Velocity of a wave of frequency 500 Hz is 300 m/s. Calculate the distance traversed by the disturbance when phase change is
a)90°
b)60°
Answers
Answer:
We are to find the distance between the two points which has
60
∘
out of phase i.e the phase difference is
ϕ
=
60
∘
=
π
3
r
a
d
As we know that for path difference
λ
there is phase difference
2
π
,
we can say , if
ϕ
is the phase difference for path difference
x
,then
ϕ
=
2
π
x
λ
x
=
ϕ
λ
2
π
=
π
3
×
0.7
2
π
=
0.7
6
m
≈
0.116
m
2) Now in
t
=
10
−
3
s
the wave moves
v
×
t
=
350
×
10
−
3
=
0.35
m
So here path difference
x
=
0.35
m
So by the relation
ϕ
=
2
π
x
λ
, the phase difference for
x
=
0.35
m
becomes
ϕ
=
2
π
×
0.35
0.7
=
π
,
Answer:
We are to find the distance between the two points which has
60
∘
out of phase i.e the phase difference is
ϕ
=
60
∘
=
π
3
r
a
d
As we know that for path difference
λ
there is phase difference
2
π
,
we can say , if
ϕ
is the phase difference for path difference
x
,then
ϕ
=
2
π
x
λ
x
=
ϕ
λ
2
π
=
π
3
×
0.7
2
π
=
0.7
6
m
≈
0.116
m
2) Now in
t
=
10
−
3
s
the wave moves
v
×
t
=
350
×
10
−
3
=
0.35
m
So here path difference
x
=
0.35
m
So by the relation
ϕ
=
2
π
x
λ
, the phase difference for
x
=
0.35
m
becomes
ϕ
=
2
π
×
0.35
0.7
=
π
,