A wave of frequency 500 Hz is travelling with a speed of 350 m/s.
i. What is the phase difference between two displacements at a certain point at times 1.0 ms apart?
ii. What will be the smallest distance between two points which are 45° out of phase at an
instant of time?
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Answers
Answer:
Given velocity of the wave
v
=
350
m
/
s
frequency of the wave
n
=
500
H
z
So wave length of the wave
λ
=
v
n
=
350
500
m
=
0.7
m
1) 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:
here it is my freind
Explanation:
Given velocity of the wave
v
=
350
m
/
s
frequency of the wave
n
=
500
H
z
So wave length of the wave
λ
=
v
n
=
350
500
m
=
0.7
m
1) 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
,
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