0
$\begingroup$

I am doing a transformation problem of getting the graph of $\sin (2x – \pi/6)$ by applying transformations to $F(x) = \sin x$

In the process, I let $f(x) = F(2x) = \sin 2x$.

Next, I then let $g(x) = f(x – \pi/12) = \sin 2[x – \pi/12] = \sin (2x – \pi/6)$. The graphs are plotted as shown.

enter image description here

From f(x) to g(x), the above equations clearly shows there is a phase-shift of $\pi/12$ and this agrees with the red and blue lines

However, if I just comparing the functions $\sin (2x)$ and $\sin (2x – \pi/6)$ directly, shouldn’t there be just a right shift of $\pi/6$?.

$\endgroup$
0
$\begingroup$

Here the graph is plotted as functions of x, taking values of x in the horizontal axis. So there is a shift of $\pi/12$.

If we plot it as a function of 2x, then we would have a shift of $$2.\frac{\pi}{12} = \frac{\pi}{6}$$

$\endgroup$
  • $\begingroup$ Right. I forgot I have to plot it as function of 2x. $\endgroup$ – Mick Apr 29 at 17:01
1
$\begingroup$

Based on your reasoning, I presume you want to use the fact that the transformation $$f(x)\to f(x+a)$$ corresponds to a horizontal shift of the function by $|a|$ (where the direction depends on the sign of $a$)

Nonetheless, when the function is of the form $$\sin(2x)$$ in order to obtain the function $$\sin(2x-\pi/6)$$ you need to find $a$ such that $$\sin({\color{red}2}(x+a))=\sin(2x+\color{red}{2}a)=\sin(2x-\pi/6)$$ and thus the needed shift is $a=\frac{\pi/6}{2}=\frac{\pi}{12}$.

But this does not contradict the property of the transformation: if you want to shift the function $f(x)=\sin(2x)$ by $\pi/6$ to the right then you do apply the transfomration as usual $$f(x-\pi/6)=\sin(\color{red}{2}(x-\pi/6))=\sin(2x-\pi/3)$$ but as you see the actual shift is hidden by the factor 2.

In general, when constructing an harmonic oscillation of the form $\sin(\omega x+\varphi)$ you need to translate/shift the function horizontally by $\varphi/\omega$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.