7.8.11 Find amplitude, period, PS, VS. graph 2 periods

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Discussion Overview

The discussion revolves around finding the amplitude, period, phase shift (PS), vertical shift (VS), and graphing two periods of the cosine function given by the equation $y=3\cos(\pi x-2)+5$. Participants explore the parameters of the function and their implications in the context of trigonometric waveforms.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • Some participants propose that the amplitude (A) is 3 and the vertical shift (VS) is 5 based on observation.
  • It is noted that the angular frequency ($\omega$) is $\pi$, leading to a calculated period (T) of 2.
  • One participant suggests that the phase shift ($\phi$) is 2, leading to a PS calculated as $\dfrac{\phi}{\omega}=\dfrac{2}{\pi}$.
  • Another participant points out a potential discrepancy in the phase shift, indicating that the general form of the cosine function may vary, which could affect the sign of $\phi$ and thus the interpretation of the phase shift.

Areas of Agreement / Disagreement

Participants generally agree on the values for amplitude and vertical shift, but there is disagreement regarding the phase shift and the interpretation of the general form of the cosine function, with competing views on the sign of $\phi$.

Contextual Notes

There are unresolved aspects regarding the definitions and forms of the cosine function being used, which may affect the calculations of phase shift. The discussion reflects different interpretations of the phase shift based on the chosen model.

karush
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$\tiny{\textbf{7.8.11 Campbell HS}}$
Find (A)mplitude, (P)eriod, PS, VS. graph 2 periods
$y=3\cos(\pi x-2)+5$

by observation we have A=3 and VS=5
ok assume $\omega=\pi$
so if period is $T=\dfrac{2\pi}{\omega}$ then $T=\dfrac{2\pi}{\pi}=2$
 
Last edited:
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karush said:
$\tiny{\textbf{7.8.11 Campbell HS}}$
Find (A)mplitude, (P)eriod, PS, VS. graph 2 periods
$y=3\cos(\pi x-2)+5$

by observation we have A=3 and VS=5...
and $\omega=\pi$ ...
Otherwise, good!

-Dan
 
ok i think $\phi =2$ then PS is $\dfrac{\phi}{\omega}=\dfrac{2}{\pi}$

really! :unsure:
 
Okay, check with your general form of the sine wave. I use
[math]y = A ~ cos( \omega x + \phi ) + y_0[/math]

Your source might be using
[math]y = A ~ cos( \omega x - \phi ) + y_0[/math]
in which case, yes, [math]\phi = 2[/math]. In this model I'm using [math]\phi = - 2[/math]. The negative sign is important because it tells which way the wave has been shifted along the x-axis.

-Dan
 

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