Higher June 2022 Unit 1 Q11
11. The wavelength of a sound wave, \(W\) metres, is inversely proportional to its frequency, \(f\) hertz (Hz).
Richard is measuring the wavelengths of different sound waves.
The wavelength is 0·5 m when the frequency is 1200 Hz.
What is the frequency when the wavelength is 10 m? [4]
| Scheme | Marks | Comments |
|---|---|---|
| \(W\ \alpha\ \dfrac{1}{f}\) OR \(W = \dfrac{k}{f}\) | B1 | Allow \(W\ \alpha\ \dfrac{k}{f}\) |
| \(0.5 = \dfrac{k}{1200}\) OR \(k = 600\) | M1 | M1 implies B1. F.T. for use of \(W\ \alpha\ \dfrac{1}{f^{\,n}}\) with \(n \gt 0\). |
| \(W = \dfrac{600}{f}\) or \(10 = \dfrac{600}{f}\) or equivalent | A1 | May be implied by further work. |
| (\(f\) =) 60 [The frequency is 60 (Hz)] | B1 | FT for ‘their k’ provided M1 awarded. |
Alternative method
| Scheme | Marks | Comments |
|---|---|---|
| 1200 ÷ 2 ÷ 10 or 1200 ÷ 20 or equivalent | M3 | A complete method (based on multiplying and dividing) M1 for W = 1 when f = 600 Hz OR W = 2 when f = 300 OR W = 5 when f = 120, i.e. where Wf = 600 provided W > 0.5 (i.e. f < 1200) |
| (f =) 60 [The frequency is 60 (Hz)] | A1 | No marks for 1200 × 20 = 24 000 Hz (using direct proportion) |