91Ó°ÊÓ

Since the pendulum has a simple harmonic motion, the maximal displacement from the center point (called amplitude) is half the distance between points A and B, i.e., \(\frac{1}{2} \times 20 \;\text{cm}\). Thus, the amplitude is \(10 \;\text{cm}\).

According to the problem, the pendulum takes 2 seconds to move from point A to point B, which is half the cycle of the motion. Therefore, the period of the motion, i.e., the time needed for the pendulum to complete one full cycle, is \(2 \times 2 = 4 \;\text{seconds}\).

Since the pendulum starts at the center line and moves to the right, we need a function that starts at 0 and has a value of 10 at \(t=1\). A cosine function is suitable for this task, as it starts at maximum value when the argument is 0. So, the function has the form: $$ d(t) = A \cos(\frac{2\pi}{T}t + p) $$ Where \(A\) is the amplitude, \(T\) is the period, and \(p\) is the phase shift. Now we can plug in the values for amplitude and period that we determined in steps 1 and 2: $$ d(t) = 10 \cos(\frac{2\pi}{4}t + p) $$ Since the pendulum starts at the center line, we know that \(d(0) = 0\). Plugging in \(t=0\), we get: $$ 0 = 10 \cos(p) $$ This means that \(p\) must be \(\pm \frac{\pi}{2}\), or in general, \(p = \pm \frac{\pi}{2} + 2n\pi\), where \(n\) is an integer. Since the pendulum is moving to the right from the center line, we can choose the positive phase shift: $$ p = \frac{\pi}{2} $$

Plugging the phase shift into the function, we get: $$ d(t) = 10 \cos(\frac{2\pi}{4}t + \frac{\pi}{2}) $$ Simplifying, we get the final rule of the function \(d(t)\): $$ d(t) = 10 \cos(\frac{\pi}{2}t + \frac{\pi}{2}) $$ This function describes the horizontal distance from the pendulum to the center line at time \(t\) seconds.