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25 votes
25 votes
A. y = 12 cos (12)B. y = cos (1) +12C. y = 12 cos (2x) + 12OD. y = 12 coB (-) +12

A. y = 12 cos (12)B. y = cos (1) +12C. y = 12 cos (2x) + 12OD. y = 12 coB (-) +12-example-1
User Allan S
by
2.8k points

1 Answer

15 votes
15 votes

To solve this question, we need to visualize the wheel. See below:

The motion f the pot is clearly a cosine motion, the minimum height is 0 instead of -12, so we can see that the cosine function has been shifted up by 12 units, so our equation is;


\begin{gathered} y=a\cos kx+12 \\ a\text{ is the amplitude of the height of the point above the centre only, so a =12} \\ y=12\cos kx+12 \\ k=(2\pi)/(\lambda) \\ \lambda=2*\pi*12=24\pi \\ so,k=(2\pi)/(24\pi)=(1)/(12) \\ \end{gathered}

Thus, the complete equation is;


y=12\cos ((1)/(12)x)+12

That is Option D

A. y = 12 cos (12)B. y = cos (1) +12C. y = 12 cos (2x) + 12OD. y = 12 coB (-) +12-example-1
User David BS
by
3.3k points
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