anonymous
  • anonymous
HEEELLLPPPP PLEASEEEE! A sound wave is modeled with the equation y =1/4 cos 2pi/3 theta a. Find the period. Explain your method. b. Find the amplitude. Explain your method
Mathematics
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anonymous
  • anonymous
HEEELLLPPPP PLEASEEEE! A sound wave is modeled with the equation y =1/4 cos 2pi/3 theta a. Find the period. Explain your method. b. Find the amplitude. Explain your method
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
note that waves can be of general eq, y=r cos wt, where r is the amplitude, w the angular velocity and t is the time parameter.
anonymous
  • anonymous
This looks like physics.
anonymous
  • anonymous
well do you know the answer?

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anonymous
  • anonymous
Yes but just so you know. This is a physics question not mathematics.
anonymous
  • anonymous
cross reference your eq with the general eq above.
anonymous
  • anonymous
ok well can i have the answer?
anonymous
  • anonymous
did you try to do it yourself first?
anonymous
  • anonymous
yes and i am beyind confused :/
anonymous
  • anonymous
what did you get? by the way, your equation didnot involve time ......
anonymous
  • anonymous
cause thats not what it asked for i just need the period of the graph and what the amplitude would be you do this by using trig and i got the amplitude is 4 and the period is 2 but i dont think that is right
anonymous
  • anonymous
\[\omega = \frac{ 2\pi }{ T }\]
anonymous
  • anonymous
amplitude, A through a mathematical analysis, the amplitude is the highest/lowest point of the curve. \(y =\frac{1}{4}\cos \frac{2\pi}{3}{ \theta}\) \(y' =-\frac{1}{4}(\frac{2\pi}{3})\sin \frac{2\pi}{3}{ \theta}\) let y'=0, \(\sin \frac{2\pi}{3}{ \theta}=0\) \(\theta=0\) y=1/4=A you follow?
anonymous
  • anonymous
kinda
anonymous
  • anonymous
yup, now, \(y'=\omega\) the period is the time for two wavecrest, i.e. two amplitude, also aka two y'=0 so you find the w for two successive max points and you'll get it.
anonymous
  • anonymous
on second thought, where is the time parameter=.=

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