AP Physics 1 Exam Question:

A loudspeaker emits a sound wave with a frequency of 1.5 kHz. The sound wave travels through air with a speed of 343 m/s. The amplitude of the sound wave is 0.02 m.

a) Calculate the wavelength of the sound wave. b) Determine the period of the sound wave. c) Find the angular frequency of the sound wave. d) Calculate the maximum displacement of air molecules from their equilibrium position due to this sound wave.

Answer and Step-by-Step Explanation:

a) To calculate the wavelength (λ) of the sound wave, we can use the formula:

λ = v/f

where v is the speed of the sound wave and f is the frequency.

Given: v = 343 m/s f = 1.5 kHz = 1500 Hz

Substituting the values into the formula, we have:

λ = 343 m/s / 1500 Hz = 0.228 m

Therefore, the wavelength of the sound wave is 0.228 m.

b) The period (T) of the sound wave is the time it takes for one complete oscillation. We can find the period using the formula:

T = 1/f

where f is the frequency.

Given: f = 1.5 kHz = 1500 Hz

Substituting the value into the formula, we have:

T = 1 / 1500 Hz = 6.67 × 10^(-4) s

Therefore, the period of the sound wave is 6.67 × 10^(-4) s.

c) The angular frequency (ω) of the sound wave is related to the period by the formula:

ω = 2π / T

where T is the period.

Given: T = 6.67 × 10^(-4) s

Substituting the value into the formula, we have:

ω = 2π / (6.67 × 10^(-4) s) ≈ 9420.6 rad/s

Therefore, the angular frequency of the sound wave is approximately 9420.6 rad/s.

d) The maximum displacement (A) of air molecules from their equilibrium position due to this sound wave is given by the formula:

A = amplitude

Given: amplitude (A) = 0.02 m

Therefore, the maximum displacement of air molecules from their equilibrium position due to this sound wave is 0.02 m.