Transcript Lecture 27 - McMaster Physics and Astronomy

Wave Motion II
•Sinusoidal (harmonic) waves
•Energy and power in sinusoidal waves
For a wave traveling in the +x direction, the
displacement y is given by
y (x,t) = A sin (kx – t)
A
with  = kv
y
x
-A
Remember: the particles in the medium move vertically.
The transverse displacement of a particle at a fixed
location x in the medium is a sinusoidal function of time –
i.e., simple harmonic motion:
y = A sin (kx – t) = A sin [ constant – t]
The “angular frequency” of the particle motion is
 ; the initial phase is kx (different for different
x, that is, particles).
ω = 2πf
ω=“angular frequency”
radians/sec
f =“frequency”
cycles/sec
(Hz=hertz)
Example
y
A
a
e
b
-A
c
x
d
Shown is a picture of a travelling wave,
y=A sin(kx- t), at the instant for time t=0.
i) Which particle moves according to y=A cos(t) ?
ii) Which particle moves according to y=A sin(t) ?
iii) If ye(t)=A cos(t+fe ) for particle e, what is
Assume ye(0)=1/2A
fe ?
Wave Velocity
The wave velocity is determined by the properties of
the medium; for example,
1) Transverse waves on a string:
v
wave
tension
T


mass/unit length

(proof from Newton’s second law and wave equation, S16.5)
2) Electromagnetic wave (light, radio, etc.)
v
1
c
 o o
(proof from Maxwell’s Equations for E-M fields, S34.3)
Example 1:
What are , k and  for a 99.7 MHz FM radio wave?
Example 2:
A string is driven at one end at a frequency of 5Hz. The
amplitude of the motion is 12cm, and the wave speed is 20
m/s. Determine the angular frequency for the wave and
write an expression for the wave equation.
Transverse Particle Velocities
Transverse particle displacement, y (x,t)
Transverse particle velocity, vy 
y
(x held
t
constant) this is called the transverse velocity
(Note that vy is not the wave speed v ! )
Transverse acceleration,
v y  2 y
ay 
 2
t t
“Standard” Traveling sine wave (harmonic wave):
y  A sin(kx  t   )
y
vy 
  A cos(kx  t   )
t
v y
ay 
 - 2 A sin(kx  t   )
t
2
 - y
maximum transverse displacement, ymax = A
maximum transverse velocity, vmax =  A
maximum transverse acceleration, amax =  2 A
These are the usual results for S.H.M
Example 3:
y
x
string: 1 gram/m; 2.5 N tension
vwave
Oscillator:
50 Hz, amplitude 5 mm, y(0,0)=0
Find: y (x, t)
vy (x, t) and maximum transverse speed
ay (x, t) and maximum transverse acceleration
Solution
Energy density in a wave
ds
dm
dx
Ignore difference between “ds”, “dx”
(this is a good approx for small A, or large  ):
dm = μ dx (μ = mass/unit length)
Each particle of mass dm in the string is executing SHM so
its total energy (kinetic + potential) is (since E= ½ mv2):
1
1
dE  dm  2 A2   dx  2 A2
2
2
The total energy per unit length is
dE 1 2 2
  A
dx 2
= energy density
Where does the potential energy in the string come from?
Power transmitted by harmonic wave with wave speed v:
A distance v of the wave travels past a fixed point in
the string in one second.
So:
P   energy density   v
For waves on a string, power transmitted is
P  12  2 A2 v
Both the energy density and the power transmitted
are proportional to the square of the amplitude and
square of the frequency. This is a general property
of sinusoidal waves.
Example 4:
A stretched rope having mass per unit length of
μ=5x10-2 kg/m is under a tension of 80 N. How much
power must be supplied to the rope to generate harmonic
waves at a frequency of 60 Hz and an amplitude of 6cm ?