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Frequency in an oscilloscope simulation

  1. Jul 24, 2011 #1
    Every time I try to find frequency on this problem I get the same answer - and according to the simulation it is incorrect.

    Here's a picture of the simulation.

    Please note: I am just trying to find frequency. I know the other two variables (.6A and 6V ; these are confirmed correct)

    [PLAIN]http://img854.imageshack.us/img854/100/screenshot20110724at104.png [Broken]

    Looking at this image, I see only 2 and 1/4 cycles.

    Frequency is, essentially, cycles per second (1/s) and period is the time it takes to complete a single cycle.

    Based on the image I attached, I see 2 and 1/4 cycles (9/4) over a time of .5e-6 s. This is the same as dividing (9/4) by (.5e-6).

    (9/4) / (.5e-6) = 4.5e6 s^(-1)

    I do not understand where my mistake is.
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Jul 24, 2011 #2
    The time scale setting for an oscillisope is usually in terms of time/division. 0.5e-6 s would correspond to the distance between vertical lines.
    Last edited: Jul 24, 2011
  4. Jul 24, 2011 #3
    Derp! I knew that but I guess I had a brainfart.

    Thanks! I got the correct answer (5e5 s^-1)
  5. Jul 24, 2011 #4


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    One period (minimum to minimum) is 4 units on the time scale. The oscilloscope shows the length of such units, it is 0.5 x 10-6 s.

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