let's go ahead and solve for the period. Now, the period for a sine function is defined as two pi over B B represents the leading coefficient of the T term in this case. And that leading coefficient is going to be three. So we can plug in three for B and that's going to give us a period of two pi over three. And that means the period of the particle is gonna be two pi over three minutes long. Now that we have the period, let's go ahead and sol for the frequency. Now, the frequency is defined as one over the period of the function. The period of the function is two pi over three in this case. So the frequency is going to be defined as 1/ pi over three. This is in the form of a complex fraction. And in order to simplify this, we're gonna multiply the denominator by its reciprocal and multiply the numerator by the reciprocal of the denominator. This is going to simplify the denominator to give us one. And what we are left with is 3/2 pi times one. That means the frequency is going to be 3/2 pi cycles per minute. So just to summarize, we calculated the amplitude which is gonna be 1/6 meters long, we calculated the period which is two pi over three minutes. And we calculated the frequency which is 3/2 pi cycles per minute. With that being said, the answer to this problem is going to be B 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 displacementb. the frequencyc. the time required for one cycle.d = 1/3 sin 2t

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