Chapter 16 - Sound Waves (Part - 1) - Physics, Solution by D C Pandey, NEET NEET Notes | EduRev

DC Pandey (Questions & Solutions) of Physics: NEET

NEET : Chapter 16 - Sound Waves (Part - 1) - Physics, Solution by D C Pandey, NEET NEET Notes | EduRev

 Page 1


16. Sound Waves
Introductory Exercise 16.1
1. P S kB
0 0
= 
Þ B
P
S k
P
S
= =
0
0
0
0
2
l
p
    =
´
´ ´ ´
-
14
2 3 14 10
6
0.35
5.5 .
    = ´ 1.4 N/m 10
5 2
2. l
n
=
v
 Þ l
max
= =
1450
20
m /s
Hz
72.5 m,
l
min
= =
1450
20000
m/s
Hz
7.25 cm
3. Pressure wave and displacement wave
has a phase difference of 
p
2
, so,
(a)When pressure is maximum,
displacement is minimum i.e., zero.
(b) S
0
 = =
×
p
kB
p
v
0 0
2
2 p
l
r
 
   = =
p
v
p v
v
0
2
0
2
2 2
l
pr
l
pnr
 =
p
v
0
2pnr
   =
´ ´ ´ ´
10
2 10 340
3
3.14 1.29
 m
   = ´
-
3.63 m 10
6
4. S
P
kB
P
v
P
v
0
0 0 0
2
= = =
pnr wr
=
Pk
0
2
rw
=
´
´
12 818
129 2700
2
.
. ( )
= ´
-
1.04 10
5
 m
Introductory Exercise 16.2
1.
v
v
T
T
2
1
2
1
2 = = Þ T T
2 1
4 4 273 = = ´ K
= ´ ° 3 273 C = ° 819 C
2. v v
t
v
t
t
= +
æ
è
ç
ö
ø
÷
= +
æ
è
ç
ö
ø
÷ 0
1 2
0
1
273
1
546
/
v v v
30 3 0
1
30
546
1
3
546
- = + - +
é
ë
ê
ù
û
ú
=
æ
è
ç
ö
ø
÷
v
0
33
546
= ´ = 332
33
546
20.06  m/s
3.  v = = ´ = nl 250 8 2000 m/s
  B v = = ´ r
2 2
900 2000 ( )
= ´ 36 10
8
N/m
 = ´ 3.6 Pa 10
9
4. v
Rt
M
= =
´ ´
´
-
g
7
5
273
32 10
3
8.314
    = 315 m/s
Page 2


16. Sound Waves
Introductory Exercise 16.1
1. P S kB
0 0
= 
Þ B
P
S k
P
S
= =
0
0
0
0
2
l
p
    =
´
´ ´ ´
-
14
2 3 14 10
6
0.35
5.5 .
    = ´ 1.4 N/m 10
5 2
2. l
n
=
v
 Þ l
max
= =
1450
20
m /s
Hz
72.5 m,
l
min
= =
1450
20000
m/s
Hz
7.25 cm
3. Pressure wave and displacement wave
has a phase difference of 
p
2
, so,
(a)When pressure is maximum,
displacement is minimum i.e., zero.
(b) S
0
 = =
×
p
kB
p
v
0 0
2
2 p
l
r
 
   = =
p
v
p v
v
0
2
0
2
2 2
l
pr
l
pnr
 =
p
v
0
2pnr
   =
´ ´ ´ ´
10
2 10 340
3
3.14 1.29
 m
   = ´
-
3.63 m 10
6
4. S
P
kB
P
v
P
v
0
0 0 0
2
= = =
pnr wr
=
Pk
0
2
rw
=
´
´
12 818
129 2700
2
.
. ( )
= ´
-
1.04 10
5
 m
Introductory Exercise 16.2
1.
v
v
T
T
2
1
2
1
2 = = Þ T T
2 1
4 4 273 = = ´ K
= ´ ° 3 273 C = ° 819 C
2. v v
t
v
t
t
= +
æ
è
ç
ö
ø
÷
= +
æ
è
ç
ö
ø
÷ 0
1 2
0
1
273
1
546
/
v v v
30 3 0
1
30
546
1
3
546
- = + - +
é
ë
ê
ù
û
ú
=
æ
è
ç
ö
ø
÷
v
0
33
546
= ´ = 332
33
546
20.06  m/s
3.  v = = ´ = nl 250 8 2000 m/s
  B v = = ´ r
2 2
900 2000 ( )
= ´ 36 10
8
N/m
 = ´ 3.6 Pa 10
9
4. v
Rt
M
= =
´ ´
´
-
g
7
5
273
32 10
3
8.314
    = 315 m/s
Introductory Exercise 16.3
1. P S kB v S
0 0 0
2 = = p nr
= ´ ´ ´ ´ ´ ´
-
2 300 344 10
6
3.14 1.2 6
= 4 67 . Pa
   I
P
v
= =
´ ´
0
2 2
2 2 344 r
( ) 4.67
1.2
 = ´
-
2.64 10
2
  W/m
2
   L
I
I
= =
´
-
-
10 10
10
10
0
2
12
log log
2.64
 
 = 104 dB
2.   2 10 10
0 0
L L
I
I
I
I
- = - log log
h
          = = 10 9 log ( ) h dB
Þ  log h = 0.9,   h = = 10
0 9 .
7.9
 3. I
r
µ
1
2
 Þ I
k
r
=
2
L L
I
I
F M
F
M
- = 10 log
   =
æ
è
ç
ç
ö
ø
÷
÷
10
2
log
r
r
M
F
 =
æ
è
ç
ö
ø
÷
= 20
3
20 log
0.3
dB
4. (a) I
P
v
=
0
2
2 r
; I
max
( )
=
´ ´
28
2 345
2
1.29
 
    = 0.881 W/m
2
L
max
log = =
-
10
12
0.881
10
119.45  dB
  I
min
( )
=
´
´ ´
-
2 10
2 345
5 2
1.29
       = ´
-
4.49 10
13
  W/m
2
  L
min
log =
´
-
-
10
10
10
13
12
4.49
 dB 
       = - 3.48 dB
(b) S
kB v
0
2
= =
Po Po
pnr
( )
max
S
0
28
2 500 345
=
´ ´ ´ ´ 3.14 1.29
= ´
-
2 10
5
 m
( )
min
S
0
5
2 10
2 500 345
=
´
´ ´ ´ ´
-
3.14 1.29
= ´ 1.43 10
11
 m
Introductory Exercise 16.4
1. ( ) 2 1
2
12 n - =
l
 cm
and ( ) 2 1
2
36 n + =
l
 cm
Þ l = - = 36 12 24 cm
n
l
= = =
v 330
1375
m/s
0.24 m
 Hz
2. D D x = f = × =
l
p
l
p
p l
2 2 3 6
= =
´
=
v
6
350
6 500 n
0.117 m = 11.7 cm
D D D f = = = ´ ´
-
2
2 2 500 10
3
p
p n p
T
t t
= p rad = ° 180
3. Dx H
d
d n
1
2
2
2
4
= + - = l
and 
Dx H h
d
d n
2
2
2
2
4
1
2
= + + - = +
æ
è
ç
ö
ø
÷
( ) l
Þ 
l
2
2
4
2
4
2
2
2
2
= + + - + ( ) H h
d
H
d
or l = + + - + 4
4
4
4
2
2
2
2
( ) H h
d
H
d
l = + + - + 2 4 2 4
2 2 2 2
( ) H h d H d
4. Dx d n
p
= = +
æ
è
ç
ö
ø
÷
sin q l
1
2
  for minima
 Sound Waves   | 23
Page 3


16. Sound Waves
Introductory Exercise 16.1
1. P S kB
0 0
= 
Þ B
P
S k
P
S
= =
0
0
0
0
2
l
p
    =
´
´ ´ ´
-
14
2 3 14 10
6
0.35
5.5 .
    = ´ 1.4 N/m 10
5 2
2. l
n
=
v
 Þ l
max
= =
1450
20
m /s
Hz
72.5 m,
l
min
= =
1450
20000
m/s
Hz
7.25 cm
3. Pressure wave and displacement wave
has a phase difference of 
p
2
, so,
(a)When pressure is maximum,
displacement is minimum i.e., zero.
(b) S
0
 = =
×
p
kB
p
v
0 0
2
2 p
l
r
 
   = =
p
v
p v
v
0
2
0
2
2 2
l
pr
l
pnr
 =
p
v
0
2pnr
   =
´ ´ ´ ´
10
2 10 340
3
3.14 1.29
 m
   = ´
-
3.63 m 10
6
4. S
P
kB
P
v
P
v
0
0 0 0
2
= = =
pnr wr
=
Pk
0
2
rw
=
´
´
12 818
129 2700
2
.
. ( )
= ´
-
1.04 10
5
 m
Introductory Exercise 16.2
1.
v
v
T
T
2
1
2
1
2 = = Þ T T
2 1
4 4 273 = = ´ K
= ´ ° 3 273 C = ° 819 C
2. v v
t
v
t
t
= +
æ
è
ç
ö
ø
÷
= +
æ
è
ç
ö
ø
÷ 0
1 2
0
1
273
1
546
/
v v v
30 3 0
1
30
546
1
3
546
- = + - +
é
ë
ê
ù
û
ú
=
æ
è
ç
ö
ø
÷
v
0
33
546
= ´ = 332
33
546
20.06  m/s
3.  v = = ´ = nl 250 8 2000 m/s
  B v = = ´ r
2 2
900 2000 ( )
= ´ 36 10
8
N/m
 = ´ 3.6 Pa 10
9
4. v
Rt
M
= =
´ ´
´
-
g
7
5
273
32 10
3
8.314
    = 315 m/s
Introductory Exercise 16.3
1. P S kB v S
0 0 0
2 = = p nr
= ´ ´ ´ ´ ´ ´
-
2 300 344 10
6
3.14 1.2 6
= 4 67 . Pa
   I
P
v
= =
´ ´
0
2 2
2 2 344 r
( ) 4.67
1.2
 = ´
-
2.64 10
2
  W/m
2
   L
I
I
= =
´
-
-
10 10
10
10
0
2
12
log log
2.64
 
 = 104 dB
2.   2 10 10
0 0
L L
I
I
I
I
- = - log log
h
          = = 10 9 log ( ) h dB
Þ  log h = 0.9,   h = = 10
0 9 .
7.9
 3. I
r
µ
1
2
 Þ I
k
r
=
2
L L
I
I
F M
F
M
- = 10 log
   =
æ
è
ç
ç
ö
ø
÷
÷
10
2
log
r
r
M
F
 =
æ
è
ç
ö
ø
÷
= 20
3
20 log
0.3
dB
4. (a) I
P
v
=
0
2
2 r
; I
max
( )
=
´ ´
28
2 345
2
1.29
 
    = 0.881 W/m
2
L
max
log = =
-
10
12
0.881
10
119.45  dB
  I
min
( )
=
´
´ ´
-
2 10
2 345
5 2
1.29
       = ´
-
4.49 10
13
  W/m
2
  L
min
log =
´
-
-
10
10
10
13
12
4.49
 dB 
       = - 3.48 dB
(b) S
kB v
0
2
= =
Po Po
pnr
( )
max
S
0
28
2 500 345
=
´ ´ ´ ´ 3.14 1.29
= ´
-
2 10
5
 m
( )
min
S
0
5
2 10
2 500 345
=
´
´ ´ ´ ´
-
3.14 1.29
= ´ 1.43 10
11
 m
Introductory Exercise 16.4
1. ( ) 2 1
2
12 n - =
l
 cm
and ( ) 2 1
2
36 n + =
l
 cm
Þ l = - = 36 12 24 cm
n
l
= = =
v 330
1375
m/s
0.24 m
 Hz
2. D D x = f = × =
l
p
l
p
p l
2 2 3 6
= =
´
=
v
6
350
6 500 n
0.117 m = 11.7 cm
D D D f = = = ´ ´
-
2
2 2 500 10
3
p
p n p
T
t t
= p rad = ° 180
3. Dx H
d
d n
1
2
2
2
4
= + - = l
and 
Dx H h
d
d n
2
2
2
2
4
1
2
= + + - = +
æ
è
ç
ö
ø
÷
( ) l
Þ 
l
2
2
4
2
4
2
2
2
2
= + + - + ( ) H h
d
H
d
or l = + + - + 4
4
4
4
2
2
2
2
( ) H h
d
H
d
l = + + - + 2 4 2 4
2 2 2 2
( ) H h d H d
4. Dx d n
p
= = +
æ
è
ç
ö
ø
÷
sin q l
1
2
  for minima
 Sound Waves   | 23
(a) \   d sin q
l
=
2
 for first minima
q
l
n
=
æ
è
ç
ö
ø
÷
=
æ
è
ç
ö
ø
÷
- -
sin sin
1 1
2 2 d
v
d
=
´ ´
æ
è
ç
ç
ö
ø
÷
÷
-
sin
1
340
2 600 2
 
= =
-
sin ( )
1
0.142 0.142 rad
= ° 8.14
(b) For, first maxima d sin q l =
Þ  q
l
=
æ
è
ç
ö
ø
÷
=
æ
è
ç
ö
ø
÷
- -
sin sin
1 1
340
1200 d
  = ° 16 46 .
(c) Dx d
max
£ Þ n d l £ , n
d
£
l
=
´
=
2 600
340
3.53
Þ n = 3  maxima.
5. (a) For coherent speakers in phase, 
I I
R
= 4
2
0
2
cos
q
D D f
p
l
p
l
l
p = = × =
2 2
2
x = q
Þ I I
R
= = 4
2
0
0
cos
p
(b) For incoherent sources, 
I I I I I I
R
= + = + =
1 2 0 0 0
2
(c) For coherent speakers with a phase
difference 180°.
D D f f p p p ¢ = ° + = + = 180 2
Þ   I I I
R
¢ = = 4
2
2
4
0
2
0
cos
p
6. 60 10
10
0
12
dB =
-
log
I
Þ 10 10
6 12
0
´ =
-
I
Þ I
0
6 2
10 =
-
W/m
D D D f
p
l
pn
= =
2 2
x
v
x
=
´
´ - = =
2 170
340
11 8 3
p
p q ( )
(a) \  I I I
R
= = = 4
2
4
3
2
0
0
2
0
2
cos cos
q p
(b) Df p p p ¢ = + = 3 4
Þ I I I
R
¢ = = 4
4
2
4
0 0
cos
p
       = ´
-
4 10
6 2
W/m
L
R
¢ =
´
-
-
10
4 10
10
6
12
log
  = 10 10
6
log dB + 10 4 log
= + 60 2 2 dB dB log
= + = 60 6 66 dB dB dB
(e) Df D ¢ ¢ = × =
´
´ -
2 2 85
340
11 8
pn p
v
x ( )
= =
3
2
p
q
       I I
R
¢¢ = 4
3
4
0
2
cos
p
= -
æ
è
ç
ö
ø
÷
4
4
0
2
I cos p
p
 = 2
0
I
Þ  L
R
¢¢ =
´
=
-
-
10
2 10
10
63
6
12
log dB
7. (a) I
1
3
2
3
10
4 2
10
16
=
´
=
- -
p p
= ´
-
19.9 W /m 10
6 2
= 19 9 . / mW m
2
   I
2
3
2
3
10
4 3
10
36
=
´
=
- -
p p
= ´
-
8.84 W /m 10
6 2
 
= 8 84 . / mW m
2
(b)  ( ) ( )
max
I I I
P
= +
1 2
2
       = + ( ) 4.46 2.97
2
       = 55.27 W/m m
2
(c)  ( ) ( )
min
I I I
P
= -
1 2
2
       = - ( ) 4.46 2.97
2
       = 2.22 W/m m
2
(d) I I I
P
= + =
1 2
2
28 7 . mW /m
24  |  Sound Waves
v
í
ì
ì
d sin q
q
P
X
d
S
1
S
2
Y
Page 4


16. Sound Waves
Introductory Exercise 16.1
1. P S kB
0 0
= 
Þ B
P
S k
P
S
= =
0
0
0
0
2
l
p
    =
´
´ ´ ´
-
14
2 3 14 10
6
0.35
5.5 .
    = ´ 1.4 N/m 10
5 2
2. l
n
=
v
 Þ l
max
= =
1450
20
m /s
Hz
72.5 m,
l
min
= =
1450
20000
m/s
Hz
7.25 cm
3. Pressure wave and displacement wave
has a phase difference of 
p
2
, so,
(a)When pressure is maximum,
displacement is minimum i.e., zero.
(b) S
0
 = =
×
p
kB
p
v
0 0
2
2 p
l
r
 
   = =
p
v
p v
v
0
2
0
2
2 2
l
pr
l
pnr
 =
p
v
0
2pnr
   =
´ ´ ´ ´
10
2 10 340
3
3.14 1.29
 m
   = ´
-
3.63 m 10
6
4. S
P
kB
P
v
P
v
0
0 0 0
2
= = =
pnr wr
=
Pk
0
2
rw
=
´
´
12 818
129 2700
2
.
. ( )
= ´
-
1.04 10
5
 m
Introductory Exercise 16.2
1.
v
v
T
T
2
1
2
1
2 = = Þ T T
2 1
4 4 273 = = ´ K
= ´ ° 3 273 C = ° 819 C
2. v v
t
v
t
t
= +
æ
è
ç
ö
ø
÷
= +
æ
è
ç
ö
ø
÷ 0
1 2
0
1
273
1
546
/
v v v
30 3 0
1
30
546
1
3
546
- = + - +
é
ë
ê
ù
û
ú
=
æ
è
ç
ö
ø
÷
v
0
33
546
= ´ = 332
33
546
20.06  m/s
3.  v = = ´ = nl 250 8 2000 m/s
  B v = = ´ r
2 2
900 2000 ( )
= ´ 36 10
8
N/m
 = ´ 3.6 Pa 10
9
4. v
Rt
M
= =
´ ´
´
-
g
7
5
273
32 10
3
8.314
    = 315 m/s
Introductory Exercise 16.3
1. P S kB v S
0 0 0
2 = = p nr
= ´ ´ ´ ´ ´ ´
-
2 300 344 10
6
3.14 1.2 6
= 4 67 . Pa
   I
P
v
= =
´ ´
0
2 2
2 2 344 r
( ) 4.67
1.2
 = ´
-
2.64 10
2
  W/m
2
   L
I
I
= =
´
-
-
10 10
10
10
0
2
12
log log
2.64
 
 = 104 dB
2.   2 10 10
0 0
L L
I
I
I
I
- = - log log
h
          = = 10 9 log ( ) h dB
Þ  log h = 0.9,   h = = 10
0 9 .
7.9
 3. I
r
µ
1
2
 Þ I
k
r
=
2
L L
I
I
F M
F
M
- = 10 log
   =
æ
è
ç
ç
ö
ø
÷
÷
10
2
log
r
r
M
F
 =
æ
è
ç
ö
ø
÷
= 20
3
20 log
0.3
dB
4. (a) I
P
v
=
0
2
2 r
; I
max
( )
=
´ ´
28
2 345
2
1.29
 
    = 0.881 W/m
2
L
max
log = =
-
10
12
0.881
10
119.45  dB
  I
min
( )
=
´
´ ´
-
2 10
2 345
5 2
1.29
       = ´
-
4.49 10
13
  W/m
2
  L
min
log =
´
-
-
10
10
10
13
12
4.49
 dB 
       = - 3.48 dB
(b) S
kB v
0
2
= =
Po Po
pnr
( )
max
S
0
28
2 500 345
=
´ ´ ´ ´ 3.14 1.29
= ´
-
2 10
5
 m
( )
min
S
0
5
2 10
2 500 345
=
´
´ ´ ´ ´
-
3.14 1.29
= ´ 1.43 10
11
 m
Introductory Exercise 16.4
1. ( ) 2 1
2
12 n - =
l
 cm
and ( ) 2 1
2
36 n + =
l
 cm
Þ l = - = 36 12 24 cm
n
l
= = =
v 330
1375
m/s
0.24 m
 Hz
2. D D x = f = × =
l
p
l
p
p l
2 2 3 6
= =
´
=
v
6
350
6 500 n
0.117 m = 11.7 cm
D D D f = = = ´ ´
-
2
2 2 500 10
3
p
p n p
T
t t
= p rad = ° 180
3. Dx H
d
d n
1
2
2
2
4
= + - = l
and 
Dx H h
d
d n
2
2
2
2
4
1
2
= + + - = +
æ
è
ç
ö
ø
÷
( ) l
Þ 
l
2
2
4
2
4
2
2
2
2
= + + - + ( ) H h
d
H
d
or l = + + - + 4
4
4
4
2
2
2
2
( ) H h
d
H
d
l = + + - + 2 4 2 4
2 2 2 2
( ) H h d H d
4. Dx d n
p
= = +
æ
è
ç
ö
ø
÷
sin q l
1
2
  for minima
 Sound Waves   | 23
(a) \   d sin q
l
=
2
 for first minima
q
l
n
=
æ
è
ç
ö
ø
÷
=
æ
è
ç
ö
ø
÷
- -
sin sin
1 1
2 2 d
v
d
=
´ ´
æ
è
ç
ç
ö
ø
÷
÷
-
sin
1
340
2 600 2
 
= =
-
sin ( )
1
0.142 0.142 rad
= ° 8.14
(b) For, first maxima d sin q l =
Þ  q
l
=
æ
è
ç
ö
ø
÷
=
æ
è
ç
ö
ø
÷
- -
sin sin
1 1
340
1200 d
  = ° 16 46 .
(c) Dx d
max
£ Þ n d l £ , n
d
£
l
=
´
=
2 600
340
3.53
Þ n = 3  maxima.
5. (a) For coherent speakers in phase, 
I I
R
= 4
2
0
2
cos
q
D D f
p
l
p
l
l
p = = × =
2 2
2
x = q
Þ I I
R
= = 4
2
0
0
cos
p
(b) For incoherent sources, 
I I I I I I
R
= + = + =
1 2 0 0 0
2
(c) For coherent speakers with a phase
difference 180°.
D D f f p p p ¢ = ° + = + = 180 2
Þ   I I I
R
¢ = = 4
2
2
4
0
2
0
cos
p
6. 60 10
10
0
12
dB =
-
log
I
Þ 10 10
6 12
0
´ =
-
I
Þ I
0
6 2
10 =
-
W/m
D D D f
p
l
pn
= =
2 2
x
v
x
=
´
´ - = =
2 170
340
11 8 3
p
p q ( )
(a) \  I I I
R
= = = 4
2
4
3
2
0
0
2
0
2
cos cos
q p
(b) Df p p p ¢ = + = 3 4
Þ I I I
R
¢ = = 4
4
2
4
0 0
cos
p
       = ´
-
4 10
6 2
W/m
L
R
¢ =
´
-
-
10
4 10
10
6
12
log
  = 10 10
6
log dB + 10 4 log
= + 60 2 2 dB dB log
= + = 60 6 66 dB dB dB
(e) Df D ¢ ¢ = × =
´
´ -
2 2 85
340
11 8
pn p
v
x ( )
= =
3
2
p
q
       I I
R
¢¢ = 4
3
4
0
2
cos
p
= -
æ
è
ç
ö
ø
÷
4
4
0
2
I cos p
p
 = 2
0
I
Þ  L
R
¢¢ =
´
=
-
-
10
2 10
10
63
6
12
log dB
7. (a) I
1
3
2
3
10
4 2
10
16
=
´
=
- -
p p
= ´
-
19.9 W /m 10
6 2
= 19 9 . / mW m
2
   I
2
3
2
3
10
4 3
10
36
=
´
=
- -
p p
= ´
-
8.84 W /m 10
6 2
 
= 8 84 . / mW m
2
(b)  ( ) ( )
max
I I I
P
= +
1 2
2
       = + ( ) 4.46 2.97
2
       = 55.27 W/m m
2
(c)  ( ) ( )
min
I I I
P
= -
1 2
2
       = - ( ) 4.46 2.97
2
       = 2.22 W/m m
2
(d) I I I
P
= + =
1 2
2
28 7 . mW /m
24  |  Sound Waves
v
í
ì
ì
d sin q
q
P
X
d
S
1
S
2
Y
Introductory  Exercise 16.5
1. (a) n
0
4
=
v
l
c
 Þ  l
v
c
= =
´ 4
345
4 220
0
n
m/s
Hz
= 0.392 m
(b) 
3
2
5
0
0
v
l
= n
Þ    l
v
0
0
3
10
3 345
10 220
= =
´
´ n
 = 0.470 m
2. (a) 
d l
A
= = 0.8 m     d
l
A
=
3
, l =
0.8
m
3
,        d
l l
l
A
=
5
3
5
, ,
=
0.8
m m 0.8 m
5
2 4
4
,
.
,
(b) 
d
A n
= 0              d
l
A
= 0
2
3
,              d
l l
A
= 0
2
5
4
5
, ,
      = 0,0.533 m      = 0 m 0.32 m, 0.64 m ,
3.
Þ HCF of the two shows, 80 and the
values, 400 Hz and 560 Hz are odd
multiples of 80. These conservative
harmonics are odd, which can be seen in
closed organ pipe only.
(b) These are 5th and 7th harmonic.
(c) n
0
4
=
v
l
c
 
Þ l
v
c
= =
´
=
4
344
4 80
0
n
1.075 m
4. v = = ´ ´ ´
-
nl 1000 2 10
2
6.77 m /s
= 135.4 m /s
      v
RT
M
=
g
 Þ  r
Mv
RT
=
2
=
´ ´ ´
´
-
n 127 10 135 4
400
3 2
( . )
8.314
 = 0.7 n
As 1 2 < < r
Þ n = 2 Þ r = ´ = = 0.7 1.4 2
7
5
 diatonic
5. n =
+
=
+ ( ) ( ) 2 1
4
2 3
4
1 2
n v
l
n v
l
2 3
2 1
100
60
2
1
n
n
l
l
+
+
= = =
5
3
 
Þ     n = 1 
\  v
l
n
=
+
=
´ ´
=
4
2 1
4 440
3
352
1
n 0.6
 m/s
Introductory Exercise 16.6
1.
Þ         n
A
= 252 Hz
       n
A
= ± ( ) 256 4 Hz 
and        n
A
n - = ± ( ) 256 6 Hz
\      256 4 256 6 ± - = ± n 
± = 4 6 m n Þ n = - + = 4 6 2
\   n
A
= - = 256 4 252 Hz 
2.
Þ        n
A
= 387 Hz
      n
A
= ± ( ) 384 3 Hz
and       n
A
n - = ± 384 m, m < 3
\  384 3 385 ± - = ± n m
 Sound Waves   | 25
'
A
1
A' B A
1
A
252Hz
ý
>4
ý
<4
256Hz 260Hz
N A N
A A N
N
A N
Fundamental
tone
First overtone Second overtone
A N A
N A
A
A N
N N A N
N
A A
2
0
400, 560
 4  20,   28
'
A
1
A' B A
1
A
387
381 384
ý
>3
ý
<3
Page 5


16. Sound Waves
Introductory Exercise 16.1
1. P S kB
0 0
= 
Þ B
P
S k
P
S
= =
0
0
0
0
2
l
p
    =
´
´ ´ ´
-
14
2 3 14 10
6
0.35
5.5 .
    = ´ 1.4 N/m 10
5 2
2. l
n
=
v
 Þ l
max
= =
1450
20
m /s
Hz
72.5 m,
l
min
= =
1450
20000
m/s
Hz
7.25 cm
3. Pressure wave and displacement wave
has a phase difference of 
p
2
, so,
(a)When pressure is maximum,
displacement is minimum i.e., zero.
(b) S
0
 = =
×
p
kB
p
v
0 0
2
2 p
l
r
 
   = =
p
v
p v
v
0
2
0
2
2 2
l
pr
l
pnr
 =
p
v
0
2pnr
   =
´ ´ ´ ´
10
2 10 340
3
3.14 1.29
 m
   = ´
-
3.63 m 10
6
4. S
P
kB
P
v
P
v
0
0 0 0
2
= = =
pnr wr
=
Pk
0
2
rw
=
´
´
12 818
129 2700
2
.
. ( )
= ´
-
1.04 10
5
 m
Introductory Exercise 16.2
1.
v
v
T
T
2
1
2
1
2 = = Þ T T
2 1
4 4 273 = = ´ K
= ´ ° 3 273 C = ° 819 C
2. v v
t
v
t
t
= +
æ
è
ç
ö
ø
÷
= +
æ
è
ç
ö
ø
÷ 0
1 2
0
1
273
1
546
/
v v v
30 3 0
1
30
546
1
3
546
- = + - +
é
ë
ê
ù
û
ú
=
æ
è
ç
ö
ø
÷
v
0
33
546
= ´ = 332
33
546
20.06  m/s
3.  v = = ´ = nl 250 8 2000 m/s
  B v = = ´ r
2 2
900 2000 ( )
= ´ 36 10
8
N/m
 = ´ 3.6 Pa 10
9
4. v
Rt
M
= =
´ ´
´
-
g
7
5
273
32 10
3
8.314
    = 315 m/s
Introductory Exercise 16.3
1. P S kB v S
0 0 0
2 = = p nr
= ´ ´ ´ ´ ´ ´
-
2 300 344 10
6
3.14 1.2 6
= 4 67 . Pa
   I
P
v
= =
´ ´
0
2 2
2 2 344 r
( ) 4.67
1.2
 = ´
-
2.64 10
2
  W/m
2
   L
I
I
= =
´
-
-
10 10
10
10
0
2
12
log log
2.64
 
 = 104 dB
2.   2 10 10
0 0
L L
I
I
I
I
- = - log log
h
          = = 10 9 log ( ) h dB
Þ  log h = 0.9,   h = = 10
0 9 .
7.9
 3. I
r
µ
1
2
 Þ I
k
r
=
2
L L
I
I
F M
F
M
- = 10 log
   =
æ
è
ç
ç
ö
ø
÷
÷
10
2
log
r
r
M
F
 =
æ
è
ç
ö
ø
÷
= 20
3
20 log
0.3
dB
4. (a) I
P
v
=
0
2
2 r
; I
max
( )
=
´ ´
28
2 345
2
1.29
 
    = 0.881 W/m
2
L
max
log = =
-
10
12
0.881
10
119.45  dB
  I
min
( )
=
´
´ ´
-
2 10
2 345
5 2
1.29
       = ´
-
4.49 10
13
  W/m
2
  L
min
log =
´
-
-
10
10
10
13
12
4.49
 dB 
       = - 3.48 dB
(b) S
kB v
0
2
= =
Po Po
pnr
( )
max
S
0
28
2 500 345
=
´ ´ ´ ´ 3.14 1.29
= ´
-
2 10
5
 m
( )
min
S
0
5
2 10
2 500 345
=
´
´ ´ ´ ´
-
3.14 1.29
= ´ 1.43 10
11
 m
Introductory Exercise 16.4
1. ( ) 2 1
2
12 n - =
l
 cm
and ( ) 2 1
2
36 n + =
l
 cm
Þ l = - = 36 12 24 cm
n
l
= = =
v 330
1375
m/s
0.24 m
 Hz
2. D D x = f = × =
l
p
l
p
p l
2 2 3 6
= =
´
=
v
6
350
6 500 n
0.117 m = 11.7 cm
D D D f = = = ´ ´
-
2
2 2 500 10
3
p
p n p
T
t t
= p rad = ° 180
3. Dx H
d
d n
1
2
2
2
4
= + - = l
and 
Dx H h
d
d n
2
2
2
2
4
1
2
= + + - = +
æ
è
ç
ö
ø
÷
( ) l
Þ 
l
2
2
4
2
4
2
2
2
2
= + + - + ( ) H h
d
H
d
or l = + + - + 4
4
4
4
2
2
2
2
( ) H h
d
H
d
l = + + - + 2 4 2 4
2 2 2 2
( ) H h d H d
4. Dx d n
p
= = +
æ
è
ç
ö
ø
÷
sin q l
1
2
  for minima
 Sound Waves   | 23
(a) \   d sin q
l
=
2
 for first minima
q
l
n
=
æ
è
ç
ö
ø
÷
=
æ
è
ç
ö
ø
÷
- -
sin sin
1 1
2 2 d
v
d
=
´ ´
æ
è
ç
ç
ö
ø
÷
÷
-
sin
1
340
2 600 2
 
= =
-
sin ( )
1
0.142 0.142 rad
= ° 8.14
(b) For, first maxima d sin q l =
Þ  q
l
=
æ
è
ç
ö
ø
÷
=
æ
è
ç
ö
ø
÷
- -
sin sin
1 1
340
1200 d
  = ° 16 46 .
(c) Dx d
max
£ Þ n d l £ , n
d
£
l
=
´
=
2 600
340
3.53
Þ n = 3  maxima.
5. (a) For coherent speakers in phase, 
I I
R
= 4
2
0
2
cos
q
D D f
p
l
p
l
l
p = = × =
2 2
2
x = q
Þ I I
R
= = 4
2
0
0
cos
p
(b) For incoherent sources, 
I I I I I I
R
= + = + =
1 2 0 0 0
2
(c) For coherent speakers with a phase
difference 180°.
D D f f p p p ¢ = ° + = + = 180 2
Þ   I I I
R
¢ = = 4
2
2
4
0
2
0
cos
p
6. 60 10
10
0
12
dB =
-
log
I
Þ 10 10
6 12
0
´ =
-
I
Þ I
0
6 2
10 =
-
W/m
D D D f
p
l
pn
= =
2 2
x
v
x
=
´
´ - = =
2 170
340
11 8 3
p
p q ( )
(a) \  I I I
R
= = = 4
2
4
3
2
0
0
2
0
2
cos cos
q p
(b) Df p p p ¢ = + = 3 4
Þ I I I
R
¢ = = 4
4
2
4
0 0
cos
p
       = ´
-
4 10
6 2
W/m
L
R
¢ =
´
-
-
10
4 10
10
6
12
log
  = 10 10
6
log dB + 10 4 log
= + 60 2 2 dB dB log
= + = 60 6 66 dB dB dB
(e) Df D ¢ ¢ = × =
´
´ -
2 2 85
340
11 8
pn p
v
x ( )
= =
3
2
p
q
       I I
R
¢¢ = 4
3
4
0
2
cos
p
= -
æ
è
ç
ö
ø
÷
4
4
0
2
I cos p
p
 = 2
0
I
Þ  L
R
¢¢ =
´
=
-
-
10
2 10
10
63
6
12
log dB
7. (a) I
1
3
2
3
10
4 2
10
16
=
´
=
- -
p p
= ´
-
19.9 W /m 10
6 2
= 19 9 . / mW m
2
   I
2
3
2
3
10
4 3
10
36
=
´
=
- -
p p
= ´
-
8.84 W /m 10
6 2
 
= 8 84 . / mW m
2
(b)  ( ) ( )
max
I I I
P
= +
1 2
2
       = + ( ) 4.46 2.97
2
       = 55.27 W/m m
2
(c)  ( ) ( )
min
I I I
P
= -
1 2
2
       = - ( ) 4.46 2.97
2
       = 2.22 W/m m
2
(d) I I I
P
= + =
1 2
2
28 7 . mW /m
24  |  Sound Waves
v
í
ì
ì
d sin q
q
P
X
d
S
1
S
2
Y
Introductory  Exercise 16.5
1. (a) n
0
4
=
v
l
c
 Þ  l
v
c
= =
´ 4
345
4 220
0
n
m/s
Hz
= 0.392 m
(b) 
3
2
5
0
0
v
l
= n
Þ    l
v
0
0
3
10
3 345
10 220
= =
´
´ n
 = 0.470 m
2. (a) 
d l
A
= = 0.8 m     d
l
A
=
3
, l =
0.8
m
3
,        d
l l
l
A
=
5
3
5
, ,
=
0.8
m m 0.8 m
5
2 4
4
,
.
,
(b) 
d
A n
= 0              d
l
A
= 0
2
3
,              d
l l
A
= 0
2
5
4
5
, ,
      = 0,0.533 m      = 0 m 0.32 m, 0.64 m ,
3.
Þ HCF of the two shows, 80 and the
values, 400 Hz and 560 Hz are odd
multiples of 80. These conservative
harmonics are odd, which can be seen in
closed organ pipe only.
(b) These are 5th and 7th harmonic.
(c) n
0
4
=
v
l
c
 
Þ l
v
c
= =
´
=
4
344
4 80
0
n
1.075 m
4. v = = ´ ´ ´
-
nl 1000 2 10
2
6.77 m /s
= 135.4 m /s
      v
RT
M
=
g
 Þ  r
Mv
RT
=
2
=
´ ´ ´
´
-
n 127 10 135 4
400
3 2
( . )
8.314
 = 0.7 n
As 1 2 < < r
Þ n = 2 Þ r = ´ = = 0.7 1.4 2
7
5
 diatonic
5. n =
+
=
+ ( ) ( ) 2 1
4
2 3
4
1 2
n v
l
n v
l
2 3
2 1
100
60
2
1
n
n
l
l
+
+
= = =
5
3
 
Þ     n = 1 
\  v
l
n
=
+
=
´ ´
=
4
2 1
4 440
3
352
1
n 0.6
 m/s
Introductory Exercise 16.6
1.
Þ         n
A
= 252 Hz
       n
A
= ± ( ) 256 4 Hz 
and        n
A
n - = ± ( ) 256 6 Hz
\      256 4 256 6 ± - = ± n 
± = 4 6 m n Þ n = - + = 4 6 2
\   n
A
= - = 256 4 252 Hz 
2.
Þ        n
A
= 387 Hz
      n
A
= ± ( ) 384 3 Hz
and       n
A
n - = ± 384 m, m < 3
\  384 3 385 ± - = ± n m
 Sound Waves   | 25
'
A
1
A' B A
1
A
252Hz
ý
>4
ý
<4
256Hz 260Hz
N A N
A A N
N
A N
Fundamental
tone
First overtone Second overtone
A N A
N A
A
A N
N N A N
N
A A
2
0
400, 560
 4  20,   28
'
A
1
A' B A
1
A
387
381 384
ý
>3
ý
<3
Þ           ± - = ± 3 n m
Þ        ± = = + 3 m m n ( ) ve 
Þ          n m = + - 3 
Þ         n
A
= + = 384 3 387 Hz
6 600
1
2
Hz Hz = =
l
T
A
m
and 600
1
2
Hz =
l
T
B
m
Þ       
606
600
= =
T
T
A
B
1.01
Þ       
T
T
A
B
= 1.02
4. 256 4
2
± =
´
v
0.25
and 256
2
=
´ -
v
x ( ) 0.25
256
252
2
2
1
1 4
=
´
-
=
-
0.25
0.25 ( ) x x
256 4 256 252 - ´ = x
Þ 4 4 256 = ´ x
x = = =
1
256
100
256
m cm 0.4 cm
Introductory Exercise 16.7
1. When source is moving, 
n n n
s
s
s
v
v v
v
v
¢ =
+
=
1
1 m
=
æ
è
ç
ö
ø
÷
-
1
1
m
v
v
s
n
= ±
æ
è
ç
ö
ø
÷
= ±
æ
è
ç
ö
ø
÷
1 1
v
v
u
v
s
n n
When observer is moving, n n
0
0
=
± v v
v
= ±
æ
è
ç
ö
ø
÷
1
0
v
v
n = ±
æ
è
ç
ö
ø
÷
1
v
v
n
So, it can be seen that, n
0
 and n
s
 are
equal if  u v << . 
2. l = =
340
200
1 7 . m
(a) l l ¢ = - = - = uT 1 7
80
200
. m 1.7 m
- = 0.4 m m 1 3 .
(b) n
l
¢ =
¢
=
v 340 m /s
1.3 m
   = 262 Hz
3. For doppler effect there has to be
relative motion between source and
receiver, but as they are at rest relative
to each other that’s why there is no shift
in wavelength and frequency.
4. l
n
= = =
v 344
500
0.688 m
(a) l l
front
0.688 = - = - uT
30
500
   = - = 0.688 0.060 0.628 m
(b) l l
behind
0.688 0.060 = + = + uT
    = 0.748 m
(c) n
front
0.628
547.8 Hz = =
344
(d) n
behind
= =
344
0748
459 9
.
. Hz
5. n n ¢ =
- -
- +
=
- -
- +
´
v w v
v w v
s
0
340 5 20
340 5 10
300 Hz
= ´ =
315
345
300 273 9 Hz Hz .
6.
26  |  Sound Waves
r
S v
s P
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