Question A sound wave with a frequency of 1000 Hz travels through air. The speed of sound in air is approximately 343 meters per second.
(a) Calculate the wavelength of the sound wave.
(b) If the amplitude of the sound wave is 0.05 meters, calculate the maximum displacement of air particles from their equilibrium position.
(c) A second sound wave with a frequency of 2000 Hz is produced. How does the wavelength of this sound wave compare to the wavelength of the first sound wave? Explain your answer.
(d) The sound wave from part (c) travels through a medium where the speed of sound is 515 meters per second. Calculate the new wavelength of the second sound wave.
Answer
(a) The speed of a wave is equal to the product of its frequency and wavelength, given by the equation v = fλ. Rearranging the equation, we find the wavelength (λ) is equal to the speed of the wave (v) divided by the frequency (f):
λ = v / f
Substituting the given values:
λ = 343 m/s / 1000 Hz
Calculating:
λ = 0.343 m
Therefore, the wavelength of the sound wave is 0.343 meters.
(b) The amplitude (A) of a wave represents the maximum displacement of particles from their equilibrium position. For a sound wave, this corresponds to the maximum displacement of air particles. Therefore, the maximum displacement is equal to the amplitude (A).
Therefore, the maximum displacement of air particles is 0.05 meters.
(c) The wavelength of a sound wave is inversely proportional to its frequency. This means that as the frequency increases, the wavelength decreases, and vice versa. Since the frequency of the second sound wave (2000 Hz) is double that of the first sound wave (1000 Hz), the wavelength of the second sound wave will be half that of the first sound wave.
Therefore, the wavelength of the second sound wave is half the wavelength of the first sound wave.
(d) Using the equation v = fλ, rearranging to solve for the new wavelength (λ):
λ = v / f
Substituting the given values:
λ = 515 m/s / 2000 Hz
Converting Hz to seconds:
λ = 515 m/s / (2000 Hz * (1 s/1 Hz))
Calculating:
λ ≈ 0.258 meters
Therefore, the new wavelength of the second sound wave is approximately 0.258 meters.