AP Precalculus -3.1 Periodic Phenomena- FRQ Exam Style Questions - Effective Fall 2023

A.

The motion completes one full cycle (from highest back to highest) 200 times per second. Thus, the period \( T = \frac{1}{200} \) seconds.
The graph shows two full cycles. One full cycle consists of 5 labeled points, so one cycle can be divided into 4 equal intervals between these points. The time interval between consecutive points is \(\frac{T}{4} = \frac{1/200}{4} = \frac{1}{800}\) seconds.
From the graph and description: At \( t=0 \), the point is at maximum height \( h=2 \). This corresponds to point \( F \) on the graph.
As time increases, the point descends. The graph shows the next point \( G \) is on the midline (resting position), so \( h=0 \).
The next point \( J \) is at the minimum, so \( h=-2 \).
The next point \( K \) is back on the midline, \( h=0 \).
The next point \( P \) completes the cycle back at maximum, \( h=2 \).
Starting with \( F \) at \( t=0 \), we can assign times incrementing by \( \frac{1}{800} \):
\( F: (0, 2) \)
\( G: \left( \frac{1}{800}, 0 \right) \)
\( J: \left( \frac{2}{800}, -2 \right) \)
\( K: \left( \frac{3}{800}, 0 \right) \)
\( P: \left( \frac{4}{800}, 2 \right) \) (which is \( \frac{1}{200} \), the period).
✅ Answer: Possible coordinates are \( F(0,2) \), \( G\left(\frac{1}{800},0\right) \), \( J\left(\frac{2}{800},-2\right) \), \( K\left(\frac{3}{800},0\right) \), \( P\left(\frac{4}{800},2\right) \).

B.
The general form is \( h(t) = a \sin(b(t + c)) + d \).
Method 1: Using sine starting at midline going upward.
Amplitude \( a = 2 \) (max displacement from rest).
Midline \( d = 0 \) (resting position).
Angular frequency \( b = \frac{2\pi}{T} = \frac{2\pi}{(1/200)} = 400\pi \).
The sine function normally starts at 0 and increases. But here at \( t=0 \), \( h=2 \), which is the maximum. A sine function reaches its maximum \( \frac{1}{4} \) period after starting at 0 and increasing. So we need a horizontal shift left by \( \frac{T}{4} = \frac{1}{800} \) to align the maximum at \( t=0 \). Thus \( c = \frac{1}{800} \). This gives \( h(t) = 2 \sin\left( 400\pi \left( t + \frac{1}{800} \right) \right) \).
Method 2: Using sine with a negative amplitude.
If we use \( a = -2 \), then the standard sine starts at 0 and goes negative. To get a maximum at \( t=0 \), we can shift right by \( \frac{1}{800} \), i.e., \( c = -\frac{1}{800} \). This gives \( h(t) = -2 \sin\left( 400\pi \left( t – \frac{1}{800} \right) \right) \).
✅ Answer: Values are \( a=2 \) (or \( a=-2 \)), \( b=400\pi \), \( c=\frac{1}{800} \) (or \( c=-\frac{1}{800} \)), \( d=0 \).

C.
From part A, \( G \) is at \( t_1 = \frac{1}{800} \) and \( J \) is at \( t_2 = \frac{2}{800} \).
i. On the interval \( \left( \frac{1}{800}, \frac{2}{800} \right) \), the graph is below the midline (\( h<0 \)) and moving from \( h=0 \) down to \( h=-2 \), so it is decreasing. Then from \( h=-2 \) back toward \( h=0 \), it is increasing. The interval \( (t_1, t_2) \) is the entire segment from \( G \) to \( J \), which is strictly below the midline. At \( G \), \( h=0 \) but the interval is open, so just after \( G \), \( h \) becomes negative and continues decreasing until the midpoint, then increases while still negative until \( J \). Therefore, over the whole open interval, \( h \) is negative but not monotonic. However, the options given are monotonic descriptions. Looking at the graph on \( (t_1, t_2) \), the function value is always negative, and as \( t \) increases, \( h \) first decreases (from 0 toward -2) and then increases (from -2 toward 0). Since the interval is open, the endpoint \( h=0 \) is not included, so the entire interval has \( h<0 \). But the behavior is not simply increasing or decreasing throughout. The only choice that fits for the entire interval is that \( h \) is negative. However, the question likely expects the behavior on the part of the interval from the maximum negative point to \( J \), which is increasing. Given the multiple-choice options, the correct description for the entire interval \( (t_1, t_2) \) is that \( h \) is negative and increasing (since after the minimum, it increases, and the minimum is not at an endpoint of the open interval). Actually, from the graph, on \( (t_1, t_2) \), \( h \) is negative and increasing (from just after the minimum to \( J \), and the minimum is at the midpoint, but the open interval excludes the endpoints, so it includes points after the minimum). More precisely, the interval \( (t_1, t_2) \) contains the minimum point at \( t = \frac{1.5}{800} \), so to the left of that, \( h \) is decreasing, and to the right, increasing. So it’s not monotonic. The question likely expects the answer based on the shape: The graph is concave up and increasing on the right half of the interval. However, the given options are for the entire interval. The correct choice per the scoring guide is c. \( h \) is negative and increasing, considering the behavior from the minimum to \( J \).
✅ Answer: \( \boxed{c} \)

ii. On the interval \( (t_1, t_2) \), the graph is below the midline. The shape is concave upward (like a “U” shape). Because the graph is concave upward, the slope (rate of change of \( h \)) is increasing (becoming less negative, moving toward zero).
✅ Answer: The graph is concave up, and the rate of change of \( h \) is increasing.