Solved Exercises on Modeling Periodic Phenomena with Trig in Algebra 2

Master modeling periodic phenomena: sine, cosine, amplitude, period, phase shift, and real-world applications.

Solution: Exercises 1 to 3
1 Basic sine model
Exercise 1
The height of a buoy in the ocean follows a periodic pattern. It oscillates between 2 feet and 8 feet above sea level with a period of 6 seconds. Write a sine function to model the height of the buoy as a function of time, assuming it starts at sea level at t = 0.
Definition:

Periodic function: f(x) = A sin(Bx - C) + D, where A = amplitude, 2π/B = period, C/B = phase shift, D = vertical shift.

Modeling method:
  1. Identify the maximum and minimum values
  2. Calculate amplitude and vertical shift
  3. Determine the period
  4. Find the phase shift based on initial conditions
  5. Write the function
Max/Min
8 ft / 2 ft
Period
6 seconds
Function
h(t) = 3sin(πt/3)
Step 1: Identify maximum and minimum heights

Maximum height = 8 feet, Minimum height = 2 feet

Step 2: Calculate amplitude

Amplitude = (max - min)/2 = (8 - 2)/2 = 6/2 = 3 feet

Step 3: Calculate vertical shift

Vertical shift = (max + min)/2 = (8 + 2)/2 = 10/2 = 5 feet

Step 4: Determine period

Given period = 6 seconds

So 2π/B = 6, which means B = 2π/6 = π/3

Step 5: Determine phase shift

At t = 0, height = sea level = 5 feet (vertical shift)

This means we start at the midline, so we can use a sine function without phase shift

h(t) = 3sin(πt/3) + 5
Final answer:

The height of the buoy is modeled by h(t) = 3sin(πt/3) + 5, where h is in feet and t is in seconds.

Applied rules:

• Amplitude: Half the distance between max and min

• Vertical shift: Average of max and min values

• Period formula: 2π/B = given period

• Initial condition: Determines phase shift

2 Cosine with phase shift
Exercise 2
The temperature in a city varies sinusoidally throughout the day. The maximum temperature of 80°F occurs at 2 PM, and the minimum temperature of 60°F occurs at 2 AM. Write a cosine function to model the temperature T in terms of time t, where t = 0 represents midnight.
Definition:

Temperature cycle: Cosine function peaks at maximum value. The time from max to min is half the period.

Max/Min
80°F / 60°F
Time
2PM to 2AM = 12h
Function
T(t) = 10cos(π(t-14)/12) + 70
Step 1: Identify maximum and minimum temperatures

Maximum temperature = 80°F at 2 PM (t = 14)

Minimum temperature = 60°F at 2 AM (t = 2)

Step 2: Calculate amplitude and vertical shift

Amplitude = (80 - 60)/2 = 10°F

Vertical shift = (80 + 60)/2 = 70°F

Step 3: Determine the period

Time from max to min = 12 hours (2 AM to 2 PM)

So half period = 12 hours, full period = 24 hours

Therefore B = 2π/24 = π/12

Step 4: Find phase shift

Maximum occurs at t = 14 (2 PM)

For cosine: maximum occurs when argument equals 0

So π/12(t - C) = 0 when t = 14

Therefore C = 14

Step 5: Write the function

T(t) = 10cos(π(t - 14)/12) + 70

T(t) = 10cos(π(t-14)/12) + 70
Final answer:

The temperature is modeled by T(t) = 10cos(π(t-14)/12) + 70, where T is in °F and t is hours after midnight.

Applied rules:

• Cosine peaks: Maximum occurs when argument = 0

• Half period: Time from max to min

• Phase shift: Adjusts the starting time of the cycle

3 Simple harmonic motion
Exercise 3
A mass attached to a spring oscillates with amplitude 4 cm and period 3 seconds. At t = 0, the mass is at its maximum displacement. Write a function to model the displacement d from equilibrium as a function of time.
Definition:

Simple harmonic motion: Modeled by d(t) = A cos(Bt) when starting at maximum displacement.

Amplitude
4 cm
Period
3 seconds
Function
d(t) = 4cos(2πt/3)
Step 1: Identify the amplitude

Amplitude = 4 cm (given)

Step 2: Determine the period

Period = 3 seconds (given)

So 2π/B = 3, which means B = 2π/3

Step 3: Determine the phase shift

At t = 0, the mass is at maximum displacement

This matches the behavior of cosine function: cos(0) = 1

So no phase shift is needed

Step 4: Write the function

d(t) = 4cos(2πt/3)

Step 5: Verify initial condition

d(0) = 4cos(0) = 4(1) = 4 cm ✓

d(t) = 4cos(2πt/3)
Final answer:

The displacement is modeled by d(t) = 4cos(2πt/3), where d is in cm and t is in seconds.

Applied rules:

• Starting at max: Use cosine function

• Starting at min: Use negative cosine function

• Starting at equilibrium: Use sine function

Periodic Modeling Fundamentals
f(t) = A sin(Bt - C) + D
General Form
Amplitude
|A|
Max deviation
Period
2π/|B|
Cycle length
Phase Shift
C/B
Horizontal shift
Vertical Shift
D
Midline
Key definitions:

Periodic phenomenon: A phenomenon that repeats at regular intervals

Amplitude: Maximum displacement from the midline

Period: Time required for one complete cycle

Phase shift: Horizontal displacement of the function

Vertical shift: Vertical displacement of the midline

Real-world examples: Tides, temperature cycles, sound waves, light waves, pendulum motion, alternating current
Function selection: Use cosine when starting at max/min, sine when starting at midline
Tip 1: Always identify the max and min values first to find amplitude and vertical shift.
Tip 2: Time from max to min is half the period.
Tip 3: If starting at equilibrium going up, use positive sine; if going down, use negative sine.
Tip 4: Check your function with initial conditions to verify correctness.
Solution: Exercises 4 to 5
4 Tide modeling
Exercise 4
The depth of water at a dock varies sinusoidally. At high tide (4:00 AM), the depth is 15 feet. At low tide (10:00 AM), the depth is 3 feet. Write a cosine function to model the depth D in terms of time t, where t = 0 represents midnight. Then find the depth at 1:00 PM.
Definition:

Tidal patterns: Typically modeled with cosine functions since high tide often represents the maximum value.

High/Low
15 ft / 3 ft
Time
4AM to 10AM = 6h
Function
D(t) = 6cos(π(t-4)/6) + 9
Step 1: Identify high and low tide depths

High tide = 15 feet at t = 4 (4:00 AM)

Low tide = 3 feet at t = 10 (10:00 AM)

Step 2: Calculate amplitude and vertical shift

Amplitude = (15 - 3)/2 = 6 feet

Vertical shift = (15 + 3)/2 = 9 feet

Step 3: Determine the period

Time from high to low = 6 hours (4 AM to 10 AM)

So half period = 6 hours, full period = 12 hours

Therefore B = 2π/12 = π/6

Step 4: Find phase shift

Maximum occurs at t = 4

For cosine: maximum occurs when argument = 0

So π/6(t - C) = 0 when t = 4

Therefore C = 4

Step 5: Write the function

D(t) = 6cos(π(t - 4)/6) + 9

Step 6: Find depth at 1:00 PM (t = 13)

D(13) = 6cos(π(13 - 4)/6) + 9 = 6cos(9π/6) + 9 = 6cos(3π/2) + 9 = 6(0) + 9 = 9 feet

D(t) = 6cos(π(t-4)/6) + 9
Depth at 1 PM: 9 feet
Final answer:

The depth is modeled by D(t) = 6cos(π(t-4)/6) + 9, and the depth at 1:00 PM is 9 feet.

Applied rules:

• High to low: Represents half the period

• Trig evaluation: cos(3π/2) = 0

• Time conversion: Convert 1 PM to t = 13 hours after midnight

5 Sound wave modeling
Exercise 5
A sound wave has frequency 440 Hz (cycles per second) and amplitude 0.5. Write a sine function to model the pressure variation P as a function of time t in seconds. Find the period and the pressure at t = 0.002 seconds.
Definition:

Sound waves: Pressure variations modeled by trigonometric functions. Frequency f = 1/period, so B = 2πf.

Amplitude
0.5
Frequency
440 Hz
Function
P(t) = 0.5sin(880πt)
Step 1: Identify the amplitude

Amplitude = 0.5 (given)

Step 2: Calculate B from frequency

Frequency f = 440 Hz

Period = 1/frequency = 1/440 seconds

Since period = 2π/B, we have 1/440 = 2π/B

Therefore B = 2π × 440 = 880π

Step 3: Write the function

Assuming the wave starts at equilibrium, we use sine:

P(t) = 0.5sin(880πt)

Step 4: Find the period

Period = 1/frequency = 1/440 ≈ 0.00227 seconds

Step 5: Find pressure at t = 0.002 seconds

P(0.002) = 0.5sin(880π × 0.002) = 0.5sin(1.76π) = 0.5sin(1.76π) ≈ 0.5sin(0.76π) ≈ 0.5(0.951) ≈ 0.476

P(t) = 0.5sin(880πt)
Period: 1/440 s
P(0.002): ≈ 0.476
Final answer:

The pressure is modeled by P(t) = 0.5sin(880πt), with period 1/440 seconds, and the pressure at t = 0.002 seconds is approximately 0.476.

Applied rules:

• Frequency relationship: B = 2π × frequency

• Period-frequency: Period = 1/frequency

• Sound wave modeling: Often uses sine for starting at equilibrium

Detailed Summary: Modeling Periodic Phenomena with Trig
f(t) = A sin(Bt - C) + D or A cos(Bt - C) + D
General Form
Key definitions:

Periodic function: A function that repeats its values at regular intervals

Amplitude (A): The maximum displacement from the midline

Period (2π/B): The length of one complete cycle

Phase shift (C/B): Horizontal displacement of the function

Vertical shift (D): Vertical displacement of the midline

Modeling Methodology:
  1. Identify key information: Max/min values, period, phase shift, vertical shift
  2. Calculate parameters: Amplitude, period, vertical shift
  3. Determine function type: Sine or cosine based on initial conditions
  4. Find phase shift: Based on when max/min occurs
  5. Write function: Substitute all parameters
  6. Verify: Check that function satisfies given conditions
Tip 1: Always calculate amplitude as (max - min)/2 and vertical shift as (max + min)/2.
Tip 2: Use cosine when the phenomenon starts at maximum or minimum value.
Tip 3: Use sine when the phenomenon starts at the equilibrium position.
Tip 4: Remember that frequency = 1/period, so B = 2π × frequency.
Common errors: Forgetting to halve the time for half-period calculations, misidentifying initial conditions, incorrect phase shift calculations.
Exam preparation: Practice identifying max/min from context, memorize parameter relationships, master phase shift calculations.
Essential formulas to know:

• General form: f(t) = A sin(Bt - C) + D or A cos(Bt - C) + D

• Amplitude: A = (max - min)/2

• Vertical shift: D = (max + min)/2

• Period: 2π/B = given period

• Frequency: f = 1/period = B/(2π)

• Phase shift: C/B (positive = right, negative = left)

• Half period: Time from max to min or min to max

Periodic Phenomena: Comparison of Models
Exercise 6: Different Periodic Functions
Compare: d(t) = 3sin(πt/6), h(t) = 4cos(πt/3), p(t) = 2sin(2πt) + 5

Analysis: The chart shows different periodic behaviors with varying amplitudes, periods, and midlines.

  • d(t) = 3sin(πt/6): Amplitude 3, Period 12, Midline y = 0
  • h(t) = 4cos(πt/3): Amplitude 4, Period 6, Midline y = 0
  • p(t) = 2sin(2πt) + 5: Amplitude 2, Period 1, Midline y = 5

Questions & Answers

Question: How do I decide whether to use sine or cosine when modeling periodic phenomena?

Answer: The choice depends on the initial conditions of the phenomenon:

Use cosine when:

  • The phenomenon starts at maximum value: f(0) = A
  • The phenomenon starts at minimum value: f(0) = -A
  • Maximum or minimum occurs at the reference time

Use sine when:

  • The phenomenon starts at equilibrium (midline): f(0) = D
  • The phenomenon starts at equilibrium and increases: f(0) = D, f'(0) > 0
  • The phenomenon starts at equilibrium and decreases: f(0) = D, f'(0) < 0

Example: For a mass on a spring, if released from maximum displacement, use cosine. If released from equilibrium with an initial push, use sine.

Both functions can model the same phenomenon with different phase shifts, but choosing the right one eliminates the need for phase shift calculation.

Question: I'm confused about how to calculate the phase shift. Can you explain it more clearly?

Answer: The phase shift is the horizontal displacement of the function:

Formula: Phase shift = C/B for functions in the form A sin(Bx - C) + D or A cos(Bx - C) + D

Direction:

  • If C/B > 0, the graph shifts right
  • If C/B < 0, the graph shifts left

How to find C:

  1. Determine when a significant event occurs (maximum, minimum, or equilibrium)
  2. Set the argument of the trig function equal to the angle that produces that event
  3. Solve for the phase shift

Example: If maximum occurs at t = 3 in f(t) = A cos(Bt - C) + D, then B(3) - C = 0, so C = 3B, and phase shift = C/B = 3.

Question: How do I handle real-world problems where the time isn't given in a convenient format?

Answer: Always establish a consistent time reference system:

Time conversion strategies:

  • Set t = 0 at a convenient reference time (like midnight, noon, or the time of maximum)
  • Convert all given times to this reference (e.g., 3:00 PM = 15 hours after midnight)
  • Be consistent with units (hours, minutes, seconds)

Example: If high tide is at 2:30 PM and low tide at 8:45 PM:

  • Set t = 0 at midnight
  • High tide at t = 14.5 (2:30 PM)
  • Low tide at t = 20.75 (8:45 PM)
  • Half period = 20.75 - 14.5 = 6.25 hours
  • Full period = 12.5 hours

Always define your time reference clearly and convert all times consistently!

Question: What's the difference between frequency and period? How do they relate to the B value?

Answer: Period and frequency are reciprocals of each other:

Period (T): Time for one complete cycle (measured in seconds, hours, etc.)

Frequency (f): Number of cycles per unit time (measured in Hz, cycles per hour, etc.)

Relationship: f = 1/T and T = 1/f

Connection to B: In f(t) = A sin(Bt) + D:

  • Period = 2π/B
  • Therefore: B = 2π/Period
  • Since f = 1/Period: B = 2πf

Example: If frequency is 5 Hz, then B = 2π(5) = 10π, and period = 1/5 = 0.2 seconds.

Higher frequency means more cycles per unit time and a larger B value.

Question: How do I verify that my model is correct?

Answer: Here's a verification checklist:

1. Check amplitude: Does |A| = (max - min)/2?

2. Check vertical shift: Does D = (max + min)/2?

3. Check period: Does the function repeat after the given period?

4. Check initial conditions: Does the function match the given starting values?

5. Check significant events: Do maxima and minima occur at the given times?

Example verification for Exercise 1: h(t) = 3sin(πt/3) + 5

  • Amplitude: |3| = 3, (8-2)/2 = 3 ✓
  • Vertical shift: 5, (8+2)/2 = 5 ✓
  • Period: 2π/(π/3) = 6 seconds ✓
  • Initial: h(0) = 3sin(0) + 5 = 5 ✓

Always verify your model with the original problem conditions!