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Sine wave problem

  1. Sep 18, 2007 #1
    I have a question that says: prepare a table of values and plot the graph for the domain (-pie,pie) for the function f(x)=2sin(2x)

    I understand how to do that for y=sinx, but for this function the wave compresses horizontally and stretches vertically so isnt the domain now (-pie/2,pie/2) and then therefore how can you make a table of values for (-pie,pie).

    thanks for any help
    andrew
     
  2. jcsd
  3. Sep 18, 2007 #2
    Are you sure you aren't confusing domain and range? It's true that the wave is now compressed horizontally, but the sine will still stretch from negative infinity to infinity. On the interval (-pi,pi), the graph will repeat itself, but there is a graph there, so you can definitely make a table of values for it. If you don't see what I mean, try graphing your function. There will be y-values for the interval (-pi,pi).
     
  4. Sep 18, 2007 #3
    The interval that you are implying is the period of the sine wave. When it is changed to 2*sin(2x), you recieve double the amplitude, but half of the period, so if your domain is reflective of the period, then cutting it in half would be valid. The period represents the repeating of the sine wave.

    To make a table of it I would first make a table of just plain sin(x). What you would then do is make a new table and start at half of the starting point of the first table and write your y values at intervals of half of the previous table until the maximum value of x at half of the first table. The table should then be the same size. You then double the y values and put them into the table because of the doubled amplitude. This plots (-pi/2,pi/2). Then for the plot of (-pi,-pi/2), you use the same y values as (0,pi/2), and for (pi/2,pi), you use the same y values as (-pi/2,0).

    I hope I answered your question.

    Da Jeans
     
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