How do you find the period and the amplitude for #y = (1/2)sin(x - pi)#?
2 Answers
Explanation:
The standard form of the
#color(blue)"sine wave function"# is
#color(red)(|bar(ul(color(white)(a/a)color(black)(y=asin(bx+c)+d)color(white)(a/a)|)))# where amplitude = |a| , period
#=(2pi)/b# c is the horizontal shift and d , the vertical shift.
here a
#=1/2,b=1,c=-pi" and "d=0# hence period
#=(2pi)/1=2pi" and amplitude"=|1/2|=1/2#
Amplitude:
The period (pitch) is
Explanation:
The given equation starts with the basis of sine. This has a maximum and minimum values of -1 to +1.
Making this into a product of
,~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
I am interpreting this to be the equivalent of pitch. Like that of a thread. That is the distance between maximum values or any other such repeat.
This standard 'untampered with' period is
As we do not have this structure the pitch remains at the standard
Suppose we had
Imagine that we have two graphs. The one we are drawing and a reference one of
As we are considering
We then turn to the graph we are drawing and plot this recorded
So multiplying