Page 1
15. Superposition of Waves
Introductory Exercise 15.1
1.
When dis place ment of all the par ti cles is
mo men tarily zero, then there is no
elas tic po ten tial en ergy stored in the
string and as the speed is max i mum at
mean po si tion, so en tire en ergy is purely
kinetic.
2. (a) v
T
=
m
Þ
v
v
2
1
1
2
=
m
m
= =
m
m
1
1
1
0.25 0.25
= =
1
05
2
.
Þ v v
2 1
2 20 = = cm/s
(b) a
v
v v
a
t i
=
+
2
2
1 2
=
´
+
=
2 20
10 20
4
3
a a
i i
and a
v v
v v
a
r i
=

+
2 1
2 1
=

+
=
20 10
20 10
1
3
a a
i i
3. (a) For fixed end, a phase change of p
takes place in re flected wave and
di rec tion be com es op po si te.
as Y x t
i
=  0 .3 cos ( ) 2 40
Þ Y x t
r
= + + 0.3 cos ( ) 2 40 p
(b) For free end, there is no change in
phase for reflected wave and
direction becomes opposite.
as Y x t
i
=  0.3 cos ( ) 2 40
Þ Y x t
r
= + 0.3 cos ( ) 2 40
4. v
k
1
1
50
2
25 = = =
w
m/s and v
2
50 = m/s
Þ a
v
v v
a a
t i i
=
+
=
´
+
2 2 50
25 50
2
1 2
= ´ ´

4
3
2 10
3
m =
8
3
mm.
a
v v
v v
a a
r i i
=

+
=

+
2 1
2 1
50 25
50 25
= ´ ´

1
3
2 10
3
m =
2
3
mm.
as v v
2 1
> Þ the boundary is rearer and
there is no phase change.
k
v
2
2
=
w
= =
50
50
p
p
\ y x t
r
= ´ +

2
3
10 50
3
cos ( ) p 0.2
and y x t
t
= ´ 

8
3
10 50
3
cos ( ) p 1.0
5. t
1
2 40
1
8 =
´
=
cm
cm/s
s, inverted
Þ
4cm
Þ
4cm
Þ
6cm
Page 2
15. Superposition of Waves
Introductory Exercise 15.1
1.
When dis place ment of all the par ti cles is
mo men tarily zero, then there is no
elas tic po ten tial en ergy stored in the
string and as the speed is max i mum at
mean po si tion, so en tire en ergy is purely
kinetic.
2. (a) v
T
=
m
Þ
v
v
2
1
1
2
=
m
m
= =
m
m
1
1
1
0.25 0.25
= =
1
05
2
.
Þ v v
2 1
2 20 = = cm/s
(b) a
v
v v
a
t i
=
+
2
2
1 2
=
´
+
=
2 20
10 20
4
3
a a
i i
and a
v v
v v
a
r i
=

+
2 1
2 1
=

+
=
20 10
20 10
1
3
a a
i i
3. (a) For fixed end, a phase change of p
takes place in re flected wave and
di rec tion be com es op po si te.
as Y x t
i
=  0 .3 cos ( ) 2 40
Þ Y x t
r
= + + 0.3 cos ( ) 2 40 p
(b) For free end, there is no change in
phase for reflected wave and
direction becomes opposite.
as Y x t
i
=  0.3 cos ( ) 2 40
Þ Y x t
r
= + 0.3 cos ( ) 2 40
4. v
k
1
1
50
2
25 = = =
w
m/s and v
2
50 = m/s
Þ a
v
v v
a a
t i i
=
+
=
´
+
2 2 50
25 50
2
1 2
= ´ ´

4
3
2 10
3
m =
8
3
mm.
a
v v
v v
a a
r i i
=

+
=

+
2 1
2 1
50 25
50 25
= ´ ´

1
3
2 10
3
m =
2
3
mm.
as v v
2 1
> Þ the boundary is rearer and
there is no phase change.
k
v
2
2
=
w
= =
50
50
p
p
\ y x t
r
= ´ +

2
3
10 50
3
cos ( ) p 0.2
and y x t
t
= ´ 

8
3
10 50
3
cos ( ) p 1.0
5. t
1
2 40
1
8 =
´
=
cm
cm/s
s, inverted
Þ
4cm
Þ
4cm
Þ
6cm
t
2
4 10 6
1
=
+ +
= 20 s upright
6.
In tr o duc tor y Ex er cis e 15.2
1. y
x
t a kx t = = 5
3
40 2 sin cos sin cos
p
p w
a = =
5
2
2.5 cm, k =

p
3
1
cm , w p =

40
1
s
v
k
= = =
w p
p/
40
3
120 cm/s
D x
k k
= = × = =
l p p p
p 2
1
2
2
9 /
cm = 3 cm
v
dy
dt
x
t
P
= =  200
3
40 p
p
p sin sin
v
P
1.5, sin
9
8
200
3
3
2
æ
è
ç
ö
ø
÷
=  ×
æ
è
ç
ö
ø
÷
p
p
sin 40
9
8
p ´
æ
è
ç
ö
ø
÷
= 
æ
è
ç
ö
ø
÷
200
2
45 p
p
p sin sin ( )
=  ´ ´ 200 1 0 p
= 0 cm/s
2. Two waves with dif fer ent am pli tudes
can pro duce par tial sta tion ary waves
with am pli t ude of antinode s be ing
a a
1 2
+ and am pli tude of nodes be ing
a a
1 2
~ . As here node is not stationary
that is why en ergy is also trans ported
through nodes.
3. (a)
l
2
2 = m Þ l = 4 m,
v
T
= =
´
= =

m
100
4 10
10
2
50
2
2
m/s
n
l
= = =
v 50
4
125 . Hz
and is fundamental tone or first
harmonic.
y x t = 0.1 sin sin
2
2
p
l
pn
= ´ 0.1 12.5 sin sin
2
4
2
p
p x t
= 0.1 sin sin
p
p
2
25 x t
(b) 3
2
2
l
= m Þ l =
4
3
m and v = 50 m/s
n
l
= =
v 50
4 3 /
Hz = 37.5 Hz and is 2nd
overtone or 3rd harmonic.
y x t = ´ 0 04
2
4 3
2 . sin
/
sin
p
p 37.5
= 0.04 sin `sin
3
2
75
p
p x t
4. v
F Fl
m
= = =
´
´

m
400 4
160 10
3
=
´
=

1600
16 10
10
2
2
= 100 m/s
(a)
l
0
4
= l Þ l
0
4 16 = = l m
Superposition of Waves  11
2 4 6 8 10
1
3
t = 2s
x
y
2 4 6 8 10
1
3
t = 3s
x
y
4
12 14 12 14
2 4 6 8 10
1
3
t = 2s
x
y
2 4 6 8 10
1
3
t = 5s
x
y
4
12 14 12 14
2 4 6 8 10
1
3
t = 2s
x
y
2 4 6 8 10
1
3
t = 3s
x
y
4
12 14 12 14
Page 3
15. Superposition of Waves
Introductory Exercise 15.1
1.
When dis place ment of all the par ti cles is
mo men tarily zero, then there is no
elas tic po ten tial en ergy stored in the
string and as the speed is max i mum at
mean po si tion, so en tire en ergy is purely
kinetic.
2. (a) v
T
=
m
Þ
v
v
2
1
1
2
=
m
m
= =
m
m
1
1
1
0.25 0.25
= =
1
05
2
.
Þ v v
2 1
2 20 = = cm/s
(b) a
v
v v
a
t i
=
+
2
2
1 2
=
´
+
=
2 20
10 20
4
3
a a
i i
and a
v v
v v
a
r i
=

+
2 1
2 1
=

+
=
20 10
20 10
1
3
a a
i i
3. (a) For fixed end, a phase change of p
takes place in re flected wave and
di rec tion be com es op po si te.
as Y x t
i
=  0 .3 cos ( ) 2 40
Þ Y x t
r
= + + 0.3 cos ( ) 2 40 p
(b) For free end, there is no change in
phase for reflected wave and
direction becomes opposite.
as Y x t
i
=  0.3 cos ( ) 2 40
Þ Y x t
r
= + 0.3 cos ( ) 2 40
4. v
k
1
1
50
2
25 = = =
w
m/s and v
2
50 = m/s
Þ a
v
v v
a a
t i i
=
+
=
´
+
2 2 50
25 50
2
1 2
= ´ ´

4
3
2 10
3
m =
8
3
mm.
a
v v
v v
a a
r i i
=

+
=

+
2 1
2 1
50 25
50 25
= ´ ´

1
3
2 10
3
m =
2
3
mm.
as v v
2 1
> Þ the boundary is rearer and
there is no phase change.
k
v
2
2
=
w
= =
50
50
p
p
\ y x t
r
= ´ +

2
3
10 50
3
cos ( ) p 0.2
and y x t
t
= ´ 

8
3
10 50
3
cos ( ) p 1.0
5. t
1
2 40
1
8 =
´
=
cm
cm/s
s, inverted
Þ
4cm
Þ
4cm
Þ
6cm
t
2
4 10 6
1
=
+ +
= 20 s upright
6.
In tr o duc tor y Ex er cis e 15.2
1. y
x
t a kx t = = 5
3
40 2 sin cos sin cos
p
p w
a = =
5
2
2.5 cm, k =

p
3
1
cm , w p =

40
1
s
v
k
= = =
w p
p/
40
3
120 cm/s
D x
k k
= = × = =
l p p p
p 2
1
2
2
9 /
cm = 3 cm
v
dy
dt
x
t
P
= =  200
3
40 p
p
p sin sin
v
P
1.5, sin
9
8
200
3
3
2
æ
è
ç
ö
ø
÷
=  ×
æ
è
ç
ö
ø
÷
p
p
sin 40
9
8
p ´
æ
è
ç
ö
ø
÷
= 
æ
è
ç
ö
ø
÷
200
2
45 p
p
p sin sin ( )
=  ´ ´ 200 1 0 p
= 0 cm/s
2. Two waves with dif fer ent am pli tudes
can pro duce par tial sta tion ary waves
with am pli t ude of antinode s be ing
a a
1 2
+ and am pli tude of nodes be ing
a a
1 2
~ . As here node is not stationary
that is why en ergy is also trans ported
through nodes.
3. (a)
l
2
2 = m Þ l = 4 m,
v
T
= =
´
= =

m
100
4 10
10
2
50
2
2
m/s
n
l
= = =
v 50
4
125 . Hz
and is fundamental tone or first
harmonic.
y x t = 0.1 sin sin
2
2
p
l
pn
= ´ 0.1 12.5 sin sin
2
4
2
p
p x t
= 0.1 sin sin
p
p
2
25 x t
(b) 3
2
2
l
= m Þ l =
4
3
m and v = 50 m/s
n
l
= =
v 50
4 3 /
Hz = 37.5 Hz and is 2nd
overtone or 3rd harmonic.
y x t = ´ 0 04
2
4 3
2 . sin
/
sin
p
p 37.5
= 0.04 sin `sin
3
2
75
p
p x t
4. v
F Fl
m
= = =
´
´

m
400 4
160 10
3
=
´
=

1600
16 10
10
2
2
= 100 m/s
(a)
l
0
4
= l Þ l
0
4 16 = = l m
Superposition of Waves  11
2 4 6 8 10
1
3
t = 2s
x
y
2 4 6 8 10
1
3
t = 3s
x
y
4
12 14 12 14
2 4 6 8 10
1
3
t = 2s
x
y
2 4 6 8 10
1
3
t = 5s
x
y
4
12 14 12 14
2 4 6 8 10
1
3
t = 2s
x
y
2 4 6 8 10
1
3
t = 3s
x
y
4
12 14 12 14
3
4
1
l
= l Þ l
1
4
3
16
3
= =
l
m = 5.33 m
and
5
4
2
l
= l Þ l
2
4
5
16
5
= =
l
m = 3.2 m
(b) n
l
0
0
100
16
= = =
v
6.25 Hz
n
l
1
1
100
16 3
= = =
v
/
18.75 Hz
n
l
3
3
100
16 5
= = =
v
/
31.25 Hz
5. l n n = =
l
2
0.54
2
= 0.27 n
and l n n = +
¢
= + ( ) ( ) 1
2 2
1
l 0.48
= + 0.24 ( ) n 1
Þ 0.27 0.24 0.24 n n = +
Þ 0.03 0.24 n = Þ n = 8
(a) These are 8th and 9th harmonic
(b) l n = = ´ = 0.27 0.27 2.16 m 8
(c)
l
0
2
= l Þ l
0
2 = = l 4.32 m
6. 5 2 54
0 0
n n  = Hz
Þ 3 54
0
n = Hz Þ n
0
18 = Hz
7. n
m
0
1
2
=
l
F
Þ
n
n
2
1
2
1
=
F
F
=
+
= =
M 2.2
2.2
260
220
13
11
Þ
M +
= = +
2.2
2.2
169
121
1
48
121
= + 1
M
2.2
Þ M =
´
= =
48
11
2.2
121
9.6
0.873 kg
8. nn
0
250 = Hz and ( ) n + = 1 300
0
n Hz
Þ n
0
50 = Hz
and n = 5 Þ So these are 5th and 6th
harmonics.
n
m
0
1
2
=
l
F
Þ F l v = = ´ 4 4 50
2
0
2 2
m
´
´
=

36 10
1
360
3
N
AIEEE Corner
¢ Subjective Question (Level 1)
1. A A A A A = + + °
1
2
1
2
1 1
2 90 cos
= = = A
1
2 4 2 cm 5.66 cm
2. v v
2 1
2 =
A
v v
v v
A
v
v
A A
r
=

+
= =
2 1
2 1
1
1
3
1
3
A
v
v v
A
v
v
A A
t
=
+
= =
2 4
3
4
3
2
2 1
1
1
I
I
A
A
r
i
r
=
æ
è
ç
ö
ø
÷
=
2
1
9
and
I
I
t
i
=  = 1
1
9
8
9
A = + + ´ ´ 10 20 2 10 20
3
2 2
cos
p
= + + = = 100 400 200 700 10 7
= 26.46 units
tan
sin
cos
q
p
p
=
+
20
3
10 20
3
= = sin
p
3
3
2
Þ q =
æ
è
ç
ö
ø
÷ =

tan
1
3
2
0.714 rad
\ Phase = + + 5 25 x t 0.714 rad.
4. y
1
1 = cm sin ( ) p p cm s
 

1 1
50 x t
12  Superposition of Waves
Page 4
15. Superposition of Waves
Introductory Exercise 15.1
1.
When dis place ment of all the par ti cles is
mo men tarily zero, then there is no
elas tic po ten tial en ergy stored in the
string and as the speed is max i mum at
mean po si tion, so en tire en ergy is purely
kinetic.
2. (a) v
T
=
m
Þ
v
v
2
1
1
2
=
m
m
= =
m
m
1
1
1
0.25 0.25
= =
1
05
2
.
Þ v v
2 1
2 20 = = cm/s
(b) a
v
v v
a
t i
=
+
2
2
1 2
=
´
+
=
2 20
10 20
4
3
a a
i i
and a
v v
v v
a
r i
=

+
2 1
2 1
=

+
=
20 10
20 10
1
3
a a
i i
3. (a) For fixed end, a phase change of p
takes place in re flected wave and
di rec tion be com es op po si te.
as Y x t
i
=  0 .3 cos ( ) 2 40
Þ Y x t
r
= + + 0.3 cos ( ) 2 40 p
(b) For free end, there is no change in
phase for reflected wave and
direction becomes opposite.
as Y x t
i
=  0.3 cos ( ) 2 40
Þ Y x t
r
= + 0.3 cos ( ) 2 40
4. v
k
1
1
50
2
25 = = =
w
m/s and v
2
50 = m/s
Þ a
v
v v
a a
t i i
=
+
=
´
+
2 2 50
25 50
2
1 2
= ´ ´

4
3
2 10
3
m =
8
3
mm.
a
v v
v v
a a
r i i
=

+
=

+
2 1
2 1
50 25
50 25
= ´ ´

1
3
2 10
3
m =
2
3
mm.
as v v
2 1
> Þ the boundary is rearer and
there is no phase change.
k
v
2
2
=
w
= =
50
50
p
p
\ y x t
r
= ´ +

2
3
10 50
3
cos ( ) p 0.2
and y x t
t
= ´ 

8
3
10 50
3
cos ( ) p 1.0
5. t
1
2 40
1
8 =
´
=
cm
cm/s
s, inverted
Þ
4cm
Þ
4cm
Þ
6cm
t
2
4 10 6
1
=
+ +
= 20 s upright
6.
In tr o duc tor y Ex er cis e 15.2
1. y
x
t a kx t = = 5
3
40 2 sin cos sin cos
p
p w
a = =
5
2
2.5 cm, k =

p
3
1
cm , w p =

40
1
s
v
k
= = =
w p
p/
40
3
120 cm/s
D x
k k
= = × = =
l p p p
p 2
1
2
2
9 /
cm = 3 cm
v
dy
dt
x
t
P
= =  200
3
40 p
p
p sin sin
v
P
1.5, sin
9
8
200
3
3
2
æ
è
ç
ö
ø
÷
=  ×
æ
è
ç
ö
ø
÷
p
p
sin 40
9
8
p ´
æ
è
ç
ö
ø
÷
= 
æ
è
ç
ö
ø
÷
200
2
45 p
p
p sin sin ( )
=  ´ ´ 200 1 0 p
= 0 cm/s
2. Two waves with dif fer ent am pli tudes
can pro duce par tial sta tion ary waves
with am pli t ude of antinode s be ing
a a
1 2
+ and am pli tude of nodes be ing
a a
1 2
~ . As here node is not stationary
that is why en ergy is also trans ported
through nodes.
3. (a)
l
2
2 = m Þ l = 4 m,
v
T
= =
´
= =

m
100
4 10
10
2
50
2
2
m/s
n
l
= = =
v 50
4
125 . Hz
and is fundamental tone or first
harmonic.
y x t = 0.1 sin sin
2
2
p
l
pn
= ´ 0.1 12.5 sin sin
2
4
2
p
p x t
= 0.1 sin sin
p
p
2
25 x t
(b) 3
2
2
l
= m Þ l =
4
3
m and v = 50 m/s
n
l
= =
v 50
4 3 /
Hz = 37.5 Hz and is 2nd
overtone or 3rd harmonic.
y x t = ´ 0 04
2
4 3
2 . sin
/
sin
p
p 37.5
= 0.04 sin `sin
3
2
75
p
p x t
4. v
F Fl
m
= = =
´
´

m
400 4
160 10
3
=
´
=

1600
16 10
10
2
2
= 100 m/s
(a)
l
0
4
= l Þ l
0
4 16 = = l m
Superposition of Waves  11
2 4 6 8 10
1
3
t = 2s
x
y
2 4 6 8 10
1
3
t = 3s
x
y
4
12 14 12 14
2 4 6 8 10
1
3
t = 2s
x
y
2 4 6 8 10
1
3
t = 5s
x
y
4
12 14 12 14
2 4 6 8 10
1
3
t = 2s
x
y
2 4 6 8 10
1
3
t = 3s
x
y
4
12 14 12 14
3
4
1
l
= l Þ l
1
4
3
16
3
= =
l
m = 5.33 m
and
5
4
2
l
= l Þ l
2
4
5
16
5
= =
l
m = 3.2 m
(b) n
l
0
0
100
16
= = =
v
6.25 Hz
n
l
1
1
100
16 3
= = =
v
/
18.75 Hz
n
l
3
3
100
16 5
= = =
v
/
31.25 Hz
5. l n n = =
l
2
0.54
2
= 0.27 n
and l n n = +
¢
= + ( ) ( ) 1
2 2
1
l 0.48
= + 0.24 ( ) n 1
Þ 0.27 0.24 0.24 n n = +
Þ 0.03 0.24 n = Þ n = 8
(a) These are 8th and 9th harmonic
(b) l n = = ´ = 0.27 0.27 2.16 m 8
(c)
l
0
2
= l Þ l
0
2 = = l 4.32 m
6. 5 2 54
0 0
n n  = Hz
Þ 3 54
0
n = Hz Þ n
0
18 = Hz
7. n
m
0
1
2
=
l
F
Þ
n
n
2
1
2
1
=
F
F
=
+
= =
M 2.2
2.2
260
220
13
11
Þ
M +
= = +
2.2
2.2
169
121
1
48
121
= + 1
M
2.2
Þ M =
´
= =
48
11
2.2
121
9.6
0.873 kg
8. nn
0
250 = Hz and ( ) n + = 1 300
0
n Hz
Þ n
0
50 = Hz
and n = 5 Þ So these are 5th and 6th
harmonics.
n
m
0
1
2
=
l
F
Þ F l v = = ´ 4 4 50
2
0
2 2
m
´
´
=

36 10
1
360
3
N
AIEEE Corner
¢ Subjective Question (Level 1)
1. A A A A A = + + °
1
2
1
2
1 1
2 90 cos
= = = A
1
2 4 2 cm 5.66 cm
2. v v
2 1
2 =
A
v v
v v
A
v
v
A A
r
=

+
= =
2 1
2 1
1
1
3
1
3
A
v
v v
A
v
v
A A
t
=
+
= =
2 4
3
4
3
2
2 1
1
1
I
I
A
A
r
i
r
=
æ
è
ç
ö
ø
÷
=
2
1
9
and
I
I
t
i
=  = 1
1
9
8
9
A = + + ´ ´ 10 20 2 10 20
3
2 2
cos
p
= + + = = 100 400 200 700 10 7
= 26.46 units
tan
sin
cos
q
p
p
=
+
20
3
10 20
3
= = sin
p
3
3
2
Þ q =
æ
è
ç
ö
ø
÷ =

tan
1
3
2
0.714 rad
\ Phase = + + 5 25 x t 0.714 rad.
4. y
1
1 = cm sin ( ) p p cm s
 

1 1
50 x t
12  Superposition of Waves
y x t
2
1 1
1 5
2
100 = 
æ
è
ç
ö
ø
÷
 
. sin cm cm s
p
p
Þ
y
1
3
5 10 1
250
1000
( , ) sin 4.5 cm 4.5 ´ = 
æ
è
ç
ö
ø
÷

p
p
= 
æ
è
ç
ö
ø
÷
1
9
2 4
sin p
p
=
æ
è
ç
ö
ø
÷
1
17
4
sin
p
= +
æ
è
ç
ö
ø
÷
1 4
4
cm sin p
p
= = 1
4
1
2
sin
p
cm and
y
2
3
5 10
9
4
500
1000
( , ) sin 4.5 1.5 cm ´ = 
æ
è
ç
ö
ø
÷

p p
= 
æ
è
ç
ö
ø
÷
1.5 cm sin
9
4 2
p p
= 1.5 sin
5
4
p
æ
è
ç
ö
ø
÷
= +
æ
è
ç
ö
ø
÷
1.5 sin p
p
4
= 
æ
è
ç
ö
ø
÷
1.5 sin
p
4
= 
1.5
cm
2
\ y y y = + = 
1 2
1
2
1 5
2
.
=  = 
0.5
cm
2
1
2 2
5. v
T
=
m
=
´ ´

16
10 10
3 2
N
0.4 kg/N
=
´
=
16 10
4
20
2
m/s
(a) For same shape, time,
t
l
v
= =
´
=
2 2
20
0.2
s 0.02 s
(b)
6. (a)
(b)
7. y x t = 1.5 0.4 sin ( ) cos ( ) 200
= 2A kx t sin cos w
l
p p
p = = =
2 2
0 4
5
k .
m = 15.7 m
n
w
p p p
= = = =
2
200
2
100
Hz 31.8 Hz
v
k
= = =
w 200
500
0.4
m /s
Superposition of Waves  13
t = 0
1 cm
1 cm/s
1 cm
Þ
t = 2s
t = 3s
1 cm 1 cm/s
1cm 1cm
1cm
1 cm
1 cm/s
1cm 1cm
Þ
t = 1s
2 cm
1cm
1 cm
1cm
t = 4s
1 cm 1 cm
1 cm
t = 0
1 cm
1 cm/3
2 cm
Þ
t = 2s t = 3s
1 cm 1 cm/s
1cm 1cm
1cm
1 cm
1 cm/s
1cm 1cm
Þ
t = 4s
1 cm
1 cm 1 cm
t = 1 s
1 cm 1 cm
Þ
t = 0 s
Þ
t = 0.01 s
t = 0.02 s
Þ
Page 5
15. Superposition of Waves
Introductory Exercise 15.1
1.
When dis place ment of all the par ti cles is
mo men tarily zero, then there is no
elas tic po ten tial en ergy stored in the
string and as the speed is max i mum at
mean po si tion, so en tire en ergy is purely
kinetic.
2. (a) v
T
=
m
Þ
v
v
2
1
1
2
=
m
m
= =
m
m
1
1
1
0.25 0.25
= =
1
05
2
.
Þ v v
2 1
2 20 = = cm/s
(b) a
v
v v
a
t i
=
+
2
2
1 2
=
´
+
=
2 20
10 20
4
3
a a
i i
and a
v v
v v
a
r i
=

+
2 1
2 1
=

+
=
20 10
20 10
1
3
a a
i i
3. (a) For fixed end, a phase change of p
takes place in re flected wave and
di rec tion be com es op po si te.
as Y x t
i
=  0 .3 cos ( ) 2 40
Þ Y x t
r
= + + 0.3 cos ( ) 2 40 p
(b) For free end, there is no change in
phase for reflected wave and
direction becomes opposite.
as Y x t
i
=  0.3 cos ( ) 2 40
Þ Y x t
r
= + 0.3 cos ( ) 2 40
4. v
k
1
1
50
2
25 = = =
w
m/s and v
2
50 = m/s
Þ a
v
v v
a a
t i i
=
+
=
´
+
2 2 50
25 50
2
1 2
= ´ ´

4
3
2 10
3
m =
8
3
mm.
a
v v
v v
a a
r i i
=

+
=

+
2 1
2 1
50 25
50 25
= ´ ´

1
3
2 10
3
m =
2
3
mm.
as v v
2 1
> Þ the boundary is rearer and
there is no phase change.
k
v
2
2
=
w
= =
50
50
p
p
\ y x t
r
= ´ +

2
3
10 50
3
cos ( ) p 0.2
and y x t
t
= ´ 

8
3
10 50
3
cos ( ) p 1.0
5. t
1
2 40
1
8 =
´
=
cm
cm/s
s, inverted
Þ
4cm
Þ
4cm
Þ
6cm
t
2
4 10 6
1
=
+ +
= 20 s upright
6.
In tr o duc tor y Ex er cis e 15.2
1. y
x
t a kx t = = 5
3
40 2 sin cos sin cos
p
p w
a = =
5
2
2.5 cm, k =

p
3
1
cm , w p =

40
1
s
v
k
= = =
w p
p/
40
3
120 cm/s
D x
k k
= = × = =
l p p p
p 2
1
2
2
9 /
cm = 3 cm
v
dy
dt
x
t
P
= =  200
3
40 p
p
p sin sin
v
P
1.5, sin
9
8
200
3
3
2
æ
è
ç
ö
ø
÷
=  ×
æ
è
ç
ö
ø
÷
p
p
sin 40
9
8
p ´
æ
è
ç
ö
ø
÷
= 
æ
è
ç
ö
ø
÷
200
2
45 p
p
p sin sin ( )
=  ´ ´ 200 1 0 p
= 0 cm/s
2. Two waves with dif fer ent am pli tudes
can pro duce par tial sta tion ary waves
with am pli t ude of antinode s be ing
a a
1 2
+ and am pli tude of nodes be ing
a a
1 2
~ . As here node is not stationary
that is why en ergy is also trans ported
through nodes.
3. (a)
l
2
2 = m Þ l = 4 m,
v
T
= =
´
= =

m
100
4 10
10
2
50
2
2
m/s
n
l
= = =
v 50
4
125 . Hz
and is fundamental tone or first
harmonic.
y x t = 0.1 sin sin
2
2
p
l
pn
= ´ 0.1 12.5 sin sin
2
4
2
p
p x t
= 0.1 sin sin
p
p
2
25 x t
(b) 3
2
2
l
= m Þ l =
4
3
m and v = 50 m/s
n
l
= =
v 50
4 3 /
Hz = 37.5 Hz and is 2nd
overtone or 3rd harmonic.
y x t = ´ 0 04
2
4 3
2 . sin
/
sin
p
p 37.5
= 0.04 sin `sin
3
2
75
p
p x t
4. v
F Fl
m
= = =
´
´

m
400 4
160 10
3
=
´
=

1600
16 10
10
2
2
= 100 m/s
(a)
l
0
4
= l Þ l
0
4 16 = = l m
Superposition of Waves  11
2 4 6 8 10
1
3
t = 2s
x
y
2 4 6 8 10
1
3
t = 3s
x
y
4
12 14 12 14
2 4 6 8 10
1
3
t = 2s
x
y
2 4 6 8 10
1
3
t = 5s
x
y
4
12 14 12 14
2 4 6 8 10
1
3
t = 2s
x
y
2 4 6 8 10
1
3
t = 3s
x
y
4
12 14 12 14
3
4
1
l
= l Þ l
1
4
3
16
3
= =
l
m = 5.33 m
and
5
4
2
l
= l Þ l
2
4
5
16
5
= =
l
m = 3.2 m
(b) n
l
0
0
100
16
= = =
v
6.25 Hz
n
l
1
1
100
16 3
= = =
v
/
18.75 Hz
n
l
3
3
100
16 5
= = =
v
/
31.25 Hz
5. l n n = =
l
2
0.54
2
= 0.27 n
and l n n = +
¢
= + ( ) ( ) 1
2 2
1
l 0.48
= + 0.24 ( ) n 1
Þ 0.27 0.24 0.24 n n = +
Þ 0.03 0.24 n = Þ n = 8
(a) These are 8th and 9th harmonic
(b) l n = = ´ = 0.27 0.27 2.16 m 8
(c)
l
0
2
= l Þ l
0
2 = = l 4.32 m
6. 5 2 54
0 0
n n  = Hz
Þ 3 54
0
n = Hz Þ n
0
18 = Hz
7. n
m
0
1
2
=
l
F
Þ
n
n
2
1
2
1
=
F
F
=
+
= =
M 2.2
2.2
260
220
13
11
Þ
M +
= = +
2.2
2.2
169
121
1
48
121
= + 1
M
2.2
Þ M =
´
= =
48
11
2.2
121
9.6
0.873 kg
8. nn
0
250 = Hz and ( ) n + = 1 300
0
n Hz
Þ n
0
50 = Hz
and n = 5 Þ So these are 5th and 6th
harmonics.
n
m
0
1
2
=
l
F
Þ F l v = = ´ 4 4 50
2
0
2 2
m
´
´
=

36 10
1
360
3
N
AIEEE Corner
¢ Subjective Question (Level 1)
1. A A A A A = + + °
1
2
1
2
1 1
2 90 cos
= = = A
1
2 4 2 cm 5.66 cm
2. v v
2 1
2 =
A
v v
v v
A
v
v
A A
r
=

+
= =
2 1
2 1
1
1
3
1
3
A
v
v v
A
v
v
A A
t
=
+
= =
2 4
3
4
3
2
2 1
1
1
I
I
A
A
r
i
r
=
æ
è
ç
ö
ø
÷
=
2
1
9
and
I
I
t
i
=  = 1
1
9
8
9
A = + + ´ ´ 10 20 2 10 20
3
2 2
cos
p
= + + = = 100 400 200 700 10 7
= 26.46 units
tan
sin
cos
q
p
p
=
+
20
3
10 20
3
= = sin
p
3
3
2
Þ q =
æ
è
ç
ö
ø
÷ =

tan
1
3
2
0.714 rad
\ Phase = + + 5 25 x t 0.714 rad.
4. y
1
1 = cm sin ( ) p p cm s
 

1 1
50 x t
12  Superposition of Waves
y x t
2
1 1
1 5
2
100 = 
æ
è
ç
ö
ø
÷
 
. sin cm cm s
p
p
Þ
y
1
3
5 10 1
250
1000
( , ) sin 4.5 cm 4.5 ´ = 
æ
è
ç
ö
ø
÷

p
p
= 
æ
è
ç
ö
ø
÷
1
9
2 4
sin p
p
=
æ
è
ç
ö
ø
÷
1
17
4
sin
p
= +
æ
è
ç
ö
ø
÷
1 4
4
cm sin p
p
= = 1
4
1
2
sin
p
cm and
y
2
3
5 10
9
4
500
1000
( , ) sin 4.5 1.5 cm ´ = 
æ
è
ç
ö
ø
÷

p p
= 
æ
è
ç
ö
ø
÷
1.5 cm sin
9
4 2
p p
= 1.5 sin
5
4
p
æ
è
ç
ö
ø
÷
= +
æ
è
ç
ö
ø
÷
1.5 sin p
p
4
= 
æ
è
ç
ö
ø
÷
1.5 sin
p
4
= 
1.5
cm
2
\ y y y = + = 
1 2
1
2
1 5
2
.
=  = 
0.5
cm
2
1
2 2
5. v
T
=
m
=
´ ´

16
10 10
3 2
N
0.4 kg/N
=
´
=
16 10
4
20
2
m/s
(a) For same shape, time,
t
l
v
= =
´
=
2 2
20
0.2
s 0.02 s
(b)
6. (a)
(b)
7. y x t = 1.5 0.4 sin ( ) cos ( ) 200
= 2A kx t sin cos w
l
p p
p = = =
2 2
0 4
5
k .
m = 15.7 m
n
w
p p p
= = = =
2
200
2
100
Hz 31.8 Hz
v
k
= = =
w 200
500
0.4
m /s
Superposition of Waves  13
t = 0
1 cm
1 cm/s
1 cm
Þ
t = 2s
t = 3s
1 cm 1 cm/s
1cm 1cm
1cm
1 cm
1 cm/s
1cm 1cm
Þ
t = 1s
2 cm
1cm
1 cm
1cm
t = 4s
1 cm 1 cm
1 cm
t = 0
1 cm
1 cm/3
2 cm
Þ
t = 2s t = 3s
1 cm 1 cm/s
1cm 1cm
1cm
1 cm
1 cm/s
1cm 1cm
Þ
t = 4s
1 cm
1 cm 1 cm
t = 1 s
1 cm 1 cm
Þ
t = 0 s
Þ
t = 0.01 s
t = 0.02 s
Þ
8. y y y x t = + = +
1 2
3 cm 0.6 sin ( ) p p
+  3 cm 0.6 sin ( ) p p x t
= = 6 cm 0.6 0.6 sin cos cos p p p x t R t
where, R x = 6 cm sin p .
(a) R ( ) sin 0. cm 25 6
1
4
= ´ p
= =
6
2
3 2 cm = 4.24 cm
(b) R ( ) sin 0.50 cm = ´ 6
1
2
p = 6 cm
(c) R ( ) sin 1.50 cm cm = =  6
3
2
6
p
Þ   R = 6 cm
(d) For antinodes, R = ± 6 cm
Þ sin px = ± 1 Þ p
p
x n = + ( ) 2 1
2
or x n = + =
1
2
0.5 cm, 1.5 cm 2.5 cm ,
9. l
p p
p
= = =
2
4
2
2
4
/
cm
(a)Distance between successive
antinodes = =
l
2
2cm
(b) Rx A kx () sin =2
= ´ ´ 2
2
p
p
cm 0.5 sin
= 2
4
p
p
sin
= =
2
2
2
p
p cm
10. n
m
n
n
l
T n
=
+
=
+
´ ´

1
2
1
2 20
20
9 10
3
=
+
´ = = +
n
n
1
60
100 2
3
5 2
9
5 2
9
1 ( )
= + 0.786 ( ) n 1
= 0.786 Hz
1.57 Hz 2.36 Hz 3.14 Hz , ,
11. (a) T v = = m mn l
2 2 2
=
´
´ ´

1.2
0.7
1.4
10
220
3
2 2
( ) ( )
= 162.6 N
(b) n n
2 0
3 3 220 660 = = ´ = Hz Hz
12. n
m
n
n
l
T n
=
+
=
+
´
1
2
1
2
50
0.6 0.01
= + = +
50 2
1 58 93 1
1.2
Hz ( ) . ( ) n n
n
n
£ 20 000 , Hz Þ n = 338
\ n
338
339 58 93 199758 = ´ = . . Hz
= 19.976 kHz
13. nn
0
420 = Hz and ( ) n + = 1 490
0
n Hz
Þ n
0
70 = Hz and n = 6
\ n
m
0
1
2
=
l
T
Þ l
T
= =
´
1
2
450
0005
2 70
0
n m
.
= =
300
140
2.143 m
14. l
n
= = =
v 400
800
1
2
m/s
Hz
m,
l = = = 4
2
2 1
l
l m
(a) 4 400
0
n = Hz Þ n
0
100 = Hz
(b) 7 700
0
n = Hz
16. n
m
0
1
2
=
l
T
Þ n
0
1
µ
l
n n n
1 2 3
1 2 3 : : : : =
=
1 1 1
1 2 3
l l l
: :
Þ l l l
1 2 3
1
1
1
2
1
3
: : : : =
= = 6 3 2 6 3 2 : : : : x x x
6 3 2 1 x x x + + = m
Þ x =
1
11
m
\ position of first bridge = = 6
6
11
x m
and position of second bridge
= + = = 6 3 9
9
11
x x x m
From the same end or 1
9
11
2
11
 = m
from other end .
14  Superposition of Waves
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