A sinusoidal wave travelling in the positive x direction has amplitude of 15 cm, a wavelength of 40 cm and a frequency of 8.0 hz. The vertical displacement of medium at t = 0 and x = 0 is also 15 cm. Find angular wave number k, period t, angular frequency and speed of the wave. Also determine phase constant and write down general expression for wave.
Answers
Answer:
Given Data:
The amplitude of the wave is,
A
=
15
c
m
=
0.15
m
The wavelength of the wave is,
λ
=
40
c
m
=
0.4
m
The frequency of the wave is,
f
=
8
H
z
At
t
=
0
and
x
=
0
, the displacement of the wave is,
15
c
m
The general equation of a wave propagating along the x-axis is:
y
(
x
,
t
)
=
A
sin
(
k
x
−
ω
t
+
ϕ
)
Here,
A
is the amplitude of the wave.
k
is the wavenumber.
ω
is the angular frequency.
ϕ
is the phase constant.
a)
The wavenumber of a wave is equal to the change in phase per unit length. Since a length equal to the wavelength
λ
corresponds to a phase change by
2
π
,
k
=
2
π
λ
=
2
π
0.4
=
5
π
r
a
d
/
m
The angular frequency is related to the normal frequency by the following equation:
ω
=
2
π
f
=
2
π
×
8
=
16
π
r
a
d
/
s
The phase velocity of a wave is related to the angular frequency and wavenumber of the wave as
v
p
=
ω
k
=
16
π
5
π
=
3.2
m
/
s
b)
The equation of the wave can now be written as
y
(
x
,
t
)
=
0.15
sin
(
5
π
x
−
16
π
t
+
ϕ
)
At
t
=
0
and
x
=
0
, the displacement of the wave is,
y
=
15
c
m
=
0.15
m
Therefore,
y
(
0
,
0
)
=
0.15
sin
(
5
π
×
0
−
16
π
×
0
+
ϕ
)
⇒
0.15
=
0.15
sin
(
ϕ
)
⇒
ϕ
=
sin
−
1
(
1
)
⇒
ϕ
=
π
/
2
Therefore the equation of the wave is
y
(
x
,
t
)
=
0.15
sin
(
5
π
x
−
16
π
t
+
π
/
2
)
=
0.15
cos
(
5
π
x
−
16
π
t
)