Problem :
Find an expression for the angular frequency of a wave in terms of the wavelength and phase
velocity.
The most general form of a harmonic wave is given by
ψ = A cos[k(x  vt)], where
v is the
phase velocity and
k is the wave number. Expanding this we have
ψ = A cos(kx  kvt).
We know that the argument of the cosine must be dimensionless, so the expression
kvt must be
dimensionless, thus
kv must be an inverse time, or the angular frequency of the wave (we know it is
an angular frequency and not a regular frequency since we want the argument of the cosine to
be in radians, which are dimensionless). Thus
σ = kv. But the wavenumber is just
k = 2Π/λ so
σ = .
Problem :
If the numbers in this problem are given in SI units, calculate the velocity of a wave given by the equation:
ψ(y, t) = (9.3×10^{4})sin[Π(9.7×10^{6}y + 1.2×10^{15}t)].
The speed is given by
v = = = 1.24×10^{8} meters per second. The direction is the along in the
yaxis in the
negative direction (since
a minus sign causes the wave to advance to the right, and we have a plus sign here).
Problem :
Write the equation for a wave with an amplitude 2.5×10^{3} V/m, a period 4.4×10^{15} seconds, and speed 3.0×10^{8} m/s, which is propagating in the negative zdirection
with value 2.5×10^{3} V/m at t = 0, z = 0.
We want a wave of the form
. The plus sign arises from the direction of
travel: when
t = 0,
z = 0 we have a peak at the origin, but as time increases (
z = 0,
t = Π/2, for
example) the peak advances to the left, and hence the wave is propagating in the negative direction as
required. We can calculate
σ, the angular frequency, from the period
T = 1/ν = 2Π/σ. Thus
σ = 2Π/T = = 1.43×10^{15} s
^{1}.
We can compute
k since we know that
v = σk hence
k = = = 4.76×10^{6} m
^{1}. The amplitude is given and the
cosine gives us the right phase (we could choose a sine an subtract a phase of
Π/2). Thus:
Problem :
Consider the wave ψ(x, t) = A cos(k(x + vt) + Π). Find an expression (in terms of A) for the
magnitude of the wave when x = 0, t = T/2, and x = 0, t = 3T/4 .
When
x = 0 we have
ψ = A cos(kvt + Π). At
t = T/2 we then have
ψ = A cos(kvT/2 + Π).
Now
k = 2Π/λ,
T = 1/ν and
v = λν so
kvT = 2Π. Thus we have
ψ = A cos(2Π/2 + Π) = A cos(2Π) = A. In the latter case we have
ψ = A cos(3×2Π/4 + Π) = A cos(5Π/2) = 0.
Problem :
Demonstrate explicitly that a harmonic function ψ(x, t) = A cos(kx  σt) satisfies the wave
equation. What condition needs to be fulfilled?
Clearly the second (partial) derivatives with respect to
y and
z are zero. The second derivative with
respect to
x is:
=  Ak^{2}cos(kx  σt) 

The second derivative with respect to time is:
=  Aσ^{2}cos(kx  σt) 

Now the onedimensional wave equation states that:
From the derivatives calculated above this gives:
 Ak^{2}cos(kx  σt) = . Canceling and rearranging this gives the required condition as:
v = ,
which is just the result we stated for the phase velocity.