- #1

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I think that the amplitude is 0.3 and my period is 0.6pi (circumference). Would there be any phase shifts left or right, up or down. How would the period be written in the function?

y=0.3sin10pi/3

- Thread starter emma3001
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- #1

- 42

- 0

I think that the amplitude is 0.3 and my period is 0.6pi (circumference). Would there be any phase shifts left or right, up or down. How would the period be written in the function?

y=0.3sin10pi/3

- #2

- 238

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I believe this problem is missing the information as to where the speck starts.

I'll assume that the speck starts at the lowest point of the wheel. There's an equation [tex]v=r\omega[/tex], where v is the velocity, r= radius, and [tex]\omega[/tex]=angular velocity. Therefore, [tex]\omega=\frac{v}{r}[/tex] and also [tex]\omega=\frac{d\theta}{dt}[/tex]

Find [tex]\omega[/tex]

Let's define the height function [tex]h(t) = r(1-cos(\theta))[/tex]. *For this height equation I use the line h(t)=0 as the ground, so you can change the equation if you use h(t)=0 as the center of the wheel. From here, find [tex]\theta [/tex] in terms of t and [tex]\omega[/tex]

I'll assume that the speck starts at the lowest point of the wheel. There's an equation [tex]v=r\omega[/tex], where v is the velocity, r= radius, and [tex]\omega[/tex]=angular velocity. Therefore, [tex]\omega=\frac{v}{r}[/tex] and also [tex]\omega=\frac{d\theta}{dt}[/tex]

Find [tex]\omega[/tex]

Let's define the height function [tex]h(t) = r(1-cos(\theta))[/tex]. *For this height equation I use the line h(t)=0 as the ground, so you can change the equation if you use h(t)=0 as the center of the wheel. From here, find [tex]\theta [/tex] in terms of t and [tex]\omega[/tex]

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