# CHAPTER16 : SOUND WAVES¶

## Example E1 : Pg 330¶

In [2]:
#example 16.1
#calculation of the audibility of a wave
#given data
v=300.#velocity(in m/s) of the wave
lambd=.60*10.**-2.#wavelength(in m) of the wave

#calculation
nu=v/lambd#frequency of the wave
nu= 50000.
print "nu=50000","Hz"
print "This is much above the audible range. It is an ultrasonic wave and  will not be audible"

nu=50000 Hz
This is much above the audible range. It is an ultrasonic wave and  will not be audible


## Example E2 : Pg 331¶

In [3]:
#example 16.2
#calculation of the amplitude of vibration of the particles of the medium
#given data
import math
lambd=40.*10.**-2.#wavelength(in m) of the wave
deltap=1.*10.**-3.#difference between the minimum and the maximum pressure(in N/m**2)
B=1.4*10.**5.#Bulk modulus(in N/m**2)

#calculation
p0=deltap/2.#pressure amplitude
s0=(p0*lambd)/(2.*math.pi*B)#from equation of Bulk modulus

print '%s %.12f %s' %('the amplitude of vibration of the particles of the medium is',s0,'m\n')

the amplitude of vibration of the particles of the medium is 0.000000000227 m



## Example E3 : Pg 334¶

In [4]:
#example 16.3
#calculation of the intensity of the sound wave

#given data
p0=2.*10.**-2.#pressure amplitue(in N/m**2)
p0dash=2.5*10.**-2.#new pressure amplitue(in N/m**2)
I=5.0*10.**-7.#intensity(in W/m**2) of the wave

#calculation
#intensity of the wave is proportional to square of the pressure amplituide
Idash=I*((p0dash/p0)**2.)

print '%s %3.2f %s' %('the intensity of the sound wave is',Idash,'W/m**2')

the intensity of the sound wave is 0.00 W/m**2


## Example E4 : Pg 334¶

In [5]:
#example 16.4
#calculation of the increase in the sound level in decibels
import math
#given data
r=20.#intensity is increase by r factor

#calculation
#using the equation.....beta = 10*log(I/I0)...we get
deltabeta=10.*math.log10(r)#increase in sound level

print '%s %.2f %s' %('the increase in the sound level in decibels is',deltabeta,'dB')

the increase in the sound level in decibels is 13.01 dB


## Example E5 : Pg 336¶

In [6]:
#example 16.5
#calculation of the nature of interference
import math
#given data
nu=1.*10**3#frequency(in Hz) of the source
deltax=83.*10**-2#difference in the length(in m) of paths
v=332.#speed(in m/s) of the sound in air
#calculation
lamba=v/nu#wavelength
delta=(2*math.pi/lamba)*deltax
n=delta/math.pi#phase difference is 'n' multiple of pi
#results
if(n%2==0):
print 'the waves will interfere constructively.'#for even values of 'n'
else:
print 'the waves will interfere destructively.'#for odd values of 'n'

the waves will interfere destructively.


## Example E6 : Pg 338¶

In [7]:
#example 16.6
#calculation of the distance of the piston from the open end,for tube to vibrate in its first overtone

#given data
nu=416.#frequency(in Hz) of the tunning fork
v=333.#speed(in m/s) of the sound in air

#calculation
lambd=v/nu#wavelength
L=3.*lambd/4.#length of the tube

print '%s %.2f %s' %('the distance of the piston from the open end,for tube to vibrate in its first overtone is',L*10**2,'cm')

the distance of the piston from the open end,for tube to vibrate in its first overtone is 60.04 cm


## Example E7 : Pg 342¶

In [8]:
#example 16.7
#calculation of the tunning frequency of fork B

#given data
nu1=384.#tunning frequency(in Hz) of fork A
n=6.#number of beats
t=2.#time(in s) taken by the beats

#calculation
deltanu=n/t#frequency of beats
nu2=nu1+deltanu#frequency of fork B
nu2dash=nu1-deltanu#another frequency of fork B

print '%s %.2f %s %.2f %s' %('the tunning frequency of fork B is',nu2dash,'Hz or',nu2,'Hz')

the tunning frequency of fork B is 381.00 Hz or 387.00 Hz


## Example E8 : Pg 343¶

In [9]:
#example 16.8
#calculation of the most dominant frequency

#given data
us=36.*10.**3./(60.*60.)#speed(in m/s) of the train
nudash=12.*10.**3.#frequency(in Hz) detected by the detector
v=340.#velocity(in m/s) of the sound in air

#calculation
#frequency detected is ......nudash = (v*nu0)/(v-us)
nu0=(1.-(us/v))*nudash#required frequency

print '%s %.2f %s' %('the most dominant frequency is',nu0*10**-3,'kHz')

the most dominant frequency is 11.65 kHz


# WORKED EXAMPLES¶

## Example E1w : Pg 346¶

In [10]:
#example 16.1w
#calculation of the depth of the sea and wavelength of the signal in the water
#given data
nu=50.*10.**3.#frequency(in Hz) of the given signal
t=0.8#time(in s)requires for reflected wave to return
v=1500.#speed(in m/s) of the sound in water

#calculation
d=v*t/2#depth of the sea
lambd=v/nu#wavelength in water

print '%s %.2f %s' %('the depth of the sea is',d,'m\n')
print '%s %.2f %s' %('the wavelength of the signal in the water is',lambd*10**2,'cm\n')

the depth of the sea is 600.00 m

the wavelength of the signal in the water is 3.00 cm



## Example E2w : Pg 346¶

In [11]:
#developed in windows XP operating system 32bit
#platform Scilab 5.4.1
#example 16.2w
#calculation of the location of the plane

#given data
v=510.*10.**3./(60.*60.)#speed(in m/s) of the plane
h=2000.#height(in m) of the plane
vs=340.#speed(in m.s) of the sound in air

#calculation
t=h/vs#time taken by the sound to reach the observer
d=v*t#location of the plane

print '%s %.2f %s' %('the plane will be',d,'m ahead of the observer on its line of motion')

the plane will be 833.33 m ahead of the observer on its line of motion


## Example E3w : Pg 346¶

In [12]:
#example 16.3w
#calculation of the frequency,wavelength,speed,maximum and minimum pressures of the sound wave
import math
#given data
#equation of the sound wave is
#p = (0.01 N/m**2)*sin((1000 s**-1)*t - (3.0 m**-1)*x)............(1)
peq=1.0*10.**5.#equilibrium pressure(in N/m**2) of the air

#calculation
#comparing equation (1) with standard equation p = p0*sin(w*(t-(x/v)))...we get
w=1000.#value of w(in s**-1)
nu=w/(2.*math.pi)#frequency
v=w/3#velocity
lambd=v/nu#wavelength
p0=0.01#pressure amplitude(in N/m**2)

print '%s %.2f %s' %('the frequency is',nu,'Hz')
print '%s %.2f %s' %('\nthe wavelength is',lambd,'m')
print '%s %.2f %s' %('\nthe speed of the sound wave is',v,'m/s')
print '%s %.2f %s' %('\nthe maximum pressure amplitude is (%3.2e + %3.2f) N/m**2',peq,p0)
print '%s %.2f %s' %('\nthe minimum pressure amplitude is (%3.2e - %3.2f) N/m**2',peq,p0)

the frequency is 159.15 Hz

the wavelength is 2.09 m

the speed of the sound wave is 333.33 m/s

the maximum pressure amplitude is (%3.2e + %3.2f) N/m**2 100000.00 0.01

the minimum pressure amplitude is (%3.2e - %3.2f) N/m**2 100000.00 0.01


## Example E4w : Pg 346¶

In [13]:
#developed in windows XP operating system 32bit
#platform Scilab 5.4.1
#example 16.4w
#calculation of the minimum separation between the two points for a given phase difference
import math
#given data
nu=10.*10.**3.#frequency(in Hz) of the sound wave
v=340.#speed(in m/s) of the wave
delta=60.#phase difference(in degree)

#calculation
lambd=v/nu#wavelength
k=2.*math.pi/lambd#wave number
d=(delta*math.pi/180.)/k

print '%s %3.2f %s' %('the minimum separation between the two points for phase difference of 60 degree is',d*10.**2.,'cm')

the minimum separation between the two points for phase difference of 60 degree is 0.57 cm


## Example E5w : Pg 347¶

In [14]:
#developed in windows XP operating system 32bit
#platform Scilab 5.4.1
#example 16.5w
#calculation of the atmospheric temperature

#given data
v1=336.#speed(in m/s) travelled by the sound
v0=332.#speed(in m/s) of the sound at O degreecelsius
T0=0+273.#temperature(in kelvin)

#calculation
T=((v1/v0)**2.)*T0#temperature (in kelvin)
t=T-273.#temperature(in degreecelsius)

print '%s %.2f %s' %('the atmospheric temperature is',round(t),'degreecelsius')

the atmospheric temperature is 7.00 degreecelsius


## Example E6w : Pg 347¶

In [15]:
#example 16.6w
#calculation of the speed of sound wave in hydrogen

#given data
gama=1.4#value of constant gama for hydrogen
voxygen=470.#speed(in m/s) of the sound wave in oxygen
import math
#calculation
#speed of sound wave in a gas is ........v = sqrt(gama*P/rho)
#at STP ,density of oxygen is 16 times density of hydrogen
vhydrogen=voxygen*math.sqrt(16.)#speed of sound in hydrogen

print '%s %.2f %s' %('the speed of sound wave in hydrogen is',vhydrogen,'m/s')

the speed of sound wave in hydrogen is 1880.00 m/s


## Example E7w : Pg 347¶

In [16]:
#example 16.7w
#calculation of the energy delivered to the microphone
import math
#given data
A=.80*10.**-4.#area(in m**2) of the cross section
U=3.#power(in W0 output of the speaker
d=2.#distance(in m) between the microphone and the speaker
t=5.#time(in s) taken

#calculation
U0=A*U/(4.*math.pi*d**2.)#energy falling on the microphone in 1 s
Udash=U0*t#energy falling on the microphone in t s

print '%s %.2f %s' %('the energy delivered to the microphone in t=5 s is',round(Udash*10**6),'microJ')

the energy delivered to the microphone in t=5 s is 24.00 microJ


## Example E8w : Pg 347¶

In [17]:
#example 16.8w
#calculation of the amplitude of vibration of the particles of the air
import math
#given data
I=2.*10.**-6.#intensity(in W/m**2) of the sound wave
nu=1.*10.**3.#frequency(in Hz) of the sound wave
rho0=1.2#density(in kg/m**3) of the air
v=330.#speed(in m/s) of the sound in the air

#calculation
s0=math.sqrt(I/(2.*math.pi**2.*nu**2.*rho0*v))#equation of displacement amplitide

print '%s %.5f %s' %('the amplitude of vibration of the particles of the air is',s0,'m')

the amplitude of vibration of the particles of the air is 0.00000 m


## Example E9w : Pg 347¶

In [18]:
#example 16.9w
#calculation of the factor by which the pressure amplituide increases
import math
#given data
n=30.#increase(in dB) of the sound level

#calculation
#m = I2/I1 = intensity ratio
m=10.**(n/10.)
#since p2/p1 = sqrt(I2/I1)
f=math.sqrt(m)#require factor

print '%s %.2f' %('the factor by which the pressure amplituide increases is',round(f))

the factor by which the pressure amplituide increases is 32.00


## Example E10w : Pg 347¶

In [19]:
#example 16.10w
#calculation of the frequency at which the maxima of intensity are detected
#given data
import math
r=20.*10.**-2.#radius(in m) of the semicircular part
v=340.#speed(in m/s) of the sound in air

#calculation
l1=2.*r#straight distance
l2=math.pi*r#curve distance
deltal=l2-l1
nu=v/deltal

print '%s %.2f %s %.2f %s' %('the frequency at which the maxima of intensity are detected are',nu,'Hz and',2*nu,'Hz\n')

the frequency at which the maxima of intensity are detected are 1489.15 Hz and 2978.30 Hz



## Example E11w : Pg 348¶

In [20]:
#example 16.11w
#calculation of the minimum distance between the source and the detector for maximum sound detection
#given data
nu=180.#frequency(in Hz)
d=2.#distance(in m)
v=360.#speed(in m/s) of the sound wave in air

#calculation
#path difference.....delta = (2*((2**2) + (x**2/4))**(1/2)) - (x)
lambd=v/nu#wavelength
delta=lambd
#solving the above equation,we get
x=4.-1.

print '%s %.2f %s' %('the minimum distance between the source and the detector for maximum sound detection is',x,'m\n')

the minimum distance between the source and the detector for maximum sound detection is 3.00 m



## Example E12w : Pg 348¶

In [21]:
#example 16.12w
#calculation of the length of the shortest closed organ pipe that will resonate with the tunning fork
#given data
nu=264.#frequency(in Hz)of the tunning fork
v=350.#speed(in m/s) of the sound in air

#calculation
#from the equation of the resonate frequency of the closed organ pipe....l = (n*v)/(4*nu)
n=1.#for l to be minimum
lmin=(v)/(4.*nu)#equation of the resonate frequency of the closed organ pipe

print '%s %.2f %s' %('the length of the shortest closed organ pipe that will resonate with the tunning fork is',lmin*10**2,'cm\n')

the length of the shortest closed organ pipe that will resonate with the tunning fork is 33.14 cm



## Example E13w : Pg 348¶

In [22]:
#example 16.13w
#calculation of the length of the closed pipe
#given data
l0=60.*10.**-2.#length(in m) of the open pipe

#calculation
#from the equation of the resonate frequency of the closed organ pipe....l=(n*v)/(4*nu)
l1=l0/4.

print '%s %.2f %s' %('the length of the closed pipe is',l1*10**2,'cm\n')

the length of the closed pipe is 15.00 cm



## Example E14w : Pg 348¶

In [23]:
#example 16.14w
#calculation of the speed of the sound in air
#given data
nu=800.#frequency(in Hz) of the tunning fork
l1=9.75*10.**-2.#distance(in m) where resonance is observed
l2=31.25*10.**-2.#distance(in m) where resonance is observed
l3=52.75*10.**-2.#distance(in m) where resonance is observed

#calculation
#from the equation of the resonate frequency ....l = (n*v)/(4*nu)
#(n*v)/(4*l1) = nu...................(1)
#((n+2)*v)/(4*l2) = nu...............(2)
#((n+4)*v)/(4*l3) = nu...............(3)
#form above equations ,we get
v=2.*nu*(l2-l1)

print '%s %.2f %s' %('the speed of the sound in air is',v,'m/s\n')

the speed of the sound in air is 344.00 m/s



## Example E15w : Pg 349¶

In [24]:
#example 16.15w
#calculation of the fundamental frequency if the air is replaced by hydrogen
#given data
import math
nu0=500.#fundamental frequency(in Hz)
rhoa=1.20#density(in kg/m**3) of air
rhoh=0.089#density(in kg/m**3) of hydrogen

#calculation
#fundamental frequency of an organ pipe is proportional to the speed of the sound
nu=nu0*math.sqrt(rhoa/rhoh)

print '%s %.2f %s' %('the fundamental frequency if the air is replaced by hydrogen is',nu,'Hz\n')

the fundamental frequency if the air is replaced by hydrogen is 1835.97 Hz



## Example E16w : Pg 349¶

In [25]:
#example 16.16w
#calculation of the speed,wavelength in the rod,frequency,wavelength in the air
#given data
import math
l=90.*10.**-2.#length(in m) of the rod
rho=2600.#density(in kg/m**3) of the aluminium
Y=7.80*10.**10.#Young modulus(in N/m**2)
vai=340.#speed(in m/s) of the sound in the air

#calculation
v=math.sqrt(Y/rho)#speed of the sound in aluminium
lambd=2.*l#wavelength....since rod vibrates with fundamental frequency
nu=v/lambd#frequency
lambdaai=vai/nu#wavelength in the air

print '%s %.2f %s' %('the speed of the sound in aluminium is',v,'m/s\n')#Textbook Correction : correct answer is 5477 m/s
print '%s %.2f %s' %('the wavelength of the sound in aluminium rod is',lambd*10.**2.,'cm\n')
print '%s %.2f %s' %('the frequency of the sound produced is',nu,'Hz\n')#Textbook Correction : correct answer is 3042 Hz
print '%s %.2f %s' %('the wavelength of the sound in air is',lambdaai*10**2,'cm\n')

the speed of the sound in aluminium is 5477.23 m/s

the wavelength of the sound in aluminium rod is 180.00 cm

the frequency of the sound produced is 3042.90 Hz

the wavelength of the sound in air is 11.17 cm



## Example E17w : Pg 349¶

In [26]:
#example 16.17w
#calculation of the frequency of the note emitted by the taut string
#given data
nu1=440.#frequency(in Hz) of the string
n=4.#number of beats per second
nuf=440.#tunning frequency(in Hz) of the fork

#calculation
fre=nuf+n#required frequncy

print '%s %.2f %s' %('the frequency of the note emitted by the taut string is',fre,'Hz\n')

the frequency of the note emitted by the taut string is 444.00 Hz



## Example E18w : Pg 349¶

In [27]:
#example 16.18w
#calculation of the apparent frequency
#given data
us=36.*10.**3./(60.*60.)#speed(in m/s)of the car
v=340.#speed(in m/s) of the sound in the air
nu=500.#frequency(in Hz)

#calculation
nudash=(v/(v+us))*nu#apparent frequency  heard by the observer

print '%s %.2f %s' %('the apparent frequency heard by the ground observer is',round(nudash),'Hz\n')
print '%s %.2f %s' %('the frequency of the reflected wave as heard by the ground observer is',nudashdash,'Hz\n')

the apparent frequency heard by the ground observer is 486.00 Hz

the frequency of the reflected wave as heard by the ground observer is 515.15 Hz



## Example E19w : Pg 349¶

In [28]:
#example 16.19w
#calculation of the frequency of the whistle of the train
#given data
us=72.*10.**3./(60.*60.)#speed(in m/s) of the train 1
u0=54.*10.**3./(60.*60.)#speed(in m/s) of the train 2
nu=600.#frequency(in Hz) of the whistle
v=340.#speed(in m/s)of sound in the air

#calculation
nudash=((v+u0)/(v-us))*nu#frequency heard by the observer before the meeting of the trains
nudashdash=((v-u0)/(v+us))*nu#frequency heard by the observer after the crossing of the trains

print '%s %.2f %s' %('the frequency heard by the observer before the meeting of the trains is',round(nudash),'Hz\n')
print '%s %.2f %s' %('the frequency heard by the observer after the crossing of the trains is',round(nudashdash),'Hz\n')

the frequency heard by the observer before the meeting of the trains is 666.00 Hz

the frequency heard by the observer after the crossing of the trains is 542.00 Hz



## Example E20w : Pg 350¶

In [29]:
#example 16.20w
#calculation of the main frequency heard by the person
#given data
us=36.*10.**3./(60.*60.)#speed(in m/s) of the person on the scooter
v=340.#speed(in m/s) of sound in the air
nu=600.#frequency(in Hz) of the siren

#calculation
nudash=(v/(v+us))*nu#main frequency

print '%s %.2f %s' %('the main frequency heard by the person is',round(nudash),'Hz\n')

the main frequency heard by the person is 583.00 Hz



## Example E21w : Pg 350¶

In [30]:
#example 16.21w
#calculation of the original frequency of the source
#given data
u0=10.#speed(in m/s) of the observer going away from the source
us=10.#speed(in m/s) of the source going away from observer
nudash=1950.#frequency(in Hz) of the sound detected by the detector
v=340.#speed(in m/s) of the sound in the air

#calculation
nu=((v+us)/(v-u0))*nudash#original frequency

print '%s %.2f %s' %('the original frequency of the source is',round(nu),'Hz\n')

the original frequency of the source is 2068.00 Hz



## Example E22w : Pg 350¶

In [31]:
#example 16.22w
#calculation of the speed of the car
#given data
nudash=440.#frequency(in Hz) emitted by the wall
nudashdash=480.#frequency(in Hz) heard by the car driver
v=330.#speed(in m/s) of the sound in the air

#calculation
#frequency received by the wall..............nudash = (v/(v-u))*nu............(1)
#frequency(in Hz) heard by the car driver....nudashdash = ((v+u)/v)*nudash....(2)
#from above two equations,we get
u=((nudashdash-nudash)/(nudashdash+nudash))*v#speed of the car

print '%s %.2f %s %.2f %s' %('the speed of the car is',u,'m/s or',round(u*10**-3*60*60),'km/h\n')

the speed of the car is 14.35 m/s or 52.00 km/h



## Example E23w : Pg 350¶

In [32]:
#example 16.23w
#calculation of the frequency of train whistle heard by the person standing on the road perpendicular to the track
#given data
import math
from math import acos,cos
v=340.#speed(in m/s) of the sound in the air
d1=300.#distance(in m) of the train from the crossing
u=120.*10.**3./(60.*60.)#speed(in m/s) of the train
nu=640.#frequency(in Hz) of the whistle
d2=400.#distance(in m) of the person from the crossing ,perpendicular to the track

#calculation
theta=acos(d1/(d1**2.+d2**2.)*57.3)*5.#pythagoras theorem
nudash=(v/(v-(u*cos(theta)))*57.3)*nu#frequency of the whistle heard

print '%s %.2f %s' %('the frequency of train whistle heard by the person standing on the road perpendicular to the track is',nudash,'Hz\n')

the frequency of train whistle heard by the person standing on the road perpendicular to the track is 37926.25 Hz