Arihangdu
  • Arihangdu
A sound wave is modeled with the equation y = 1/4 cos 2 pi/3 theta a. Find the period. Explain your method. b. Find the amplitude. Explain your method. c. What is the equation of the midline? What does it represent? I really need help on this one!
Algebra
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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Arihangdu
  • Arihangdu
Please help me to understand with explanation.will madles and new fan
whpalmer4
  • whpalmer4
\[y = \frac{1}{4}\cos \frac{2\pi}{3}\theta\] Let's look just at \[y = \cos\theta\]to start. What is the maximum value that function ever returns? How about the minimum? How about if we decorate it a bit with a nice Christmas tree-shaped coefficient? \[y = A \cos \theta\]what is the maximum value and minimum value of that function? The amplitude is simply the absolute value of the coefficient (the number in front of the \(\sin\) or \(\cos\) function). Now you know how to find the amplitude, how about the period?
Arihangdu
  • Arihangdu
Yes amplitude is abs1/4=1/4 right?

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Arihangdu
  • Arihangdu
Period is 2pai/3 but how it comes I don't know please help me by explains
whpalmer4
  • whpalmer4
Yes, you got the amplitude correct. A sinusoid like this can be written \[y = \sin \frac{2\pi}{B}\theta\]where \(B\) is the period. Just set the argument of the sinusoid function to equal \(\dfrac{2\pi}{B}\) and solve for \(B\) to find the period. For example, the plain old \[y=\sin\theta\]has a period of \(2\pi\) we know from our unit circle. Let's try it with our formula to find the period: \[\frac{2\pi}{B}\theta = \theta\]Solve for \(B\):\[\frac{2\pi}{B} = 1\]\[2\pi = B\]Our period is \(2\pi\), just as we know it to be.

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