a pendulum is swinging next to a wall. the distance from the bob of the swinging pendulum to the wall varies…

Explanation:

Step1: Determine the general form

The general form of a periodic trig - function is $H(t)=A\sin(B(t - C))+D$ or $H(t)=A\cos(B(t - C))+D$. The mid - line gives $D$, the amplitude gives $A$, and the period is used to find $B$. Given amplitude $A = 6$, mid - line $D=15$, and period $T = 0.8$. Since $T=\frac{2\pi}{B}$, then $B=\frac{2\pi}{T}=\frac{2\pi}{0.8}=\frac{5\pi}{2}$.

Step2: Choose the function type and find the phase - shift

We know that at $t = 0.5$ seconds, the bob is at its mid - line, moving towards the wall. A cosine function $y = A\cos(B(t - C))+D$ starts at its maximum or minimum value. A sine function $y=A\sin(B(t - C))+D$ starts at the mid - line. Let's use the sine function $H(t)=A\sin(B(t - C))+D$. Substitute $A = 6$, $B=\frac{5\pi}{2}$, $D = 15$ into it: $H(t)=6\sin(\frac{5\pi}{2}(t - C))+15$. When $t = 0.5$, $H(0.5)=15$, so $15=6\sin(\frac{5\pi}{2}(0.5 - C))+15$. Then $\sin(\frac{5\pi}{2}(0.5 - C))=0$. We want the bob to be moving towards the wall at $t = 0.5$. For $y=\sin x$, when $x = 0$, the function is increasing. So $\frac{5\pi}{2}(0.5 - C)=0$, which gives $C = 0.5$.

Answer:

$H(t)=6\sin\left(\frac{5\pi}{2}(t - 0.5)\right)+15$