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 10 ms cm⁻¹. Determine $f$, the frequency of the alternating pd. Figure 17 shows the first harmonic stationary wave produced when the distance $x$ between the bridges is adjusted. Determine the wavelength $\lambda$ of the stationary wave shown in Figure 17. The stationary wave is formed by two waves of frequency $f$ and wavelength $\lambda$ travelling with speed $c$ in opposite directions. Determine $c$. Determine, in kg m⁻¹, the mass per unit length of the wire.

A-Level AQA Physics Equations of Motion: The output potential difference

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

Question 3

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 $f$ , 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:

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

Using the formula for frequency:

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

Step 2

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

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Answer

To determine the wavelength $\lambda$ , 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:

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

Step 3

Determine $c$.

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Answer

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

$c = f \lambda$

Substituting our previously found values:

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

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 $\mu$ , 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:

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

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

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

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

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

Determine $c$.

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

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