Dogs can hear higher frequencies than humans because their hearing range is 40-60,000 Hz while humans can generally hear 20-20,000 Hz. This means dogs can hear sounds that are too high in frequency for humans to detect. The document then provides information on sound waves, terms related to sound, and works through an example problem to calculate the frequency of a whistle and determine if a human, dog, or bat could hear it (24.2 kHz frequency, which humans cannot hear but dogs and bats can).

1. Why can dogs hear things
humans cannot?
SOUND
WAVES

2. A SOUND WAVE IS A LONGITUDINAL
PRESSURE WAVE CONSISTING OF
COMPRESSIONS AND RAREFRACTIONS

3.  Frequency of a longitudinal wave refers to number of cycle
(wavelengths) per second
 Humans can generally hear 20-20 000 Hz
 Dogs can generally hear 40-60 000 Hz
 That means dogs can hear approximately 40 000 frequencies
that are inaudible to us
THE FREQUENCY OF SOUND IS
MEASURED IN HERTZ (HZ)

4.  Pitch refers to frequency. A higher frequency produces a high
pitch
 Loudness depends on amplitude. A greater amplitude
produces a louder sound
 Wave speed depends on the medium through which the wave
is transmitted. Sound travels faster in a solid than a liquid,
and a liquid than a gas.
 The bulk modulus (B) refers to the amount which a volume is
changed by a pressureΔ
 The speed of sound can also be given with respect to B
SOME IMPORTANT TERMS RELATING TO
SOUND WAVES
v=λ f
B= -V*ΔP
ΔV
v=√(B/ρ)

5.  Q) Amanda blows a whistle in her backyard. The package for
the whistle gave the wavelength as 16 cm, and it also gave
the frequency but it was scratched off. Knowing that the bulk
modulus for air (at 20 C) is 1.01x105 Pa, and the density of air
is 1.2 kg/m3,
 A) What is the frequency of the whistle?
 B) Will Amanda hear it?
 C) Will Amanda’s dog, who is sitting at her feet, hear it?
 D) Would a bat hear it? (Bats hear a frequency of app. 15-90
kHz)
Try it now before looking at the walkthrough on the next slides!
TEST YOUR KNOWLEDGE

6.  Using our equation v=√(B/ρ) combined with v=λ f we can use
substitution to create an equation for frequency that uses
only the information we have:
 v=λ f  f=v/λ where v=√(B/ρ) therefor f=(√(B/ρ) )/λ
 Now we can sub in our values:
 *Remember to convert centimeters to meters for your
wavelength
 f=(√(1.01x105 Pa/1.2 kg/m3) )/0.012m
 =24 176.2 Hz  24.2 kHz
A) WHAT IS THE FREQUENCY OF THE
WHISTLE

7.  No, 24.2 kHz is outside of the average humans hearing
capacity of 20-20 000 Hz.
B) CAN AMANDA HEAR IT?
I can’t hear
anything!

8.  Probably, 24.2 Hz is well within the average canine hearing
capacity of 40-60 000 Hz
C) WILL AMANDA’S DOG, WHO IS SITTING
AT HER FEET, HEAR IT?

9.  Yes, it is also within the range of the bats hearing capacity
D) WOULD A BAT HEAR IT?

10. Image and information sources:
 http://revisionworld.com/gcse-
revision/physics/waves/describing-waves
 http://en.wikipedia.org/wiki/Hearing_range#Bats
 http://www.privateline.com/TelephoneHistory/soundwaves.ht
ml
THANKS FOR WATCHING