IB Mathematics: Applications and Interpretation Quiz: Periodic Function Modeling

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Periodic Function ModelingQuestion 1 of 20

A complex sound wave is created by combining two pure tones. Its pressure variation over time t is modelled by S(t) = 5\sin(220 $\pi$ t) + 3\sin(440 $\pi$ t).

What is the fundamental period of the combined sound wave S(t)?

\frac{1}{440}

\frac{1}{220}

\frac{1}{110}

\frac{3}{440}

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IB Mathematics: Applications and Interpretation Quiz

IB Mathematics: Applications and Interpretation Quiz: Periodic Function Modeling

Practice Periodic Function Modeling in IB Mathematics: Applications and Interpretation with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Periodic Function Modeling, giving you a quick way to practice the rules, question types, and explanations that matter most for IB Mathematics: Applications and Interpretation.

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Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

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Question 1

A complex sound wave is created by combining two pure tones. Its pressure variation over time t is modelled by S(t) = 5\sin(220 $\pi$ t) + 3\sin(440 $\pi$ t).

What is the fundamental period of the combined sound wave S(t)?

$\frac{1}{440}$ s
$\frac{1}{220}$ s
$\frac{1}{110}$ s (correct answer)
$\frac{3}{440}$ s

Explanation: The period of the first component,

5\sin(220\pi t)

, is

T_1 = \frac{2\pi}{220\pi} = \frac{1}{110}

s. The period of the second component,

3\sin(440\pi t)

, is

T_2 = \frac{2\pi}{440\pi} = \frac{1}{220}

s. The period of the combined wave is the least common multiple (LCM) of the individual periods. LCM

(\frac{1}{110}, \frac{1}{220}) = \frac{1}{110}

s. The combined pattern repeats every

\frac{1}{110}

seconds.

Question 2

The flow of current in an AC circuit is given by I(t) = 15\sin(100 $\pi$ t), where I is in amperes and t is in seconds.

What is the average value of the current over the first positive half-cycle of the function?

0 A
7.5 A
15 A
9.55 A (correct answer)

Explanation: When you encounter AC circuit problems involving sinusoidal functions, you need to distinguish between instantaneous values and average values over specific intervals. The key insight is that while a complete sine cycle averages to zero, a half-cycle has a non-zero average. To find the average value of

I(t) = 15\sin(100\pi t)

over the first positive half-cycle, you need to determine when this half-cycle occurs and then integrate. The function completes one full cycle when

100\pi t = 2\pi

, so

t = \frac{1}{50}

seconds. The first positive half-cycle runs from

t = 0

t = \frac{1}{100}

seconds. The average value formula gives us:

\text{Average} = \frac{1}{\frac{1}{100} - 0} \int_0^{1/100} 15\sin(100\pi t) \, dt

Evaluating this integral:

100 \cdot 15 \int_0^{1/100} \sin(100\pi t) \, dt = 1500 \left[-\frac{\cos(100\pi t)}{100\pi}\right]_0^{1/100} = \frac{15}{\pi}[\cos(0) - \cos(\pi)] = \frac{30}{\pi} \approx 9.55

A. Choice A (0 A) would be correct for a complete cycle, not a half-cycle. Choice B (7.5 A) incorrectly assumes the average is half the amplitude. Choice C (15 A) confuses the average with the maximum amplitude. Choice D is correct at 9.55 A. Remember: For sinusoidal functions, the average over a half-cycle is

\frac{2A}{\pi}

where A is the amplitude—a useful formula to memorize for AC circuit problems.

Question 3

The depth of water in a harbour varies sinusoidally with time. The minimum depth is 4 m, which occurs at 3:00 am. The subsequent maximum depth is 12 m, which occurs at 9:00 am. Let t be the number of hours after midnight.

Which of the following functions best models the water depth, d(t)?

$d(t) = 4\cos\left(\frac{\pi}{6}(t-9)\right) + 8$ (correct answer)
$d(t) = 4\cos\left(\frac{\pi}{12}(t-9)\right) + 8$
$d(t) = 8\cos\left(\frac{\pi}{6}(t-9)\right) + 4$
$d(t) = 4\cos\left(\frac{\pi}{6}(t-3)\right) + 8$

Explanation: The vertical shift (mean depth) is D = (12+4)/2 = 8 m. The amplitude is A = (12-4)/2 = 4 m. The time from a minimum to a maximum is half a period, so T/2 = 9 - 3 = 6 hours. The period is T = 12 hours. The angular frequency is B = 2π/12 = π/6. Using a cosine model, d(t) = Acos(B(t-C)) + D, the phase shift C corresponds to the time of a maximum. Since the maximum is at t=9, C=9. Thus, the model is d(t) = 4cos(π/6(t-9)) + 8.

Question 4

The voltage in an alternating current (AC) circuit is described by the function V(t) = 170 \cos(120 $\pi$ t - $\frac{$ \pi $}{4}$ ), where V is the voltage in volts and t is the time in seconds.

What is the first positive time t, in seconds, at which the voltage is exactly +100 V?

0.00121 s
0.00458 s (correct answer)
0.00917 s
0.01667 s

Explanation: Set V(t) = 100 and solve for t:

170 \cos(120\pi t - \frac{\pi}{4}) = 100 \implies \cos(120\pi t - \frac{\pi}{4}) = \frac{10}{17}

Let

u = 120\pi t - \frac{\pi}{4}

. The principal value is

u = \arccos(\frac{10}{17}) \approx 0.9416

radians. The general solutions for u are

\pm 0.9416 + 2n\pi

. We test the two cases for the first positive t. Case 1:

120\pi t - \frac{\pi}{4} = 0.9416 \implies 120\pi t \approx 1.7270 \implies t \approx 0.00458

. Case 2:

120\pi t - \frac{\pi}{4} = -0.9416 \implies 120\pi t \approx -0.1562 \implies t \approx -0.00041

. The first positive time is approximately 0.00458 s.

Question 5

When two sound waves of slightly different frequencies are played together, a phenomenon known as 'beats' is produced. This is a periodic variation in volume whose frequency is the 'beat frequency'. Two tuning forks with frequencies of 254 Hz and 258 Hz are struck simultaneously.

What is the frequency of the beats produced?

2 Hz
4 Hz (correct answer)
256 Hz
512 Hz

Explanation: The beat frequency is the absolute difference between the two individual frequencies. Given

f_1 = 254

Hz and

f_2 = 258

Hz, the beat frequency is

f_{beat} = |f_2 - f_1| = |258 - 254| = 4

Hz. This means the sound will get louder and softer 4 times per second.

Question 6

A weight is attached to a spring and released from a point 10 cm above its equilibrium position. It oscillates with a period of 2 seconds. Due to air resistance, the amplitude of oscillation is reduced by a constant factor each second, such that after 1 second the amplitude is 85% of its previous value.

Which function best models the displacement d(t) in cm from the equilibrium position at time t seconds? (Note: $e^{-0.1625} \approx 0.85$ )

$d(t) = 10e^{-0.15t} \cos(2\pi t)$
$d(t) = 10e^{-0.1625t} \cos(\pi t)$ (correct answer)
$d(t) = 10(1-0.15t) \cos(\pi t)$
$d(t) = 10e^{-0.1625t} \sin(\pi t)$

Explanation: The initial displacement is +10 cm from equilibrium, so a cosine model is appropriate, with d(0)=10. The period is T=2s, so the angular frequency is B = 2π/T = π. The amplitude decays exponentially. If the amplitude is multiplied by 0.85 each second, the amplitude function is

A(t) = 10(0.85)^t

. This is equivalent to

A(t) = 10(e^{\ln(0.85)})^t = 10e^{\ln(0.85)t}

. Since

\ln(0.85) \approx -0.1625

, the model is

d(t) = 10e^{-0.1625t} \cos(\pi t)

Question 7

In an ecosystem, the population of rabbits, R, and foxes, F, are modelled by the functions R(t) = 1000 \sin\left( $\frac{$ \pi $}{6}$ t\right) + 5000 and F(t) = 150 \sin\left(\frac{\pi}{6}(t-1.5)\right) + 400, where t is the time in months.

According to the models, what is the approximate fox population when the rabbit population is at its first peak?

250
400
506 (correct answer)
550

Explanation: The rabbit population, R(t), is at its first peak when the sine term is 1. This occurs when

\frac{\pi}{6}t = \frac{\pi}{2}

, which gives t = 3 months. To find the fox population at this time, we substitute t=3 into the function F(t):

F(3) = 150 \sin\left(\frac{\pi}{6}(3-1.5)\right) + 400 = 150 \sin\left(\frac{\pi}{4}\right) + 400

F(3) = 150\left(\frac{\sqrt{2}}{2}\right) + 400 \approx 106.07 + 400 \approx 506

Question 8

The height of a buoy oscillating in the sea is modelled by h(t) = A\cos(B(t-C)) + D, where A, B, C, and D are all positive constants.

Which of the following correctly interprets the quantity D - A?

The average height of the buoy.
The maximum height of the buoy.
The minimum height of the buoy. (correct answer)
The amplitude of the buoy's oscillation.

Explanation: In the model h(t) = A\cos(B(t-C)) + D, D represents the vertical shift or the mean height. A represents the amplitude, which is the maximum displacement from the mean height. The maximum height is D + A. The minimum height occurs when the cosine term is -1, so h_min = A(-1) + D = D - A.

Question 9

The height of a tide is modelled by H(t) = A\cos(B(t-4)) + 6.5. Time t is in hours after midnight. The time between a high tide and the next low tide is exactly 6 hours. The height of the high tide is 9 metres.

Find the first time after midnight (t=0) that the height of the tide is exactly 8 metres.

2.23 hours (correct answer)
4.00 hours
5.77 hours
6.00 hours

Explanation: The time between high and low tide is half a period, so T/2 = 6 hours, which means the period T = 12 hours. B = 2π/T = 2π/12 = π/6. The maximum height is D+A = 6.5+A = 9, so the amplitude A=2.5. The model is H(t) = 2.5cos(π/6(t-4)) + 6.5. We solve H(t) = 8:

2.5\cos(\frac{\pi}{6}(t-4)) + 6.5 = 8 \implies \cos(\frac{\pi}{6}(t-4)) = 0.6

Let u = π/6(t-4). Then u = arccos(0.6) ≈ ±0.9273 + 2kπ. For the first positive t, we test u = -0.9273:

\frac{\pi}{6}(t-4) = -0.9273 \implies t-4 \approx -1.771 \implies t \approx 2.229

. The other solution (from u = +0.9273) gives t ≈ 5.771. The first time is t ≈ 2.23 hours.

Question 10

The distance of a moon from its planet is periodic. The maximum distance is 350,000 km and the minimum distance is 150,000 km. It completes one orbit every 20 days. At time t=0, the moon is at its mean distance from the planet and moving towards its maximum distance. Let d(t) be the distance in thousands of km.

Which of the following functions correctly models the moon's distance from the planet?

$d(t) = 100\sin\left(\frac{\pi}{10}t\right) + 250$ (correct answer)
$d(t) = 100\cos\left(\frac{\pi}{10}t\right) + 250$
$d(t) = -100\sin\left(\frac{\pi}{10}t\right) + 250$
$d(t) = 250\sin\left(\frac{\pi}{10}t\right) + 100$

Explanation: The mean distance (vertical shift) is D = (350+150)/2 = 250. The amplitude is A = (350-150)/2 = 100. The period is T=20 days, so B = 2π/20 = π/10. The model is of the form d(t) = 100f(π/10(t-C)) + 250. The condition 'at t=0, moon is at mean distance' means the sinusoidal term is zero. This suggests a sine function with C=0. The condition 'moving towards its maximum distance' means the function is increasing at t=0. For d(t) = 100sin(π/10 t) + 250, d(0)=250 and the derivative d'(0) > 0, so it is increasing. This matches the conditions.

Question 11

The height of a rider on a Ferris wheel is modelled by $h(t) = 25\sin\left(\frac{\pi}{20}(t-10)\right) + 28$ , where h is the height in metres and t is the time in seconds.

During the first two full revolutions, what is the total amount of time that the rider is above 40.5 metres?

13.33 s
20.00 s
40.00 s
26.67 s (correct answer)

Explanation: When you encounter a Ferris wheel problem involving trigonometric functions, you're dealing with periodic motion where you need to find when the function exceeds a certain threshold value. To find when the rider is above 40.5 meters, set up the inequality:

25\sin\left(\frac{\pi}{20}(t-10)\right) + 28 > 40.5

Solving this:

25\sin\left(\frac{\pi}{20}(t-10)\right) > 12.5

, which gives

\sin\left(\frac{\pi}{20}(t-10)\right) > 0.5

Since

\sin(\theta) > 0.5

when

\frac{\pi}{6} < \theta < \frac{5\pi}{6}

, you have:

\frac{\pi}{6} < \frac{\pi}{20}(t-10) < \frac{5\pi}{6}

Solving for t:

\frac{10}{3} < t-10 < \frac{50}{3}

, so

\frac{40}{3} < t < \frac{80}{3}

This means the rider is above 40.5m from

t = \frac{40}{3}

t = \frac{80}{3}

seconds in each revolution. The duration per revolution is

\frac{80}{3} - \frac{40}{3} = \frac{40}{3}

seconds. Since the period is

\frac{2\pi}{\pi/20} = 40

seconds, two full revolutions take 80 seconds. The pattern repeats in the second revolution, so the total time above 40.5m is

2 \times \frac{40}{3} = \frac{80}{3} = 26.67

seconds. Answer D (26.67s) is correct. Answer A (13.33s) gives only one revolution's worth. Answer B (20.00s) might come from incorrect trigonometric calculations. Answer C (40.00s) represents the full period duration, not the time above the threshold. Remember: in periodic motion problems, always identify the period first, then solve for one cycle and multiply by the number of cycles requested.

Question 12

The height, h metres, of a passenger on a Ferris wheel at time t seconds is modelled by h(t) = -20 \cos\left( $\frac{$ \pi $}{15}$ t\right) + 22. The wheel completes one full revolution in 30 seconds.

During one full revolution, for how long is a passenger at least 32 metres above the ground?

5.0 seconds
10.0 seconds (correct answer)
15.0 seconds
20.0 seconds

Explanation: We need to solve the inequality h(t) ≥ 32.

-20 \cos\left(\frac{\pi}{15} t\right) + 22 \ge 32 \implies -20 \cos\left(\frac{\pi}{15} t\right) \ge 10 \implies \cos\left(\frac{\pi}{15} t\right) \le -0.5

Using a GDC or inverse trigonometry, the interval for the argument

\frac{\pi}{15} t

within

[0, 2\pi]

[2\pi/3, 4\pi/3]

. Solving for t:

\frac{2\pi}{3} \le \frac{\pi}{15} t \le \frac{4\pi}{3} \implies 10 \le t \le 20

The duration is 20 - 10 = 10 seconds.

Question 13

A complex sound wave is created by combining two pure tones. Its pressure variation over time t is modelled by S(t) = 5\sin(220 $\pi$ t) + 3\sin(440 $\pi$ t).

What is the fundamental period of the combined sound wave S(t)?

$\frac{1}{440}$ s
$\frac{1}{220}$ s
$\frac{1}{110}$ s (correct answer)
$\frac{3}{440}$ s

Explanation: The period of the first component,

5\sin(220\pi t)

, is

T_1 = \frac{2\pi}{220\pi} = \frac{1}{110}

s. The period of the second component,

3\sin(440\pi t)

, is

T_2 = \frac{2\pi}{440\pi} = \frac{1}{220}

s. The period of the combined wave is the least common multiple (LCM) of the individual periods. LCM

(\frac{1}{110}, \frac{1}{220}) = \frac{1}{110}

s. The combined pattern repeats every

\frac{1}{110}

seconds.

Question 14

The number of daylight hours, H, in a particular city is modelled by the function H(t) = 3.5\cos\left(\frac{2\pi}{365}(t-172)\right) + 12.2, where t is the day number of the year (t=1 for January 1st).

What is the approximate range of the number of daylight hours in this city over one year?

[3.5, 12.2] hours
[8.7, 15.7] hours (correct answer)
[12.2, 15.7] hours
[0, 15.7] hours

Explanation: The function is a cosine wave with amplitude A=3.5 and vertical shift D=12.2. The maximum value is D + A = 12.2 + 3.5 = 15.7 hours. The minimum value is D - A = 12.2 - 3.5 = 8.7 hours. Therefore, the range of daylight hours is [8.7, 15.7].

Question 15

Find the first time after midnight (t=0) that the height of the tide is exactly 8 metres.

2.23 hours (correct answer)
4.00 hours
5.77 hours
6.00 hours

2.5\cos(\frac{\pi}{6}(t-4)) + 6.5 = 8 \implies \cos(\frac{\pi}{6}(t-4)) = 0.6

Let u = π/6(t-4). Then u = arccos(0.6) ≈ ±0.9273 + 2kπ. For the first positive t, we test u = -0.9273:

\frac{\pi}{6}(t-4) = -0.9273 \implies t-4 \approx -1.771 \implies t \approx 2.229

. The other solution (from u = +0.9273) gives t ≈ 5.771. The first time is t ≈ 2.23 hours.

Question 16

The monthly sales of air conditioners in a Northern Hemisphere city are modelled by $S_N(t) = 200\cos\left(\frac{\pi}{6}(t-7)\right) + 350$ . In a Southern Hemisphere city, the sales are modelled by $S_S(t) = 150\cos\left(\frac{\pi}{6}(t-1)\right) + 250$ . In both models, t is the month number, with t=1 representing January.

Which statement accurately describes the relationship between the peak sales months in the two cities?

Peak sales in the South occur 6 months before peak sales in the North. (correct answer)
Peak sales in the South occur 6 months after peak sales in the North.
Peak sales occur in the same month in both hemispheres.
Peak sales in the South occur 1 month after peak sales in the North.

Explanation: Peak sales occur when the cosine term is 1. For the North, this is when

\frac{\pi}{6}(t-7) = 0

, so t=7 (July). For the South, this is when

\frac{\pi}{6}(t-1) = 0

, so t=1 (January). The difference in peak times is 7 - 1 = 6 months. Since January (t=1) comes before July (t=7), the peak sales in the South occur 6 months before the peak in the North, reflecting the opposite seasons.

Question 17

Which function best models the displacement d(t) in cm from the equilibrium position at time t seconds? (Note: $e^{-0.1625} \approx 0.85$ )

$d(t) = 10e^{-0.15t} \cos(2\pi t)$
$d(t) = 10e^{-0.1625t} \cos(\pi t)$ (correct answer)
$d(t) = 10(1-0.15t) \cos(\pi t)$
$d(t) = 10e^{-0.1625t} \sin(\pi t)$

A(t) = 10(0.85)^t

. This is equivalent to

A(t) = 10(e^{\ln(0.85)})^t = 10e^{\ln(0.85)t}

. Since

\ln(0.85) \approx -0.1625

, the model is

d(t) = 10e^{-0.1625t} \cos(\pi t)

Question 18

During one full revolution, for how long is a passenger at least 32 metres above the ground?

5.0 seconds
10.0 seconds (correct answer)
15.0 seconds
20.0 seconds

Explanation: We need to solve the inequality h(t) ≥ 32.

-20 \cos\left(\frac{\pi}{15} t\right) + 22 \ge 32 \implies -20 \cos\left(\frac{\pi}{15} t\right) \ge 10 \implies \cos\left(\frac{\pi}{15} t\right) \le -0.5

Using a GDC or inverse trigonometry, the interval for the argument

\frac{\pi}{15} t

within

[0, 2\pi]

[2\pi/3, 4\pi/3]

. Solving for t:

\frac{2\pi}{3} \le \frac{\pi}{15} t \le \frac{4\pi}{3} \implies 10 \le t \le 20

The duration is 20 - 10 = 10 seconds.

Question 19

A swinging pendulum's displacement from its central position, d cm, is modelled by a damped harmonic motion function d(t) = 15e^{-0.05t} \cos(4 $\pi$ t), where t is the time in seconds.

The amplitude of the pendulum's swing decreases over time. To the nearest hundredth of a second, how long does it take for the amplitude to be halved?

7.50 seconds
13.86 seconds (correct answer)
20.00 seconds
27.73 seconds

Explanation: The amplitude of the oscillation is given by the function A(t) = 15e^{-0.05t}. The initial amplitude (at t=0) is 15 cm. We need to find the time t when the amplitude is halved, i.e., A(t) = 7.5.

15e^{-0.05t} = 7.5 \implies e^{-0.05t} = 0.5

Taking the natural logarithm of both sides:

-0.05t = \ln(0.5) \implies t = \frac{\ln(0.5)}{-0.05} \approx 13.86 \text{ seconds.}

Question 20

The voltage in an alternating current (AC) circuit is described by the function V(t) = 170 \cos(120 $\pi$ t - $\frac{$ \pi $}{4}$ ), where V is the voltage in volts and t is the time in seconds.

What is the first positive time t, in seconds, at which the voltage is exactly +100 V?

0.00121 s
0.00458 s (correct answer)
0.00917 s
0.01667 s

Explanation: Set V(t) = 100 and solve for t:

170 \cos(120\pi t - \frac{\pi}{4}) = 100 \implies \cos(120\pi t - \frac{\pi}{4}) = \frac{10}{17}

Let

u = 120\pi t - \frac{\pi}{4}

. The principal value is

u = \arccos(\frac{10}{17}) \approx 0.9416

radians. The general solutions for u are

\pm 0.9416 + 2n\pi

. We test the two cases for the first positive t. Case 1:

120\pi t - \frac{\pi}{4} = 0.9416 \implies 120\pi t \approx 1.7270 \implies t \approx 0.00458

. Case 2:

120\pi t - \frac{\pi}{4} = -0.9416 \implies 120\pi t \approx -0.1562 \implies t \approx -0.00041

. The first positive time is approximately 0.00458 s.