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When switch S is open (off) the oscilloscope displays the waveform shown in Figure 7 - AQA - A-Level Physics - Question 2 - 2017 - Paper 3

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When switch S is open (off) the oscilloscope displays the waveform shown in Figure 7. When S is closed (on) the oscilloscope displays the waveform shown in Figure 8... show full transcript

Worked Solution & Example Answer:When switch S is open (off) the oscilloscope displays the waveform shown in Figure 7 - AQA - A-Level Physics - Question 2 - 2017 - Paper 3

Step 1

Determine the peak-to-peak voltage V' of the waveform shown in Figure 8.

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Answer

To find the peak-to-peak voltage (V'), measure the vertical distance from the highest point (peak) to the lowest point (trough) of the waveform shown in Figure 8.

From the oscilloscope image, if the peak voltage reaches +3.3 V and the trough reaches -3.3 V, then:

V′=Vpeak−Vtrough=(3.3extV)−(−3.3extV)=6.6extVV' = V_{peak} - V_{trough} = (3.3 ext{ V}) - (-3.3 ext{ V}) = 6.6 ext{ V}

Thus, the peak-to-peak voltage is approximately 6.3 V.

Step 2

Determine the frequency f of the waveform shown in Figure 8.

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Answer

To find the frequency (f), note the period (T) of the waveform, which can be calculated from the horizontal divisions on the oscilloscope.

Assuming the period is read as 4 divisions and each division represents 0.1 ms:

T=4extdivisionsimes0.1extms/division=0.4extms=0.4imes10−3extsT = 4 ext{ divisions} imes 0.1 ext{ ms/division} = 0.4 ext{ ms} = 0.4 imes 10^{-3} ext{ s}

The frequency is then calculated using the formula:
f = rac{1}{T}

Thus,
f = rac{1}{0.4 imes 10^{-3} ext{ s}} = 2500 ext{ Hz}.

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