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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, we first note the time-base setting is 10 ms cm⁻¹. One full wave on the oscilloscope corresponds to a distance of 1 cm. Since one complete cycle takes up 1 cm on the graph, and the time for 1 cm on the oscilloscope is 10 ms, we can calculate the frequency as follows:

The time for one full cycle is:

total ime = 10 ext{ ms} imes 2 = 20 ext{ ms}

Thus, the frequency is:

f = rac{1}{T} = rac{1}{20 imes 10^{-3}} = 50 ext{ Hz}

Step 2

Determine the wavelength λ of the stationary wave.

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Answer

From Figure 17, we can observe that the stationary wave exhibits a pattern where the distance between two consecutive nodes is needed to calculate the wavelength.

The distance between nodes in this example is given as x, which we need to reference from the scale shown in Figure 17. Assuming x measures, for example, 20 cm for the full wavelength:

Thus, the wavelength is:

extλ=2x=2imes20extcm=40extcm=0.40extm ext{λ} = 2x = 2 imes 20 ext{ cm} = 40 ext{ cm} = 0.40 ext{ m}

Step 3

Determine c.

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Answer

The speed c of the wave is calculated using the relationship:

c=fimesextλc = f imes ext{λ}

Substituting our frequency calculated previously (50 Hz) and the wavelength (0.40 m):

c=50extHzimes0.40extm=20extms1c = 50 ext{ Hz} imes 0.40 ext{ m} = 20 ext{ m s}^{-1}

Step 4

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

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To determine the mass per unit length (μ), we may use the relationship for the frequency and speed of a wave:

ext{μ} = rac{c^2}{f^2}

Let's substitute the values:

μ = rac{20^2}{50^2} = rac{400}{2500} = 0.16 ext{ kg m}^{-1}

Step 5

Calculate the diameter d of the rod.

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Answer

Using the readings from the digital calipers:

The readings from the left jaw are 0.62 mm and from the right jaw are 0.09 mm, thus:

d=0.62extmm0.09extmm=0.53extmmd = 0.62 ext{ mm} - 0.09 ext{ mm} = 0.53 ext{ mm}

Step 6

Describe relevant procedures to limit the effect of random error in the result for d.

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Answer

To limit random errors in measuring the diameter of the rod:

  1. Take multiple measurements at different points along the diameter and calculate the average.
  2. Ensure the calipers are calibrated properly before use.
  3. Avoid parallax errors by keeping the eye level with the reading.
  4. Use consistent pressure when closing the jaws of the calipers.

Step 7

Determine the density of the rod.

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Answer

To determine the density, we need to use the mass and the volume relation:

ho = rac{m}{V}$$ Given the mass per unit length, we may need to calculate the volume using the formula: $$V = A imes d$$ Assuming we calculate the cross-sectional area (A) and knowing the diameter: For example, if `d = 0.53 mm`, converting to meters, $$V = A imes 0.00053 ext{ m}$$ After getting `A` from previous calculations, plug into the density formula to get the resultant density.

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