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Guest Jan 19, 2018

Best Answer 

 #1
avatar+12227 
+2

First One:  Looks like a cosine wave inverted (multiplied by -1)      and the AMPLITUDE is reduced by 1/2

 soooo:   -1/2 cos x

 

2:  Frequency is   1/period    period is pi     frequency then would be  1/pi

 

3: Looks like a sine wave  but the period is pi .....so the frequency is DOUBLED

    sooooo:      sin 2x

 

4: Similar to #3   the frequency is doubled (2x)  so the period is HALVED   to  pi

ElectricPavlov  Jan 19, 2018
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2+0 Answers

 #1
avatar+12227 
+2
Best Answer

First One:  Looks like a cosine wave inverted (multiplied by -1)      and the AMPLITUDE is reduced by 1/2

 soooo:   -1/2 cos x

 

2:  Frequency is   1/period    period is pi     frequency then would be  1/pi

 

3: Looks like a sine wave  but the period is pi .....so the frequency is DOUBLED

    sooooo:      sin 2x

 

4: Similar to #3   the frequency is doubled (2x)  so the period is HALVED   to  pi

ElectricPavlov  Jan 19, 2018
 #2
avatar+92194 
+1

 

\(y=a* cos[n(\theta+p) ]+ L\)

 

 

Amplitude =a

phase shift = p (units in the NEGATIVE direction - opposite direction to what most people expect)

 

wave length  \(\lambda   = \frac{2\pi}{n}\)

 

L is the vertical shift   

 

SO CONSIDER

 

\(y=1.8cos(3\theta+\frac{\pi}{2})-1.5\\ rewrite\;\; as \\ y=1.8cos(3[\theta+\frac{\pi}{6}])-1.5\\\)

 

It has the basic \(y=cos(\theta) \)shape.

wavelength = \(\frac{2\pi}{3}\)

Phase (horizontal) shift =\( \frac{\pi}{6}\;\)units  in the negative direction

Amplitude =1.8

Vertical shift is 1.5 units DOWN

 

check

Here is the graph.

You can play iwth the circles on the left to see how I 'developed' the graph 

https://www.desmos.com/calculator/bluugr6nna

Melody  Jan 22, 2018

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