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Byju's Answer
Standard XII
Physics
Expression for Standing Waves
If u is ins...
Question
If
→
u
is instantaneous velocity of particle and
→
v
is velocity of wave , then
A
→
u
is perpendicular
→
v
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B
→
u
is parallel to
→
v
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C
∣
∣
→
u
∣
∣
is equal to
∣
∣
→
v
∣
∣
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D
∣
∣
→
u
∣
∣
is equal to (slope of wave form )
∣
∣
→
v
∣
∣
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Solution
The correct option is
D
∣
∣
→
u
∣
∣
is equal to (slope of wave form )
∣
∣
→
v
∣
∣
We know that equation of a wave with velocity v.
y
=
a
s
i
n
ω
(
t
−
x
v
)
...(i)
differentiating (i) with respect to time 't' at some instant
x
=
x
0
d
y
d
t
=
A
w
c
o
s
w
(
t
−
x
0
v
)
differentiate (i) with respect to x
d
y
d
x
=
−
A
w
v
c
o
s
w
(
t
−
x
v
)
(
d
y
d
x
)
x
0
=
−
A
w
v
c
o
s
w
(
t
−
x
0
v
)
...(iii)
from (ii)
(
d
y
d
x
)
x
0
=
−
1
v
(
d
y
d
x
)
⇒
d
y
d
x
=
−
(
d
y
d
x
)
x
0
v
⇒
|
→
u
|
=
(slope of wave from)
|
→
v
|
Suggest Corrections
1
Similar questions
Q.
If
u
is perpendicular to
A
B
and
v
be the velocity of the second particle, find
u
v
to the nearest integer.
Q.
Let
→
u
,
→
v
, and
→
w
be such that
|
→
u
|
=
1
,
|
→
v
|
=
2
,
and
|
→
w
|
=
3.
If the projection of
→
v
along
→
u
is equal to that
→
w
along
→
u
, and
→
v
and
→
w
are perpendicular to each other, then |
→
u
-
→
v
+
→
w
| equals to
Q.
Let
u
,
v
and
w
be such that
|
u
|
=
1
,
|
v
|
=
3
and
|
w
|
=
2
. If the projection of
v
along
u
is equal to that of
w
along
u
and vectors
v
and
w
are perpendicular to each other, then
|
u
−
v
+
w
|
equals
Q.
Let u,v,w be such that |u|=1, |v|=2, |w|=3. If the projection v along u is equal to that of w along u and v, w are perpendicular to each other then |u-v+w| =
Q.
Let
→
u
,
→
v
,
→
w
be such that
|
→
u
|
=
1
,
|
→
v
|
=
2
,
|
→
w
|
=
3
. If the projection of
→
v
along
→
u
is equal to that of
→
w
along
→
u
and
→
v
,
→
w
are perpendicular to each other then
|
→
u
−
→
v
+
→
w
|
equals
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