An object oscillates along a vertical line, and its position i...

Question

An object oscillates along a vertical line, and its position in centimeters is given by y(t) = 30(sin t - 1), where t ≥ 0 is measured in seconds and y is positive in the upward direction.

The acceleration of the oscillator is a(t) = v′(t). Find and graph the acceleration function.

Accepted Answer

Hi everyone. Let's take a look at this practice problem. This problem says a wave travels along a string and its position in millimeters is described by H of T, which is equal to 15 multiplied by the quantity of 2 multiplied by cosine of T, plus 1 in quantity, or T greater than or equal to 0, where T represents time in seconds. Determine the acceleration function AFT of the wave in graphites function. Consider only one period in graphene. Below the problem, we're given an empty graph, where we have on the Y axis our acceleration A in units of millimeters per second squared, and on the X-axis, we have our time T in seconds. Now, the first part of this problem wants us to determine our acceleration function AFT. So recall that our acceleration function is the second derivative with respect to time of our position equation. So we're gonna need to calculate two derivatives of our function H of T with respect to T. So we're gonna start off by looking at our first derivative, which also happens to be our velocity. So we'll have V of T is equal to the derivative of H with respect to T. Which is equal to the derivative with respect to T of the quantity of 15 multiplied by the quantity of 2 multiplied by cosine of T plus 1 in quantity. So taking this derivative, this is going to be equal to. The 15 out front as a constant multiplying a different quantity can be moved through derivatives by our constant multiplication rule for derivatives, so we'll have 15, multiplied by the quantity of the derivative with respect to T of the quantity of 2 cosine of T plus 1. So taking the derivative of 2 cosine of T, we will get minus 2 multiplied by sine of T. recall that the derivative with respect to T of cosine of T is equal to minus sine of T, and again, our uh constant out front, which is 2, can move through the derivative. So that's why we get -2 multiplied by sin of T. Then we will have less, the derivative of 1 with respect to T, which is zero, since we're taking the derivative of a constant. So, simplifying this expression, this is going to be equal to -30, multiplied by thine of T. So now we need to calculate our acceleration. So, our acceleration, AFT. Is going to be equal to the 2nd derivative of H with respect to T. Which is also equal to the derivative of our velocity V with respect to T. And so this is going to be equal to the derivative with respect to T. Of the quantity of minus 30 multiplied by sine of t. So, taking this derivative, this is going to be equal to -30, multiplied by cosine of T. So recall that the derivative of sine of T is cosine of T, and the minus 30 there is constant, so it will move through the derivative. So this here is our acceleration function, AFT, which is -30 multiplied by cosine of T. So now we need to plot this function. So we only need to plot this for one period and recall that the period of our cosine function is to pi. So we're going to plot this from 0 to 2 pi. So we'll just choose some points to place on our graph. So the first point is when T is equal to 0, and the cosine of 0 is 1, so our AFT is going to be equal to -30. Cosine of T crosses the X-axis, when T is equal to pi divided by 2, so we'll have that point plotted as well on a graph. When we have T equal to pi, cosine of pi is equal to -1, so we'll get -30 multiplied by -1, which will give me 30. So we'll plot that point on a graph as well. Our cosine function will again cross the x axis when T is equal to 3 pi divided by 2. So we'll plot that point, and then finally we'll set T equal to 2 pi. And again, cosine of 2 pi is equal to 1, then we'll get -34 our acceleration. So now I'll need to do is Draw a smooth curve connecting these points. And this here is just gonna be a cosine curve that has an amplitude of 30. So making sure that we draw a line through every points, and that Every point is the curve there is smooth at every point. So now we have drawn this, this is the graph that the question was asking for. So I hope that this explanation has been useful, and I'll see you in the next video.

An object oscillates along a vertical line, and its position in centimeters is given by y(t) = 30(sin t - 1), where t ≥ 0 is measured in seconds and y is positive in the upward direction.
The acceleration of the oscillator is a(t) = v′(t). Find and graph the acceleration function.

Key Concepts

Differentiation

Acceleration

Graphing Functions

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