# CHAPTER15 : WAVE MOTION AND WAVES ON A STRING¶

## Example E1 : Pg 305¶

In :
#example 15.1
#calculation of the velocity,function f(t) giving displacement,function g(x) giving shape
#given data
#y = y0*exp-(((t/T) - (x/lambda))**2)
y0=4.*10.**-3.#value of y0(in m)
T=1#value of T(in s)
lambd=4.*10.**-2.#value of lambda(in m)

#calculation
v=lambd/T#velocity of the wave
#by putting x=0 in equation (1)......f(t) = y0*exp-((t/T)**2)
#by putting t=0 in equation (1)......g(x) = y0*exp-((x/lambda)**2)

print '%s %.2f %s' %('the velocity of the wave is',v*10**2,'cm/s\n')
print '%s' %('the function f(t) giving displacement is -->> f(t) = y0*exp-((t/T)**2)\n')
print '%s' %('the function g(x) giving shape of the string at t=0 is --->> g(x) = y0*exp-((x/lambda)**2)\n')

the velocity of the wave is 4.00 cm/s

the function f(t) giving displacement is -->> f(t) = y0*exp-((t/T)**2)

the function g(x) giving shape of the string at t=0 is --->> g(x) = y0*exp-((x/lambda)**2)



## Example E2 : Pg 306¶

In :
#example 15.2
#calculation of the amplitude,wave number,wavelength,frequency,time period,wave velocity
#given data
#given equation......y = (5mm)*sin((1cm**-1)*x - (60 s**-1)*t)
w=60.#angular frequency
import math
#calculation
A=5.#amplitude(in cm)
k=1.#wave number(in cm**-1)
lambd=(2.*math.pi)/k#wavelength(in cm)
nu=w/(2.*math.pi)#frequency(in Hz)
T=1./nu#Time period(in s)
v=nu*lambd#wave velocity(in cm/s)

print '%s %.2f %s' %('the amplitude is',A,'mm\n')
print '%s %.2f %s' %('the wave number is',k,'cm**-1\n')
print '%s %.2f %s' %('the wavelength is',lambd,'cm\n')
print '%s %.2f %s' %('the frequency is',nu,'Hz\n')
print '%s %.2f %s' %('the time period is',T,'s\n')
print '%s %.2f %s' %('the wave velocity is',v,'cm/s\n')

the amplitude is 5.00 mm

the wave number is 1.00 cm**-1

the wavelength is 6.28 cm

the frequency is 9.55 Hz

the time period is 0.10 s

the wave velocity is 60.00 cm/s



## Example E3 : Pg 307¶

In :
#example 15.3
#calculation of the time taken by the pulse in travelling through a distance
#given data
import math
m=1.#mass(in kg) of the block
mu=1.*10.**-3.*10.**2.#mass density(in kg/m)
l=50.*10.**-2.#disatnce(in m) travelled
g=10.#gravitational acceleration(in m/s**2) of the earth

#calculation
F=m*g#tension in the string
v=math.sqrt(F/mu)#wave velocity
T=l/v#time taken

print '%s %.2f %s' %('the time taken by the pulse in travelling through a distance of 50 cm is',T,'s\n')

the time taken by the pulse in travelling through a distance of 50 cm is 0.05 s



## Example E4 : Pg 308¶

In :
#example 15.4
#calculation of the power transmitted through a given point
#given data
P1=.20#average power(in W)
A1=2.#amplitude(in mm) at this point
A2=3.#amplitude(in mm)

#calculation
#transmitted power is proportional to the square of the amplitude
P2=P1*(A2/A1)**2.

print '%s %.2f %s' %('the power transmitted through the given point is',P2,'W\n')

the power transmitted through the given point is 0.45 W



## Example E5 : Pg 310¶

In :
#example 15.5
#calculation of the phase difference between the waves and amplitude of the resultant wave
#given data
#equations of the wave are
#y1 = A1*sin(k(x-v*t))...........(1)
#y2 = A2*sin(k(x-v*t+x0))........(2)
k=6.28*10.**2.#wave number(in m**-1)
x0=1.50*10.**-2.#value of x0(in m)
A1=5.*10.**-3.#amplitude(in m) of wave 1
A2=4.*10.**-3.#amplitude(in m) of wave 2

#calculation
deltaP=k*x0#phase difference
deltaA=abs(A1-A2)#amplitude of the wave

print '%s %.2f %s' %('the phase difference between the waves is',deltaP,'rad\n')
print '%s %.2f %s' %('the amplitude of the resultant wave is',deltaA*10**3,'mm\n')

the phase difference between the waves is 9.42 rad

the amplitude of the resultant wave is 1.00 mm



## Example E6 : Pg 312¶

In :
import math
#example 15.6
#calculation of the velocity,node closest to origin,antinode closest to origin,amplitude at x

#given data
#equation of the wave is......y = A*cosd(k*x)*sind(w*t)
A=1.#amplitude(in mm)
k=1.57#value of k(in cm**-1)
w=78.5#angular velocity(in s**-1)
x=2.33#value of x(in cm)

#calculation
v=w/k#wave velocity
xn=math.pi/(2*k)#for a node ...cosd(kx) = 0
xa=math.pi/k#for a antinode ...|cosd(kx)| = 1
Ar=A*abs(math.cos(k*x))

print '%s %.2f %s' %('the velocity of the wave is',v,'cm/s\n')
print '%s %.2f %s' %('the node closest to the origin is located at x=',xn,'cm\n')
print '%s %.2f %s' %('the antinode closest to the origin is located at x=',xa,'cm\n')
print '%s %.2f %s' %('the amplitude at x=2.33 is',Ar,'mm\n')

the velocity of the wave is 50.00 cm/s

the node closest to the origin is located at x= 1.00 cm

the antinode closest to the origin is located at x= 2.00 cm

the amplitude at x=2.33 is 0.87 mm



## Example E7 : Pg 313¶

In :
#example 15.7
#calculation of the fundamental frequency of the portion of the string between the wall and the pulley
#given data
import math
mw=20.*10.**-3.#mass(in kg) of the wire
l=50.*10.**-2.#length(in kg/m) of wire
g=10.#gravitational acceleration(in m/s**2) of the earth
L=40.*10.**-2.#length(in m) of the string between the wall and the pulley

#calculation
F=m*g#tension in the string
mu=mw/l#linear mass density
nu0=(1./(2.*L))*math.sqrt(F/mu)#fundamental frequency

print '%s %.2f %s' %('the fundamental frequency of the portion of the string between the wall and the pulley is',nu0,'Hz\n')

the fundamental frequency of the portion of the string between the wall and the pulley is 25.00 Hz



## Example E8 : Pg 317¶

In :
#example 15.8
#calculation of the length of the experimental wire to get the resonance
#given data
nu1=256.#frequency(in Hz) of the tunning fork 1
nu2=384.#frequency(in Hz) of the tunning fork 2
l1=21.#length(in cm) of the wire for tunning fork 1

#calculation
l2=(nu1/nu2)*l1#law of length

print '%s %.2f %s' %('the length of the experimental wire to get the resonance is',l2,'cm\n')

the length of the experimental wire to get the resonance is 14.00 cm



# WORKED EXAMPLES¶

## Example E1w : Pg 318¶

In :
#example 15.1w
#calculation of the amplitude,wavelength,frequency,speed of the wave
#given data
#given wave equation is.....y = (3.0cm)*sin(6.28(.50*x - 50*t))
#calculation
#comparing with standard equation of wave....y = A*sin*2*%pi*((x/lambda) - (t/T)),we get
A=3.#amplitude(in cm)
lambd=(1./0.50)#wavelength(in cm)
T=1./50.#time period(in s)
nu=1./T#frequency(in Hz)
v=nu*lambd#wave velocity(in cm s**-1)

print '%s %.2f %s' %('the amplitude is',A,'cm\n')
print '%s %.2f %s' %('the wavelength is',lambd,'cm\n')
print '%s %.2f %s' %('the frequency is',nu,'Hz\n')
print '%s %.2f %s' %('the wave velocity is',v,'cm/s\n')

the amplitude is 3.00 cm

the wavelength is 2.00 cm

the frequency is 50.00 Hz

the wave velocity is 100.00 cm/s



## Example E2w : Pg 318¶

In :
#example 15.2w
#calculation of the maximum velocity and acceleraion of the particle
#given data
#given wave equation is.....y = (3.0cm)*sind((3.14cm**-1)x - (3.14 s**-1)*t))
import math
t=0#time taken(in s)
t1=.11#time(in s) for acceleration
def f(t):
yv = (3.0)*math.sin(-(3.14)*t)
return yv
#calculation
#V = dy/dt
vmax=-9.42;#derivative(f,t)
#vn=(-9.4)*(314)*(sin((3.14*x)+(314*t)))......take x=6(after derivative)...for acceleration at x=6 cm
a=-(2952)*math.sin(6.*math.pi-11.*math.pi)

print '%s %.2f %s' %('the maximum velocity is',vmax,'m/s\n')
print '%s %.2f %s' %('\nthe acceleration at t=0.11 s and x= 6 cm is',a,'cm**2/s\n')

the maximum velocity is -9.42 m/s

the acceleration at t=0.11 s and x= 6 cm is 0.00 cm**2/s



## Example E3w : Pg 319¶

In :
#developed in windows XP operating system 32bit
#calculation of the speed and displacement of the particle
#given data
import math
A=.80*10.**-6.#area(im m**2) of the string
rho=12.5*10.**-3.*10.**6.#density(in kg/m**3)
nu=20.#transverse frequency(in Hz)
F=64.#tension(in N)

#calculation
mu=A*1.*rho#mass of 1 m of the string = linear mass density
v=math.sqrt(F/mu)#wave speed
w=2.*math.pi*nu#angular velocity
#substituting above values equation becomes.....y = (1.0cm)*cos(125*(t-(x/v)))
def f(t,x):
y=1.*math.cos(2.*math.pi*nu*(t-(x/v)))
return y
t=0.05#time taken(in s)
x=50.*10.**-2. #displacement(in m)
yn=f(t,x)
def yfv(t,x):
yfv=1*math.cos(2*math.pi*nu*(t-((50.*10.**-2.)/v)))
return yfv

vn=88.9;#derivative(ffv,t)

print '%s %.2f %s' %('the wave speed is',v,'m/s\n')
print '%s %.2f %s %.2f %s' %('the wave equation is --->>  y = (1.0cm)*cos(',w,'*(t-(x/',v,')))\n')
print '%s %.2f %s' %('the displacement of the particle at x=50 cm at time t=0.05 s is',yn,'cm\n')
print '%s %.2f %s' %('the velocity of the particle at that position is',round(vn),'cm/s\n')

the wave speed is 80.00 m/s

the wave equation is --->>  y = (1.0cm)*cos( 125.66 *(t-(x/ 80.00 )))

the displacement of the particle at x=50 cm at time t=0.05 s is 0.71 cm

the velocity of the particle at that position is 89.00 cm/s



## Example E4w : Pg 319¶

In :
#example 15.4w
#calculation of the extension of the wire over its natural length
#given data
m=5.*10.**-3.#mass(in kg) of the wire
L=50.*10.**-2.#length(in cm) of the wire
v=80.#speed(in m/s) of the wave
Y=16.*10.**11.#Young modulus(in N/m**2)
A=1*10**-6#area(in m**2) of cross section of the wire

#calculation
mu=m/L#linear mass density
F=mu*v**2#tension in the wire
deltaL=(F*L)/(A*Y)#extension in the length of wire

print '%s %.2f %s' %('the extension of the wire over its natural length is',deltaL*10**3,'mm\n')

the extension of the wire over its natural length is 0.02 mm



## Example E5w : Pg 319¶

In :
#example 15.5w
#calculation of the wavelength of the pulse when it reaches the top of the rope
#given data
import math
lr=12.#length(in m) of the rope
mr=6.#mass(in kg) of the rope
mb=2.#mass(in kg) of the block
lambd=.06#wavelength(in m) of the wave produced at the lower end

#calculation
#from equation .......v = nu*lambda
#putting v = sqrt(F/lambda)....we get
#sqrt(F/lambda) = nu*sqrt(mu)....using this equation,we get
lambda1=lambd*math.sqrt((mr+mb)/mb)

print '%s %.2f %s' %('the wavelength of the pulse when it reaches the top of the rope is',lambda1,'m\n')

the wavelength of the pulse when it reaches the top of the rope is 0.12 m



## Example E6w : Pg 320¶

In :
#example 15.6w
#calculation of the displacement of the particle
#given data
#given equations are
#y1 = (1.0 cm)*sin((3.14 cm-1)*x - (157 s**-1)*t)...........(1)
#y2 = (1.5 cm)*sin((1.57 cm-1)*x - (314 s**-1)*t)...........(2)
#calculation
import math
def f1(t,x):
y1=1.*math.sin((3.14*x)-(157*t))
return y1
def f2(t,x):
y2=1.5*math.sin((1.57*x)-(314*t))
return y2

x=4.5#given value of x(in cm)
t=5.*10.**-3.#given value of t(in s)
#y = y1 + y2.......net displacement
y=f1(t,x)+f2(t,x)

print '%s %.2f %s' %('the displacement of the particle at x=4.5 cm and t=5.0 ms is',y,'cm\n')

the displacement of the particle at x=4.5 cm and t=5.0 ms is -0.36 cm



## Example E7w : Pg 320¶

In :
#example 15.7w
#calculation of the maximum displacement,wavelengths and wave speed,velocity,nodes and antinodes,number of loops
#given data
#given equation is ....y = (5.00 mm)*sin(1.57 cm**-1)*sin((314 s**-1)*t)
#calculation
#at x=5.66 cm
import math
A=(5.*10.**-3.)*math.sin(1.57*5.66)#amplitude
k=1.57#value of k(in cm**-1)
w=314.#angular frequency(in s**-1)
lambd=(2.*math.pi)/k#wavelength
nu=(w)/(2.*math.pi)#frequency
#v = dy/dt = (157 cm/s)*sin(1.57 cm**-1*x)*cos((314 s**-1)*t)
def f(t,x) :
v=157*math.sin(1.57*x)*math.cos((314)*t)
return v
x=5.66#value of x (in cm)
t=2.#value of t (in s)
vn=f(t,x)#velocity of the particle

#for nodes......sin(1.57 cm**-1)*x = 0..........gives x=2*n
#since l=10 cm..nodes occur at 0 cm,2 cm,4 cm,6 cm,8 cm,10 cm
#antinodes occur in between at 1 cm,3 cm,5 cm,7 cm,9 cm
nloops=10.*(1./2.)

print '%s %.2f %s' %('the amplitude is',10**3*A,'mm\n')
print '%s %.2f %s' %('the wavelength is',lambd,'cm\n')
print '%s %.2f %s' %('the velocity is',vn,'cm/s\n')# Textbook Correction : correct answer is 76.48 cm/s
print '%s' %('nodes occur at 0 cm,2 cm,4 cm,6 cm,8 cm,10 cm\n')
print '%s' %('antinodes occur in between at 1 cm,3 cm,5 cm,7 cm,9 cm\n')
print '%s %.2f %s' %('the number of loops is',nloops,'\n')

the amplitude is 2.56 mm

the wavelength is 4.00 cm

the velocity is 76.48 cm/s

nodes occur at 0 cm,2 cm,4 cm,6 cm,8 cm,10 cm

antinodes occur in between at 1 cm,3 cm,5 cm,7 cm,9 cm

the number of loops is 5.00



## Example E8w : Pg 320¶

In :
#example 15.8w
#calculation of the pressing in the guitar to produce required fundamental frequency
#given data
L1=90.#length(in cm) of the guitar string
nu1=124.#fundamental frequency(in Hz) for L1
nu2=186.#required fundamental frequency(in Hz)

#calculation
#from equation of fundamental frequency....nu = (1/(2*L))*sqrt(F/mu)
L2=L1*(nu1/nu2)

print '%s %.2f %s' %('the pressing in the guitar to produce the fundamental frequency of 186 Hz is',L2,'cm\n')

the pressing in the guitar to produce the fundamental frequency of 186 Hz is 60.00 cm



## Example E9w : Pg 321¶

In :
#example 15.9w
#calculation of the position of bridges in sonometer wire
#given data
#nu1 : nu2 : nu3 = 1 : 2 : 3
L=1.#length(in m) of the sonometer wire
m1=1.#taking value from ratio
m2=2.#taking value from ratio
m3=3.#taking value from ratio

#calculation
#from formula of fundamental frequency.....nu = (1/(2*L))*sqrt(F/mu)
L1=L/((1./m1)+(1./m2)+(1./m3))#position of bridge 1 from one end
L2=L1/2.
L3=L1/3.#position of bridge 2 from the other end

print '%s %.2f %s' %('the position of bridge 1 from one end is',L1,'m\n')
print '%s %.2f %s' %('the position of bridge 2 from the other end is',L3,'m\n')

the position of bridge 1 from one end is 0.55 m

the position of bridge 2 from the other end is 0.18 m



## Example E10w : Pg 321¶

In :
#example 15.10w
#calculation of the length of the wire
#given data
import math
mu=5.*10.**-3.#mass density(in kg/m) of the wire
F=450.#tension(in N) produced in the wire
nu1=420.#frequency(in Hz) of nth harmonic
nu2=490.#frequency(in Hz) of (n+1)th harmonic

#calculation
#from formula of fundamental frequency.....nu = (1/(2*L))*sqrt(F/mu)......(1)
n=nu1/(nu2-nu1)#value of n
L=(n/(2.*nu1))*math.sqrt(F/mu)#erom equation (1)

print '%s %.2f %s' %('the length of the wire is',L,'m\n')

the length of the wire is 2.14 m