Physics 214 Problem 1 Week 1
Solution
Wave Properties
The graph below shows the transverse displacement of a string segment at the position
x = 0.75 m as a function of time.
-1.0
-0.5
0.0
0.5
1.0
0.000 0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008
displacement vs time x = 0.75 m
displacement in mm
time in seconds
Suppose that the function that describes the wave on the string is
yx,t
()
=Acos 2
π
x
λ
−
ω
t+
φ
⎛
⎝
⎜
⎞
⎠
⎟
.
Measurements show that the speed of propagation of waves on the string is 250 m/s.
From data in the graph above and some calculations, find:
a) the amplitude, A
The cosine oscillates between –1 and +1. Thus y oscillates between –A and +A. So, A is
the maximum value of y: A = 0.7 mm.
b) the wavelength,
λ
The graph tells us the time dependence (the period), not the spatial dependence (the
wavelength). However, they are related: λ = vT. We are told that v = 250 m/s, and we
read from the graph, T = 0.005 s. Thus, λ = 1.25 m.
c) the angular frequency,
ω
= 2πf
f = 1/T, so
ω
= 2π/T = 1250 radians/sec. Note that radians are dimensionless.
d) the phase,
φ
.
Here you have to be a little careful. Remember that equations such as cos(x) = 0.5 have
