Help on a simple oscilloscope tracing...

I am having trouble understanding what I am doing wrong with this question, my answer is wrong in the mark scheme. The value is supposed to be 100 Hz
While i'm coming to a value of 95Hz.

Thanks.

Heres the question:

I am having trouble understanding what I am doing wrong with this question, my answer is wrong in the mark scheme. The value is supposed to be 100 Hz
While i'm coming to a value of 95Hz.

Thanks.

Heres the question:

Mark schemes are a guide for the examiner and not meant to be a model/exact answer. In questions like these where the student is required to make a physical measurement, there will always be an allowance for judgment error. Did the mark scheme give a tolerance limit like 100Hz, 95 to 105Hz acceptable or something like that?

I calculated the period (t) to be 6.8 divisions x 1.5ms/div = 10.2ms.

f=1/t

f = 98Hz.

Rounding up would give 100Hz. So 95Hz should give you full marks since the examiner will know you used the correct method for the calculation with a margin of error for the measurement.

1.5T=(10)(1.5 X 10^-3)
1.5T=1.5 X 10^-2
1.5T=0.015
T=0.01s

f=1/T
f=1/0.01
=100Hz

Happy?

I'm guessing you measure T. The problem here is a loss of accuracy.

Edit: if you are confused about how I figured out that 10 boxes is for 1.5T, then think about the y=sinx curve and the x values.

Yeah i'm really baffled at how there are 10 boxes between?, assuming both peaks are around halfway the number of boxes should amount to something around 7 right?

This should help

So you don't measure time period from peak to peak? :s

So you don't measure time period from peak to peak? :s

I'm not measuring a single time period though, I'm measuring 1.5T as it improves accuracy.

A bit hard to explain, but I'll try.

Remember time period is the time taken for a single oscillation per second. Frequency is the number of oscillations per second, so it's the inverse of time period, so f=1/T and T=1/f.

540/360=1.5

Hence for 0<x<540 y=sinx has 1.5 oscillations. Comparing the shape of y=sinx for 0<x<540 with the shape of the oscilloscope trace, we can say that there are 1.5 oscillations, start to end of the trace. You should be able to see it visually anyway - a semicircle with its peak, top or bottom, is half an oscillation, so three semicircles, as in the trace, shows 1.5 oscillation. The time it takes for these 1.5 oscillations is found by counting up the boxes to get 10. You know that each box represents 1.5ms.

Total time for 1.5T=10(1.5 X 10^-3)
=1.5 X 10^-2
=0.015

So for 1.5 oscillations it takes a total time of 0.015s. From the definition of frequency,

f=number of oscillations/total time taken
=1.5/0.015
=100Hz

You can do that, but there will be a greater measurement error.

It's more accurate to do it from points measured at the zero crossing threshold.

Krishs' method works only if you assume the waveform is perfectly symmetrical. If it is not, his method gives an 'average' frequency because the zero crossing points are not all at accurate 3.33ms intervals (at least as far as I measured them). Meaning the waveform is composite and contains more than one frequency component.

At A-level, it is safe to assume the waveform will be symmetrical. In real life, this is often not the case.

As I said, if you show your working and method, you will not be penalised.