A-Level Physics October/November 2024 Q4(b)(ii): The speed of sound in air is 340 m s-1. Determine the frequency of the sound wave.

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  The relationship between speed $v$, frequency $f$, and wavelength $\lambda$ is given by the wave equation: $v = f\lambda$

  Rearranging for frequency, $f$: $f = \frac{v}{\lambda}$

  From the previous part of the question, the stationary wave is the fundamental mode in a column of length $L = 4.5 \times 10^{-2}$ m. For this mode, the length is one-quarter of the wavelength ($L = \lambda/4$).

  Therefore, the wavelength is: $\lambda = 4L = 4 \times 4.5 \times 10^{-2} \text{ m}$ $\lambda = 0.18 \text{ m}$

  Now, substitute the values for speed and wavelength into the frequency equation: $f = \frac{340}{0.18}$ $f = 1888.8... \text{ Hz}$

  Rounding to two significant figures gives: $f = 1900 \text{ Hz}$

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  How the marks are awarded

  • C1 — Correctly stating or using the wave equation rearranged for frequency, $f = v / \lambda$.
  • C1 — Correctly determining the wavelength by identifying the relationship $\lambda = 4L$ and substituting the length, as in $\lambda = 4 \times 4.5 \times 10^{-2}$ m.
  • A1 — Correctly calculating the final answer of 1900 Hz, including the correct substitution of values.

  Common mistakes

  • Using an incorrect relationship for the wavelength, such as $\lambda = 2L$ (for an open pipe or string) or $\lambda = L$, which leads to an incorrect frequency.
  • Forgetting to convert a length given in centimetres into metres before calculating the wavelength.
  • Incorrectly rearranging the wave speed equation, for example using $f = v \times \lambda$.
  • A calculation error, or providing the final answer to an inappropriate number of significant figures (e.g., writing the full calculator display of 1888.88... Hz).

  Examiner tip: Always start by identifying the type of stationary wave and the harmonic shown to correctly relate the system's length to the wavelength before applying the main wave equation.

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