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1 November, 13:35

In each case, you should demonstrate how you worked out your answer, as well as giving the answer.

You must show your working to gain full marks. In each case, about 50% of the marks are awarded for a clear explanation of your working.

a. A sinewave has a period (duration of one cycle) of 6.5ms. What is its corresponding frequency expressed to 3 significant figures?

If the frequency of this sinewave is increased by a factor of 3, calculate the new period value, expressing your answer again to 3 significant figures. Clearly, show all your working.

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Answers (2)
  1. 1 November, 16:07
    0
    First frequency f = 153.846 Hz

    Second period T = 1.625 ms

    Step by stepplanation:

    T = 6.5ms = 6.5 * (10^-3)

    But T = 1/f

    f = 1/T

    f = 1 (6.5 * (10^-3))

    f = 153.846 Hz

    Now the frequency f is increased by a factor of 3

    = 3*153.846

    = 461.538

    New frequency = 461.538+153.846

    = 615.384 Hz

    New period T = 1/615.384

    Period T = 1.625*10^-3

    T = 1.625 ms
  2. 1 November, 17:14
    0
    (a)

    Time period, T1 = 6.5 ms

    = 6.5 * 10^-3 s

    Frequency is the reciprocal of time period.

    f1 = 1 / T1 = 1 / (6.5 * 10^-3) = 153.85 Hz

    Now the frequency becomes 3 times

    f2 = 3 * f1 = 3 * 153.85 = 461.55 Hz

    The new period is

    T2 = 1 / f2 = 1 / 461.55

    = 0.0022 second = 2.22 ms
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