Relating to period of a function.

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  • #1
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If we have f(k)=Asin([itex]\frac{xa}{2}[/itex]). Then it was mentioned that f(x) is a periodic function with period [itex]\frac{Δx.a}{2}[/itex]=π. How come?

Thanks!

Please note, A and a are constants.
 

Answers and Replies

  • #2
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If we have f(k)=Asin([itex]\frac{xa}{2}[/itex]).
This should be f(x) = A sin(##\frac{xa}{2}##). The only variable here is x. k doesn't appear at all in the formula for this function.
Then it was mentioned that f(x) is a periodic function with period [itex]\frac{Δx.a}{2}[/itex]=π. How come?
This is wrong, and I have no idea where you got this.

The sine and cosine functions are periodic. The period of both sin(x) and cos(x) is 2##\pi##. The period of sin(Kx) and cos(Kx) is ##\frac{2 \pi}{K}##.

What then would be the period of sin((a/2)x)?
Thanks!

Please note, A and a are constants.
 
  • #3
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Sorry it is x, I typed it by mistake. It should be 4π/a. But here they related the variable of the sin function "x" to the period in some way I didn't understand.. (i.e, the formula that I wrote in my first post and that you quoted second).
Thanks
 
  • #4
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Sorry it is x, I typed it by mistake. It should be 4π/a. But here they related the variable of the sin function "x" to the period in some way I didn't understand.. (i.e, the formula that I wrote in my first post and that you quoted second).
Thanks
If the period of sin(Kx) is ##2\pi/K##, what is the period of sin((a/2)x)?
 
  • #5
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I answered you previously, it would be 4π/a
 
  • #6
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You said it "should be 4π/a", which I interpreted to mean that you knew that was the answer, but didn't know how it was obtained.

You asked about the formula in your first post (and that I quoted). I have no idea what they mean by that formula, especially the part with Δx.
 
  • #7
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Yes. Neither do I. Thank you anyway!
 

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