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anonymous
 4 years ago
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
anonymous
 4 years ago
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

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0note 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
 4 years ago
Best ResponseYou've already chosen the best response.0This looks like physics.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0well do you know the answer?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yes but just so you know. This is a physics question not mathematics.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0cross reference your eq with the general eq above.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok well can i have the answer?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0did you try to do it yourself first?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yes and i am beyind confused :/

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0what did you get? by the way, your equation didnot involve time ......

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0cause 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
 4 years ago
Best ResponseYou've already chosen the best response.0\[\omega = \frac{ 2\pi }{ T }\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0amplitude, 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
 4 years ago
Best ResponseYou've already chosen the best response.0yup, 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
 4 years ago
Best ResponseYou've already chosen the best response.0on second thought, where is the time parameter=.=
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