A student measures the speed of water waves in a ripple tank at different frequencies. The results are shown in the table. Frequency / Hz Wavelength / cm Sp...
Question 1 Report
A student measures the speed of water waves in a ripple tank at different frequencies. The results are shown in the table.
| Frequency / Hz | Wavelength / cm | Speed / cm/s |
|---|---|---|
| 4.0 | 6.0 | ........ |
| 6.0 | 4.0 | ........ |
| 8.0 | 3.0 | ........ |
| 10.0 | 2.4 | ........ |
| 12.0 | 2.0 | ........ |
(a) Complete the speed column using the wave equation. [2]
(b) State the relationship between frequency and wavelength shown by the data, given that the speed is constant. [1]
(c) State the wave equation. [1]
(d) Explain why the wave speed is approximately constant even though the frequency changes. [2]
(e) Describe how the wavelength could be measured accurately. [1]
Answer Details
(a) Completed speed column using \(v = f \times \lambda\):
| Frequency / Hz | Wavelength / cm | Speed / cm/s |
|---|---|---|
| 4.0 | 6.0 | 4.0 \(\times\) 6.0 = 24 |
| 6.0 | 4.0 | 6.0 \(\times\) 4.0 = 24 |
| 8.0 | 3.0 | 8.0 \(\times\) 3.0 = 24 |
| 10.0 | 2.4 | 10.0 \(\times\) 2.4 = 24 |
| 12.0 | 2.0 | 12.0 \(\times\) 2.0 = 24 |
[1 for correct formula applied, 1 for all five values correct]
Every calculation gives 24 cm/s, confirming the wave speed is constant.
(b) As frequency increases, the wavelength decreases. They are inversely proportional (when speed is constant). [1]
Doubling the frequency from 4.0 to 8.0 Hz halves the wavelength from 6.0 to 3.0 cm, consistent with \(\lambda = v/f\).
(c) The wave equation is: wave speed = frequency \(\times\) wavelength, or \(v = f\lambda\). [1]
(d)
- The wave speed in a ripple tank depends on the depth of the water, not on the frequency. [1]
- Since the depth does not change when the motor frequency is adjusted, the speed remains constant at 24 cm/s. [1]
The source determines how many waves are produced per second (frequency), but the medium determines how fast each wave travels. Changing one does not affect the other.
(e) Measure the distance across several wavelengths on the screen and divide by the number of wavelengths. [1]
This technique averages out any measurement error, giving a more precise value than trying to measure a single wavelength, which may be only a few centimetres across.
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