1) The document discusses sound waves, including equations for speed of sound, pressure variations, and intensity. 2) It provides an example problem of calculating the displacement amplitude, intensity, and power delivered to the ear for a 1000 Hz sound wave with a pressure amplitude of 10.0 μPa. 3) The solutions show the displacement amplitude is 5.4 x 10-12 m, the intensity is 2.37 x 10-13 W/m2, and the power delivered to a 0.5 cm radius eardrum is 1.86 x 10-17 W.

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