The distance from the bob of the pendulum to the wall varies in a periodic way that can be modeled by a trigonometric function. The modeling function has period $0.8$ seconds, amplitude $6\,\text{cm}$ , and midline $H=15\,\text{cm}$ . At time $t=0.2$ , the bob is at its midline, moving towards the wall. Find the formula of the trigonometric function that models the distance $H$ between the bob and the wall after $t$ seconds. Define the function using radians. $\newline$ $H(t)=-\left(\frac{\pi}{2}\right)+1$ ? $\newline$ What is the distance from the pendulum to the wall after $0.5$ seconds? Round your answer, if necessary, to two decimal places.

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Q. The distance from the bob of the pendulum to the wall varies in a periodic way that can be modeled by a trigonometric function. The modeling function has period

0.8

seconds, amplitude

6\,\text{cm}

, and midline

H=15\,\text{cm}

. At time

t=0.2

, the bob is at its midline, moving towards the wall. Find the formula of the trigonometric function that models the distance

H

between the bob and the wall after

t

seconds. Define the function using radians.

\newline

H(t)=-\left(\frac{\pi}{2}\right)+1

\newline

What is the distance from the pendulum to the wall after

0.5

seconds? Round your answer, if necessary, to two decimal places.

Identify Amplitude and Midline: To find the formula of the trigonometric function that models the distance $H$ between the bob and the wall after $t$ seconds, we need to consider the given information: the period is $0.8$ seconds, the amplitude is $6$ cm, the midline is at $H=15$ cm, and at $t=0.2$ seconds the bob is at its midline, moving towards the wall. Since the bob is moving towards the wall, we need a cosine function that starts at the midline and goes towards the maximum. The general form of a cosine function is $H(t) = A\cos(B(t - C)) + D$ , where $A$ is the amplitude, $B$ is related to the period, $C$ is the phase shift, and $t$ $0$ is the vertical shift (midline).
Determine Vertical Shift: First, we identify the amplitude $A$ as $6\,\text{cm}$ , because that's the maximum distance from the midline the bob reaches.
Find B for Period: Next, we determine the vertical shift $D$ , which is the midline of the function. The midline is given as $H=15\,\text{cm}$ , so $D=15\,\text{cm}$ .
Calculate Phase Shift: Now, we need to find $B$ , which is related to the period of the function. The period of a cosine function is given by $(2\pi)/B$ . Since the period is $0.8$ seconds, we have $(2\pi)/B = 0.8$ , which we can solve for $B$ .
Adjust Phase Shift: Calculating $B$ , we get $B = \frac{(2\pi)}{0.8} = \frac{(2\pi)}{0.8} = 2.5\pi$ .
Combine Values for Function: Since the bob is at its midline at $t=0.2$ seconds and moving towards the wall, we know that the phase shift $C$ must be $0.2$ seconds. However, we need to adjust this value because the cosine function starts at its maximum, not the midline. Since the cosine function reaches its midline at $\pi/2$ radians, we need to subtract $\pi/2$ from the phase shift to account for this.
Substitute $t=0.5$ : Adjusting for the phase shift, we get $C = 0.2 - (\pi/2)/(2.5\pi) = 0.2 - 0.2 = 0$ seconds. So there is no horizontal shift in the function.
Evaluate $H(0.5)$ : Putting all the values together, we get the function $H(t) = 6\cos(2.5\pi t) + 15$ .
Substitute $\cos(1.25\pi)$ : To find the distance from the pendulum to the wall after $0.5$ seconds, we substitute $t=0.5$ into the function $H(t)$ .
Calculate Final Distance: Calculating $H(0.5)$ , we get $H(0.5) = 6\cos(2.5\pi\cdot 0.5) + 15 = 6\cos(1.25\pi) + 15$ .
Calculate Final Distance: Calculating $H(0.5)$ , we get $H(0.5) = 6\cos(2.5\pi\cdot0.5) + 15 = 6\cos(1.25\pi) + 15$ . Evaluating the cosine of $1.25\pi$ , we know that $\cos(\pi)$ is $-1$ and $\cos(1.5\pi)$ is $0$ , so $\cos(1.25\pi)$ is between $-1$ and $0$ . Specifically, $H(0.5) = 6\cos(2.5\pi\cdot0.5) + 15 = 6\cos(1.25\pi) + 15$ $0$ since $1.25\pi$ is the angle where the cosine function reaches its minimum in the second period.
Calculate Final Distance: Calculating $H(0.5)$ , we get $H(0.5) = 6\cos(2.5\pi\cdot0.5) + 15 = 6\cos(1.25\pi) + 15$ . Evaluating the cosine of $1.25\pi$ , we know that $\cos(\pi)$ is $-1$ and $\cos(1.5\pi)$ is $0$ , so $\cos(1.25\pi)$ is between $-1$ and $0$ . Specifically, $H(0.5) = 6\cos(2.5\pi\cdot0.5) + 15 = 6\cos(1.25\pi) + 15$ $0$ since $1.25\pi$ is the angle where the cosine function reaches its minimum in the second period. Substituting the value of $\cos(1.25\pi)$ into the function, we get $H(0.5) = 6\cos(2.5\pi\cdot0.5) + 15 = 6\cos(1.25\pi) + 15$ $3$ cm.
Calculate Final Distance: Calculating $H(0.5)$ , we get $H(0.5) = 6\cos(2.5\pi\cdot0.5) + 15 = 6\cos(1.25\pi) + 15$ . Evaluating the cosine of $1.25\pi$ , we know that $\cos(\pi)$ is $-1$ and $\cos(1.5\pi)$ is $0$ , so $\cos(1.25\pi)$ is between $-1$ and $0$ . Specifically, $H(0.5) = 6\cos(2.5\pi\cdot0.5) + 15 = 6\cos(1.25\pi) + 15$ $0$ since $1.25\pi$ is the angle where the cosine function reaches its minimum in the second period. Substituting the value of $\cos(1.25\pi)$ into the function, we get $H(0.5) = 6\cos(2.5\pi\cdot0.5) + 15 = 6\cos(1.25\pi) + 15$ $3$ cm. Therefore, the distance from the pendulum to the wall after $H(0.5) = 6\cos(2.5\pi\cdot0.5) + 15 = 6\cos(1.25\pi) + 15$ $4$ seconds is $H(0.5) = 6\cos(2.5\pi\cdot0.5) + 15 = 6\cos(1.25\pi) + 15$ $5$ cm.