For a sine function, how do I know whether the shift in the x direction was positive or negative?
And the solution here is y= 3*sin(π*(x-1)) or 3*sin(π*(x+1))
Thank you for your help in advance
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Usually vibrations are seen as a function of time. Since time does not run backwards, the shift is only positive.
Actually, that’s not important when you look at such a vibration as “swinging”.
Physically, a sinusoidal vibration must be considered as if it had begun an infinitely long time ago, it is “swinging”.
What you do not need to know and know yet can be vibrations that are not so beautifully sinusoidal from multiple sinusoidal vibrations of other frequencies. A start like in the picture is something like “not curved”.
Remember when you should study. Furier analysis is a keyword.
Thank you.
In your case, you have two possible functions:
y=3⋅sin(π(x-1)y=3⋅sin(π(x−1))y=3⋅sin(π(x+1)) y=3⋅sin(π(x(1))
For the first function y=3⋅sin(π(x−1))
y=3⋅sin(π(x−1) is c=1
c)=1, which means that the function is shifted by 1 unit to the right.
For the second function y
=3⋅sin(pi(x+1))
y=3⋅sin(π(x+1) is c=−1
c)=−1, which means that the function is shifted by 1 Einiz to the left.
In order to understand it correctly, I would like to ask again. So both functional terms are correct?
The function f(x)=a*sin(π*(x – c)) has the period 2. For this reason, the functional argument x can be displaced by a multiple of 2 both in the positive and in the negative direction. This does not change the function.
a*sin(π*(x – c)) = a*sin(π*(x – c + 2*n )), n € Z
In the question, the primary shift was set to 1, c=1
a*sin(π*(x – 1)) = a*sin(π*(x – 1 + 2*n ), n € Z
Examples:
n = 0 –> f(x) = a*sin(π*(x – 1)
n = 1 –> f(x) = a*sin(π*(x + 1))
n = -3 –> f(x) = a*sin(π*(x – 7))
n = 5 –> f(x) = a*sin(π*(x + 9))
The assertion that, in the case of vibrations, only temporally positive shifts are to be assumed may be valid in the field of physics. It doesn’t matter purely mathematically.
No. both facts are correct, because they show the same graph, right?
you can even get them with cos so
but normally, as unwritten commandment, it is thought to be shifted to the right, that is, away from (0/0)
That’s different than just simple. At (x-1) one would think it is shifted to the left, but that is not true, it is shifted to the right. Reason: from every pi * x is deducted from x what. The sine curve reaches the expected values only later. So moved to the right
y= 3*sin(π*(x-1)
Thanks for your help
Applies to All Functions:
f(x) → f(x-c) is a shift to the right about the value c
f(x) → f(x+c) is a shift to the left about the value c
Example:
f(x) = sin(x) ……. g(x) = sin(x-1) → g the function f is shifted by 1 to the right.
Use Geogebra for illustration: