AP Physics 2 Exam Question:

A speaker emits a sound wave with a frequency of 1000 Hz and a wavelength of 0.34 meters. The sound wave travels through air with a speed of 340 m/s.

(a) Determine the period and the angular frequency of the sound wave.

(b) Calculate the amplitude of the wave if the maximum displacement of air particles is found to be 0.2 mm.

(c) If the sound wave is incident on a rigid barrier, explain what happens to the wave.

Answer:

(a) The period of a wave (T) is the time taken for one complete wave to pass a certain point. The relationship between the period and the frequency (f) of a wave is given by the equation:

T = 1/f

Given the frequency of the sound wave as 1000 Hz, we can substitute it into the equation to find the period:

T = 1/(1000 Hz) = 0.001 s

The angular frequency (ω) of a wave is the rate at which the wave oscillates in radians per second. It is related to the frequency by the equation:

ω = 2πf

Substituting the given frequency into the equation, we can find the angular frequency:

ω = 2π(1000 Hz) = 2000π rad/s

(b) The amplitude of a wave (A) represents the maximum displacement of the wave from its equilibrium position. In this case, the maximum displacement of air particles is given as 0.2 mm.

To convert the displacement to meters, we divide the given value by 1000:

A = 0.2 mm / 1000 = 0.0002 meters

Therefore, the amplitude of the wave is 0.0002 meters.

(c) When a sound wave is incident on a rigid barrier, it reflects back without any change in its properties. The barrier effectively acts as an impedance mismatch, causing the wave to reverse its direction and travel back in the opposite direction. The reflected wave has the same frequency, wavelength, and amplitude as the incident wave, but its direction of propagation is opposite.