1. Learning Object: Sound Waves
Introduction + Relevant Equations
• Sound waves can propagate in 3 dimensions.
• Sound is a longitudinal wave; when the medium is compressed, the
pressure is elevated and when the medium is stretched, the pressure is
lowered.
• Bulk Modulus, B, is given by: 𝐵 = −
∆!
∆!
!
= (−𝑉)(
∆!
∆!
), where ∆𝑃= change
in pressure and
∆!
!
= fractional change in volume. The negative sign
indicates that the sign of
∆!
!
always opposes the sign of ∆𝑃.
• Speed of sound given by: 𝑣 = √
!
!
• S will be used to denote the sound wave function:
𝑠 𝑥, 𝑡 = 𝑆! cos 𝑘𝑥 − 𝜔𝑡 + 𝜙
• Relationship between displacement, pressure, and intensity:
∆𝑃 = 𝐵𝑘𝑆! sin 𝑘𝑥 − 𝜔𝑡 + 𝜙
• We can find the amplitude of the pressure variations as the coefficient of
the sine factor: ∆𝑃! = 𝐵𝑘𝑆!
• Intensity: 𝐼 = 𝑃/𝐴 (where P is power and A is area) or 𝐼 = (
!
!
)𝜌𝑣𝜔!
𝑆!
!
Problem
a) Determine the displacement amplitude of a 1000 Hz sound wave whose
pressure amplitude is 10.0 𝜇Pa.
b) Calculate the intensity of the sound wave.
c) Assuming the human eardrum has a radius of 0.50 cm, find how much
power is delivered to the ear by that wave.
Solutions + Explanations
a) Re-arrange ∆𝑃! = 𝐵𝑘𝑆! for 𝑆!: 𝑆! = ∆𝑃/𝐵𝑘= (∆𝑃)/(𝐵
!!"
!
)=
(0.00001 𝑃𝑎)/( 1.01 𝑥 10!
𝑃𝑎
!! !""" !"
!"!!
!
)= 5.4 x 10-12 m
b) We know that 𝐼 = (
!
!
)𝜌𝑣𝜔!
𝑆!
!
, so
𝐼 = (
!
!
)(
!.!!"
!!
)(343
!
!
)(2𝜋 1000𝐻𝑧 !
)(5.4𝑥10!!"
𝑚)!
= 2.37 x 10-13 W/m2
c) Re-arrange 𝐼 = 𝑃/𝐴 for power ! 𝑃 = 𝐼𝐴=
(2.37𝑥10!!"
𝑊/𝑚!
)(𝜋 0.005𝑚!
)= 1.86 x 10-17 W
