In [2]:

```
# Given
v = 34500 # speed of sound in cm/sec
f = 20 # lower limit of frequency for human hearing ear in Hz
f_ = 20000 # upper limit of frequency for human hearing ear in Hz
#Clculations
l1 = v / f
l2 = v / f_
#Results
print "Wavelength range of the sound wave is %.f cm to %.f cm."%(l1,l2)
```

In [3]:

```
from math import sqrt
# Given
T = 373. # temperature in kelvin
d = 1.293e-3 # density of air at S.T.P. in gm/cm^3
d_ = 13.6 # density of mercury in gm/cm^3
Cp = 0.2417 # specific heat of air at constant pressure
Cv = 0.1715 # specific heat of air at constant volume
g = 980 # gravitational constant i dynes/cm^3
#calculations
p = 76 * d_ * g
gama = Cp / Cv
v = sqrt(gama * (p / d))
v_ = v * sqrt(T / 273)
#Result
print "Velocity of sound in the air in %.f cm/sec."%v_
#Incorrect answer in textbook
```

In [4]:

```
from math import sqrt
# Given that
n = 512. # frequency of tuning fork in Hz
T = 290. # temperature in kelvin
l = 66.5 # wavelength of the gas emitted by tuning fork in cm
d = 1.293e-3 # density of air at S.T.P. in gm/cm^3
d_ = 13.6 # density of mercury in gm/cm^3
g = 980 # gravitational constant i dynes/cm^3
#Calculations
p = 76 * d_ * g# calculation for pressure
v_ = n * l# calculation for velocity of sound in air at temperature 17 c
v = v_ * sqrt(273 / T)# calculation for velocity of sound in air at temp 0 c
gama = v**2 * (d / p)# calculation for ratio of two specific heat
#Result
print "Ratio of two principal specific heats of air is %.2f"%gama
```

In [6]:

```
# Given
A = 15 * 30 # area of the floor in square meter
h = 6 # height of hall in meter
N = 500 # no. of people
t = 1.36 # optimum time for orchestral music in sec
k = 0.44 # absorption coefficient per person
#Calculations
aS = 0.161 * ((A * h) / t)
a = N * k
a_ = aS - a
w = a_ + (N / 2) * k + (N / 2) * 0.02
t = (0.161 * (A * h)) / w
#Results
print "Coefficient of absorption to be provided by the walls, floor and ceiling when the hall is fully occupied is %.f SI unit."%a_
print "Reverberation time if only half upholstered seats are occupied is %.2f sec."%t
#Answer varies due to rounding-off
```

In [10]:

```
# Given
V = 8000 # volume of hall in meter^3
t = 1.8 # reverberation time in sec
#Calculation
aS = (0.161 * V) / t# calculation for the total absorption constant
#Result
print "The total absorption constant = %.3f O.W.U."%aS
#Incorrect answer in the textbook
```

In [11]:

```
# Given
V = 1700 # volume in meter^3
a1 = 98 # area of plastered wall in m^2
a2 = 144 # area of plastered ceiling in m^2
a3 = 15 # area of wooden door in m^2
a4 = 88 # area of cushioned chairs in m^2
a5 = 150 # area of audience (each person) in m^2
k1 = 0.03 # coefficient of absorption for plastered wall in O.W.U.
k2 = 0.04 # coefficient of absorption for plastered ceiling in O.W.U.
k3 = 0.06 # coefficient of absorption for wooden door in O.W.U.
k4 = 1 # coefficient of absorption for cushioned chair in O.W.U.
k5 = 4.7 # coefficient of absorption for audience (each person) in O.W.U.
#Calculations
A1 = a1 * k1# calculation for the absorption by the plaster wall
A2 = a2 * k2# calculation for the absorption by the plastered ceiling
A3 = a3 * k3# calculation for wooden door
A4 = a4 * k4# calculation for cushioned chairs
A = A1 + A2 + A3 + A4# calculation for total absorption
T = 0.161 * (V / A)# calculation for reverberation time
#Result
print "Reverberation time is %.2f sec"%T
```

In [12]:

```
# Given
V = 1400 # volume of hall in meter^3
C = 110 # seating capacity of hall
a1 = 98 # area of plastered wall in m^2
a2 = 144 # area of plastered ceiling in m^2
a3 = 15 # area of wooden door in m^2
a4 = 88 # area of cushioned chairs in m^2
a5 = 150 # area of audience (each person) in m^2
k1 = 0.03 # coefficient of absorption for plastered wall in O.W.U.
k2 = 0.04 # coefficient of absorption for plastered ceiling in O.W.U.
k3 = 0.06 # coefficient of absorption for wooden door in O.W.U.
k4 = 1 # coefficient of absorption for cushioned chair in O.W.U.
k5 = 4.7 # coefficient of absorption for audience (each person) in O.W.U.
#Calculations
A1 = a1 * k1# calculation for the absorption by the plaster wall
A2 = a2 * k2# calculation for the absorption by the plastered ceiling
A3 = a3 * k3# calculation for wooden door
A4 = a4 * k4# calculation for cushioned chairs
A5 = C*k5 # the absorption due to persons
A = A1 + A2 + A3 + A4 + A5 # calculation for total absorption
T = (0.161 * V) / A# calculation for the reverberation time
#Result
print "Reverberation time is %.3f sec"%T
```

In [13]:

```
# Given
V = 980 # volume in meter^3
a1 = 150 # area of wall in m^2
a2 = 95 # area of ceiling in m^2
a3 = 90 # area of floor in m^2
k1 = 0.03 # coefficient of absorption for wall in O.W.U.
k2 = 0.80 # coefficient of absorption for ceiling in O.W.U.
k3 = 0.06 # coefficient of absorption for floor in O.W.U.
#calculations
A1 = a1 * k1
A2 = a2 * k2
A3 = a3 * k3
A = A1 + A2 + A3
T = 0.161 * (V / A)
#Result
print "Reverberation time = %.2f sec"%T
```

In [14]:

```
# Given
V = 980 # volume in meter^3
a = 1.58 # area of window in m^2
I_ = 1e-12 # standard intensity level of sound wave in W/m^2
l = 60 # intensity level in dB
#calculations
I = I_ * 10**(l / 10)# calculation for intensity
AP = I * a# calculation for acoustic power
#Result
print "Acoustic power = %.2e watt"%AP
```