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"
```

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')
```

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')
```

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')
```

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'
```

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')
```

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')
```

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')
```

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')
```

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')
```

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)
```

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')
```

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')
```

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')
```

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')
```

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')
```

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))
```

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')
```

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')
```

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')
```

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')
```

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')
```

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')
```

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')
```

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')
```

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
nudashdash=(v/(v-us))*nu#frequency received by the wall
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')
```

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')
```

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')
```

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')
```

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')
```

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')
```