2/10/2016 HW1
Timothy Gao
1
2.10 Table Construction
x
kx
cos(kx- π/4)
cos(kx+3π/4)
-λ/2
-π
cos(-5π/4)≈-0.71
cos(-π/4) ≈0.71
-λ/4
-π/2
cos(-3π/4) ≈-0.71
cos(π/4) ≈0.71
0
0
cos(- π/4) ≈0.71
cos(3π/4) ≈-0.71
λ/4
π/2
cos(π/4) ≈0.71
cos(5π/4) ≈-0.71
λ/2
π
cos(3π/4) ≈-0.71
cos(7π/4) ≈0.71
3λ/4
3π/2
cos(5π/4) ≈-0.71
cos(9π/4) ≈0.71
λ
2π
cos(7π/4) ≈0.71
cos(11π/4) ≈-0.71
The main point of this exercise is to understand the effect of phase lead/lag,
which is apparent in the plot below:
Clearly, the two functions have a phase difference of π, which makes it ambiguous
which leads and which lags.
2/10/2016 HW1
2.16
Timothy Gao
2
It is helpful to put the wavefunctions in the form
where A is the amplitude, k is the propagation number, and ω is the angular
temporal frequency. In this form,
A table of the relevant quantities for parts a) through e) is given below:
Quantity
Formula
ψ1(x,t)
ψ2(x,t)
Frequency
f=|ω|/2π
3s-1
0.56s-1
Wavelength
λ=2π/|k|
5m
0.9m
Period
T=1/f=2π/|ω|
0.3s
1.8s
Amplitude
A
4m
0.4m
Phase velocity
v= ω/k
15m/s
0.5m/s
f) The wavefunction ψ1 is travelling in the positive x-direction because x has to increase
as t increases in order to keep the argument in the sine constant (the coefficients of x
and t have opposite sign).
By the same logic, the wavefunction ψ2 is travelling in the negative x-direction
because x has to decrease as t increases in order to keep the argument in the sine
constant (the coefficients of x and t have the same sign).
2/10/2016 HW1
Timothy Gao
3
2.23) The profiles at t=0 and t=2 are
If we want the profile to keep the same shape for all times (and just move along the xaxis with speed v in the negative x-direction), then we need to make the pulse have the
equation
where
Thus, since the pulse at t=0 has f(x,t=0)=x, the constant term is 0 and the pulse should
have the equation
2/10/2016 HW1
Timothy Gao
2.25) This does not represent any physical wave, since it goes off to infinity at y=±∞.
Plotting this function at various times (t=0, t=2/v) makes it clear that it is just a line
with slope A that travels with speed v along the y=axis:
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2/10/2016 HW1
Timothy Gao
2.33) Factoring the exponent gives
The coefficients of x and t have the same sign (assuming a>0 and b>0). Thus, if t
increases, x must decrease to preserve the term in the exponent. Therefore, the wave is
propagating in the negative x-direction. If a=25, the wave looks like
at t=0. The speed of the wave is
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2/10/2016 HW1
Timothy Gao
2.41) The point of this problem is to show that rigid “everyday objects” (those with
masses around a kg or so) have a de Broglie wavelength so small that they are much
easier described as particles than as waves.
The 6kg stone has momentum p=6 kg m/s. Thus, the de Broglie wavelength associated
with this object is
λ=(6.6x10-34)/6 m=10-34m.
Light in the visible spectrum has wavelengths in the hundreds of nanometers (~10-7m).
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