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 = − 1/2 sin(πt/4 − π/2)

Accepted Answer

Hello, today we're going to be solving the following problem. The problem says 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 amplitude, the frequency, the period and the phase shift of the function. So let's take a look at the equation that is given to us, we are told that H is equal to negative 1/ sign of pi over two X minus pi over six. Now, a helpful way to look at this function is to look at the generalized form of a sign function. The generalized form is going to be given to us as a times sine of B X plus C plus K. Now A is the leading coefficient in the sine function and this is going to represent the value of the amplitude. So negative 1/3 takes up the spot of A in the generalized form of the sine function. So what we wanna do for the amplitude is you want to take the absolute value of the leading coefficient which is negative 1/3. Now, the reason why we take the absolute value is because the amplitude represents the deviation from the center of the graph that the particle is traveling. So amplitude can never be negative. So absolute value of negative 1/3 is going to be positive 1/3. And that means that the amplitude or at least the deviation that the particle is traveling is going to be 1/3 feet. Next, let's go ahead and find the period of the function. So the period for a sine function is defined as two pi over B B is the coefficient in front of the X term of the sine function. And looking at the function that is given to us, the coefficient in front of X is going to be pi over two. So we can represent the period as two pi over pi over two. Now, in order to simplify the period, we're gonna multiply the denominator by its reciprocal which is two over pi and we're gonna multiply the numerator by the denominators reciprocal. The period then simplifies to leave us with two pi multiplied by two over pi we can cancel out the pies in the numerator and denominator to just one. And what we are left with is two times two, which is four. So what that means is that the period of the particle is going to be four hours. So it's going to take the particle four hours to complete one period of its movement. Next, let's take a look at the phase shift. I'm sorry, the frequency. Now we always want to get the frequency after we get the period because the frequency is dependent on the period, frequency is defined as one over our period. Now recall that we said the period was four hours. So that means the frequency is going to be 1/4. And that means that the particle is gonna complete 1/4 cycles per hour. Now, finally, we need to find the phase shift of the particle. Now, the phase shift is represented by any horizontal shift of the graph which is represented by X minus H. So let's take a look at the X quantity of the given function we are given pi over two times X minus pi over six. Now, what we wanna do is we want to simplify this quantity to be in the form of X minus H. What we can do is we wanna get a coefficient of one in front of X. So we need to factor out a pi over two from this entire quantity factory. Now pi over two is going to give us pi over two times the quantity X minus pi over six divided by pi over two, this quantity alone represents X minus H. So now what we wanna do is we want to simplify the H term. Now, the H term in this case is going to equal to pi over six divided by pi over two. In order to simplify this quantity, we're gonna multiply the denominator by its reciprocal and multiply the numerator by the reciprocal of the denominator. The denominator is going to simplify to just one. And what we are left with is pi over six, multiplied by two over pi both of these pies are gonna simplify to just one leaving us with 2/6. So the quantity is going to simplify to X minus 2/6. This is where H is equal to positive 2/6. And that means that we have a phase shift of 2/ units of the function. And this can simplify to 1/3. So we have a total phase shift of 1/3 radiant. So just to summarize what we found, we've identified the amplitude as 1/3 feet, we've identified the period as four hours. The frequency, the frequency is 1/5 4th cycles per hour and the phase shift is going to be one third radiance. With that being said, the answer to this problem is going to be B so I hope this video helps you in understanding and under and 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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