AP Physics 1 Exam Question - Sound Waves
A loudspeaker emits a pure tone sound wave with a frequency of 500 Hz. The speaker is positioned 2.0 meters away from a stationary observer. The speed of sound in air is 343 m/s. Assume that the source and observer are at rest.
Solution:
The speed of sound in air, v, is given as 343 m/s.
The frequency of the sound wave, f, is given as 500 Hz.
The wavelength of the sound wave, λ, can be calculated using the formula:
λ = v / f
Substituting the given values:
λ = 343 m/s / 500 Hz = 0.686 m
Therefore, the wavelength of the sound wave emitted by the loudspeaker is 0.686 m.
Solution:
The period of a wave, T, is the reciprocal of its frequency.
Given the frequency of the sound wave emitted by the loudspeaker is 500 Hz.
T = 1 / f = 1 / 500 Hz = 0.002 s
Therefore, the period of the sound wave emitted by the loudspeaker is 0.002 seconds.
Solution:
The amplitude of a wave refers to the maximum displacement of a particle from its equilibrium position.
The information about the amplitude is not given directly in the question. However, it is important to note that the amplitude is not related to the frequency or wavelength of a sound wave.
Therefore, without additional information about the amplitude, it is not possible to calculate its value.
Solution:
Given the new distance between the loudspeaker and observer is 4.0 meters.
The speed of sound in air, v, is still 343 m/s.
The frequency of the sound wave, f, is still 500 Hz.
Since the speed of sound and frequency remain constant, the wavelength, λ, can be calculated using the formula:
λ = v / f
Substituting the given values:
λ = 343 m/s / 500 Hz = 0.686 m
Therefore, the wavelength of the sound wave emitted by the loudspeaker remains the same at 0.686 m.
The new frequency, f', can be calculated using the formula:
f' = v / λ
Substituting the given values:
f' = 343 m/s / 0.686 m = 500 Hz
Therefore, the new frequency of the sound wave emitted by the loudspeaker remains the same at 500 Hz.