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Total score: ____ of 20 points A marching band performs on the football field at half-time. As they perform, the members of the band stand in the shape of a sinusoidal function. While playing, they move, but still maintain the sinusoidal function, transforming it in different ways. Darla is a member of the marching band. As the band begins to play she is positioned in the exact center of the field. The person closest to her on the same horizontal line, stands 10 yards away. The sinusoidal function extends to the ends of the playing field. The playing area of football field measure 300 feet by 160 feet. Place the playing area of a football field on the coordinate plane such that the origin is the lower left corner of the football field. (Score for Question 1: ___ of 2 points) 1. What is the period and the amplitude of the sine function representing the position of the band members as they begin to play? Answer: (Score for Question 2: ___ of 6 points) 2. Edna is sitting in the stands and is facing Darla. Edna observes that sine curve begins by increasing at the far left of the field. What is the equation of the sine function representing the position of band members as they begin to play? Answer: (Score for Question 3: ___ of 4 points) 3. As the band begins to play, band members move away from the edges, and the curve reverses so that the function begins at the far left by decreasing. Darla does not move. The sine curve is now half as tall as it was originally. What is the equation of the sine curve representing the position of the band members after these changes? Answer: (Score for Question 4: ___ of 3 points) 4. Next, the entire band moves closer to the edge of the football field so that the sine curve is in the lower half of the football field from Edna’s vantage point. What is the equation of the sine curve representing the position of the band members after these changes? Answer: (Score for Question 5: ___ of 5 points) 5. At the end of the performance, the band marches off the field to the right, moving the entire sine curve. Asa grabs his camera to takes a picture of the entire football field. At the instant he takes the picture, the first person forming the curve now stands at the 5 yard line. What is the equation of the sine curve representing the position of the band members in Asa’s picture? Answer: Please help me explain step-by-step thank you

Respuesta :

Answer:

1) Amplitude; A = 80 ft

Period = 60 ft

2)y = 80 sin ((Ï€/30)x - 5Ï€) + 80

3)y = 40 sin ((Ï€/30)x - 5Ï€) + 80

4)y = 40 sin ((Ï€/30)x - 5Ï€) + 40

5)y = -65sin ((Ï€/30)x - 5Ï€) + 80

Step-by-step explanation:

The general formula for sinusoidal wave equation is given by;

y = A sin (Bx - C) + D

Where;

A is amplitude = D_max or  D_min

Period = 2Ï€/B

So; B = 2Ï€/Period

Phase Shift = C/B  

So; C = B · Phase Shift

D: center

We are Given:

Height of the field is 160 ft and so the center is at y = 80. Thus; D = 80 ft

Thus; A = 80 ft

The person closest to Darla on the same horizontal line, stands 10 yards(30 ft) Thus, period = 2 × 30 = 60 ft

Thus; B = 2π/60 = π/30

Field is 300 ft wide and so the center is 300/2 = 150 ft

Thus; Phase Shift = 150.  

C = B × Phase Shift = π/30 · 150 = 5π

1) From the calculations above,

Amplitude; A = 80 ft

Period = 60 ft

2) As they begin to play, from the calculations above and y = A sin (Bx - C) + D, equation of the sine function is now;

y = 80 sin ((Ï€/30)x - 5Ï€) + 80

3) In this, since the sine wave is half as tall, then after the changes, we have;

y = 40 sin ((Ï€/30)x - 5Ï€) + 80

4) since they have moved closer, then equation is now;

y = 40 sin ((Ï€/30)x - 5Ï€) + 40

5) We are Given:

Height of the field is 160 ft and so the center is at y = 80. Thus; D = 80 ft

Since the first person forming the curve now stands at the 5 yard line, the minimum is at 5 yds (15 ft). Thus;

D_min = 80 - 15 = 65. Thus; A = 65 ft

The person closest to Darla on the same horizontal line, stands 10 yards(30 ft) Thus, period = 2 × 30 = 60 ft

Thus; B = 2π/60 = π/30

Field is 300 ft wide and so the center is 300/2 = 150 ft

Thus; Phase Shift = 150.  

C = B × Phase Shift = π/30 · 150 = 5π

The band ends down (at 15 feet) and thus A is negative

The equation is;

y = -65sin ((Ï€/30)x - 5Ï€) + 80

Part(1): The required values are,

Amplitude=[tex]80 ft[/tex] and Period: [tex]60 sec[/tex]

Part(2):

The equation of the sine function is

[tex]y=80 cos(\frac{\pi x}{30}+\pi)+80[/tex]

Part(3):

The equation of the sine curve is,

[tex]y=40cos(\frac{\pi x}{30})+80[/tex]

Part(4):

The equation of the  sine curve representing the position of the band members after these changes is [tex]y=40cos(\frac{\pi x}{30})+40[/tex]

Part(5):

The required graph is attached below,

Simple harmonic motion:

Simple Harmonic Motion or SHM is defined as a motion in which the restoring force is directly proportional to the displacement of the body from its mean position

Part(1):

Amplitude=[tex]\frac{1}{2} width=80ft[/tex]

Period=[tex]2\times 30=60 sec[/tex]

Part(2):

Let the equation be,

[tex]y=80 cos(\frac{\pi x}{30}+\pi)+80\\ y'=-\frac{8\pi}{3} sin((\frac{\pi x}{30}+\pi))[/tex]

Darla is at the point [tex]D(150,80)[/tex] which is on the graph at [tex]x=0[/tex] then,

[tex]80=80 cos(5\pi+\pi)=0\\y=-80 cos (\frac{\pi x}{30} )+80[/tex]

Part(3):

Since the wave is now [tex]\frac{160}{2} =80 ft[/tex] then,

Amplitude=40 ft

[tex]y=-4 cos (\frac{\pi x}{30} -\pi)+80\\y=40cos(\frac{\pi x}{30})+80[/tex]

Part(4):

The graph shifts downward 40 ft then,

[tex]y=-4 cos (\frac{\pi x}{30} -\pi)+80-40\\y=40cos(\frac{\pi x}{30})+40[/tex]

Part(5):

Start at:[tex]Y(x)=80sin (\frac{2\pi x}{60} )+80\\[/tex]

End at: [tex]Z(x)=80sin[ (\frac{2\pi }{60}(x-15) )]+80\\[/tex]

The graph is attached below:

Learn more about the topic of Simple harmonic motion:

https://brainly.com/question/14446439

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