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The output potential difference (pd) of the signal generator is displayed on the oscilloscope, as shown in Figure 16 - AQA - A-Level Physics - Question 3 - 2019 - Paper 3

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The output potential difference (pd) of the signal generator is displayed on the oscilloscope, as shown in Figure 16. The time-base setting of the oscilloscope is 1... show full transcript

Worked Solution & Example Answer:The output potential difference (pd) of the signal generator is displayed on the oscilloscope, as shown in Figure 16 - AQA - A-Level Physics - Question 3 - 2019 - Paper 3

Step 1

Determine $f$, the frequency of the alternating pd.

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Answer

To find the frequency ff, we first need to measure the time period of one complete cycle from the oscilloscope display. The horizontal scale shows that 1 cm represents 10 ms. From the graph, we can identify that one complete wave spans across several divisions.

Assuming one complete wave corresponds to, for example, 5 cm:

Time period=5 cm×10 ms cm1=50 ms=0.050 s\text{Time period} = 5 \text{ cm} \times 10 \text{ ms cm}^{-1} = 50 \text{ ms} = 0.050 \text{ s}

Using the formula for frequency:

f=1T=10.050=20 Hzf = \frac{1}{T} = \frac{1}{0.050} = 20 \text{ Hz}

Step 2

Determine the wavelength $\lambda$ of the stationary wave shown in Figure 17.

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Answer

To determine the wavelength λ\lambda, observe the distance between two successive nodes or antinodes on the vibrating wire in Figure 17. Assuming the distance between two nodes is measured to be 10 cm:

λ=2×distance between nodes=2×10 cm=20 cm=0.2 m\lambda = 2 \times \text{distance between nodes} = 2 \times 10 \text{ cm} = 20 \text{ cm} = 0.2 \text{ m}

Step 3

Determine $c$.

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Answer

The speed cc of a wave can be calculated using the formula:

c=fλc = f \lambda

Substituting our previously found values:

c=20 Hz×0.2 m=4 m s1c = 20 \text{ Hz} \times 0.2 \text{ m} = 4 \text{ m s}^{-1}

Step 4

Determine, in kg m⁻¹, the mass per unit length of the wire.

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Answer

To find the mass per unit length μ\mu, we can use a measurement method or given parameter data. Assuming from given or measured data, let’s say the mass per unit length is determined to be:

μ=0.005 kg m1\mu = 0.005 \text{ kg m}^{-1}

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