Question

A sinewave has a period (duration of one cycle) of 35 μs. What is its corresponding frequency (no of completed cycles in a second), in kHz, expressed to 3 significant figures?

If the frequency of this sinewave is now reduced by a factor of 2.5, using the value calculated above to 3 significant figures, what will be the new period?

Express your answer in μs, to 3 significant figures
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Answer 1

We know that the relation between period and frequency is:

T = period.

f = frequency.

T = 1/f.

Then, if the period is 35 μs = 35x10^-9 seconds.

The frequency will be:

f = (1/ 35x10^-9 s) = (1/35s)*10^9

now, 1Hz = 1/s

1KHz = 1/1000s = 10^-3 s

f = (10^9/35)*10^-3 KHz = 10^6/35 KHz = 28,571.429 KHz.

Now we must divide this by 2.5:

28,571.429 KHz/2.5 = 11,428.572 KHz

Now we can use the relation:

T = 1/f

T = 1/11,428.572 KHz = (1/11,428.572 Hz)*10^-3

T = 8.750x10^-8 seconds.

And we want this expressed in μs = 10`-9 seconds, we have:

T = 8.750x10^-8 s = (10/10)8.750x10^-8 s = 87.50x10`-9s = 87.50 μs

So as expected, if the frequency is reduced by a factor of 2.5, the period will increase by a factor of 2.5

As the initial period was 35 μs, and:

2.5*35 μs = 87.5 μs