DC Pandey Solutions: Sound Waves- 2 | DC Pandey Solutions for JEE Physics PDF Download

 Page 1


¢ Objective Questions (Level 1)
1. Sound cannot travel in vacuum, as it is
mechanical wave.
2. Longitudinal waves can travel through
all mechanical mediums.
3.
g g RT R
32
288
28
=
´
T = ´ = ´ = °
32
28
288
8
7
288 56 K K C
4. Third overtone is 7th harmonic ie, there
4 nodes and 4  antinodes.
5. n µ
v
l
 Þ n µ T so with increase in
temperature, frequency increases.
6. For sound water is rarer medium and air 
is densor medium so, it bends towards
normal while going from water to air.
7. n n
c
c
o
o
v
l
v
l
= = =
4 2
 Þ 
l
l
c
o
= =
2
4
1 2 :
8.
n
n
2
1
2
1
=
F
F
 
Þ  F F
2
2
1
2
1
=
æ
è
ç
ç
ö
ø
÷
÷
n
n
Þ  M M
2
2
1
2
1
2
256
320
10 =
æ
è
ç
ç
ö
ø
÷
÷
=
æ
è
ç
ö
ø
÷
´
n
n
kg
        = 6.4 kg
\    OM M M = - = - = -
2 1
10 6.4 3.6 kg
i.e., Mass has to be decreased by 3.6 kg
9. n n
direct 
=
-
v
v v
s
 and n n
reflected
=
-
v
u v
s
as n n
D R
=  so there will be no beats i e . .,
beat frequency will be zero.
10. Dn n n
l l
l l
l l
= - = - =
-
2 1
2 1
1 2
1 2
v v
v
( )
Þ     v =
l l n
l
1 2
D
D
\      v =
´ ´
=
1
10
3
1.01
0.01
337 m/s
11. n n n n ¢ =
+
= =
v
v v
n
1
2
 Þ n = =
1
2
0.5
12. I I I I NI
max
( ) = + = =
2
4 Þ N = 4
13. n =
+
=
+
v
l e
v
l e 4
3
4
1 2
( ) ( )
Þ l e l e
2 1
3 3 + = +
Þ e
l l
=
-
2 1
3
2
\ e =
- ´
=
42 3 17
2
cm 0.5 cm
v l e = + = ´ + ´
-
4 4 500 17 10
1
2
n ( ) ( ) 8.5
= ´ = 20 350 17.5 m/ s
14. At the moment when velocity of source is 
perpendicular to the line joining source
and observer then there is no Doppler
effect i.e.,  n n n + =
1
 Þ n
1
0 =
15. n =
+
= +
´
= +
( )
( ) ( )
n v
l
n n
1
4
2 1
340
4 1
85 2 1
= 85 255 425 595 765 935 , , , , ,
\  6 frequencies below 1 kHz.
16. Dn n n n =
-
-
-
-
+
= -
+
+
æ
è
ç
ç
ö
ø
÷
÷
v v
v v
v v
v v
v v
v v
s s s
0 0 0
1
= ×
+
+
= ´ = n
v v
v v
s
s
0
10
360
180 5 Hz
17. Dn n n
l l
= = - = - n
v v
1 2
1 2
=
- v ( ) l l
l l
2 1
1 2
 Þ v
n
=
-
l l
l l
1 2
1 2
18.
 Sound Waves   | 31
S
A
o
A'
A B
345Hz
ý
ý
Dn>5 Dn<5
250Hz
355Hz
N
N N A A A N
A
Page 2


¢ Objective Questions (Level 1)
1. Sound cannot travel in vacuum, as it is
mechanical wave.
2. Longitudinal waves can travel through
all mechanical mediums.
3.
g g RT R
32
288
28
=
´
T = ´ = ´ = °
32
28
288
8
7
288 56 K K C
4. Third overtone is 7th harmonic ie, there
4 nodes and 4  antinodes.
5. n µ
v
l
 Þ n µ T so with increase in
temperature, frequency increases.
6. For sound water is rarer medium and air 
is densor medium so, it bends towards
normal while going from water to air.
7. n n
c
c
o
o
v
l
v
l
= = =
4 2
 Þ 
l
l
c
o
= =
2
4
1 2 :
8.
n
n
2
1
2
1
=
F
F
 
Þ  F F
2
2
1
2
1
=
æ
è
ç
ç
ö
ø
÷
÷
n
n
Þ  M M
2
2
1
2
1
2
256
320
10 =
æ
è
ç
ç
ö
ø
÷
÷
=
æ
è
ç
ö
ø
÷
´
n
n
kg
        = 6.4 kg
\    OM M M = - = - = -
2 1
10 6.4 3.6 kg
i.e., Mass has to be decreased by 3.6 kg
9. n n
direct 
=
-
v
v v
s
 and n n
reflected
=
-
v
u v
s
as n n
D R
=  so there will be no beats i e . .,
beat frequency will be zero.
10. Dn n n
l l
l l
l l
= - = - =
-
2 1
2 1
1 2
1 2
v v
v
( )
Þ     v =
l l n
l
1 2
D
D
\      v =
´ ´
=
1
10
3
1.01
0.01
337 m/s
11. n n n n ¢ =
+
= =
v
v v
n
1
2
 Þ n = =
1
2
0.5
12. I I I I NI
max
( ) = + = =
2
4 Þ N = 4
13. n =
+
=
+
v
l e
v
l e 4
3
4
1 2
( ) ( )
Þ l e l e
2 1
3 3 + = +
Þ e
l l
=
-
2 1
3
2
\ e =
- ´
=
42 3 17
2
cm 0.5 cm
v l e = + = ´ + ´
-
4 4 500 17 10
1
2
n ( ) ( ) 8.5
= ´ = 20 350 17.5 m/ s
14. At the moment when velocity of source is 
perpendicular to the line joining source
and observer then there is no Doppler
effect i.e.,  n n n + =
1
 Þ n
1
0 =
15. n =
+
= +
´
= +
( )
( ) ( )
n v
l
n n
1
4
2 1
340
4 1
85 2 1
= 85 255 425 595 765 935 , , , , ,
\  6 frequencies below 1 kHz.
16. Dn n n n =
-
-
-
-
+
= -
+
+
æ
è
ç
ç
ö
ø
÷
÷
v v
v v
v v
v v
v v
v v
s s s
0 0 0
1
= ×
+
+
= ´ = n
v v
v v
s
s
0
10
360
180 5 Hz
17. Dn n n
l l
= = - = - n
v v
1 2
1 2
=
- v ( ) l l
l l
2 1
1 2
 Þ v
n
=
-
l l
l l
1 2
1 2
18.
 Sound Waves   | 31
S
A
o
A'
A B
345Hz
ý
ý
Dn>5 Dn<5
250Hz
355Hz
N
N N A A A N
A
As beat frequency between A and B
decreases on loading A.
i e . ., n n
B A
< Þ n
B
= 345 Hz
After loading A,  n
A
¢ = + = 345 2 247 Hz
and n n
A c
¢ - = ± 6  Þ n l
c A
= ¢ m 6
         = 347 6 m
         = 341 or  353 Hz.
As possible frequency of C are 341 Hz
and 249 Hz then only 341 Hz is justified.
19. e
l l
=
-
=
- ´
=
2 1
3
2
122 3 40
2
1 cm cm
So, 
v
l l e
v
l e ( ) ( )
1 1
5
4 +
=
+
Þ l l l
3 1
5 4 = +
= ´ + ´ = 5 40 4 1 204 cm
20.
D D n
n
=
1
2
F
F
  Þ D
D
n n =
1
2
F
F
= ´ ´ =
1
2
200
1
100
1 Hz
21. n =
+ 2 1
4
n
l
v Þ l
n
v =
+ 2 1
4n 
 
= +
´
=
+
( ) 2 1
340
4 340
2 1
4
n
n
m
=
1
4
3
4
5
4
m m m , , .
As, l
max
= 120 cm Þ l = 25 cm 75 cm.
\ Height of water column 
 = - = 120 75 45 cm cm cm
22. 7
4
105
l
= cm Þ l =
´
=
105 4
7
60 cm
Þ 
l
4
60
4
15 = = cm
So, nodes are at, 
l l l
4
3
4
5
4
, , and 7
4
l
 from
closed end ie ..,  they are at, 
15 45 75 cm cm cm , , and 105 cm.
23. n n
c o
v
l
v
l
v
l
= = = =
4
512
2
2
4
Hz, 
= = ´ = 2 2 512 1024 n
c
Hz Hz
24. M
min
=
´ + ´
+
=
1 32 1 2
1 1
17
v
v
M
M
min
min H
2
2
2
17
= =
H
25. n
a
f >  and  n
r
f <   but n
a
= constant and 
n
r
= cosntant.
So, curve in (b) represents correctly.
26. u
n v
l
m v
l
=
+
-
+ ( ) ( ) 2 1
4
1
2
How, 
( ) ( ) 2 1
4 2
1
4 2
n v
l
m v
l
+
´
-
+
´
   = =
4
2
2  beat/s
27. Dn n n n n = - =
-
-
+
a r
v
v v
v
v v
1 1
=
- +
2
1
1 1
vv
v v v v
n
( ) ( )
=
´ ´ ´
´
=
2 320 4 243
316 324
6 Hz
28. n
c
c
n v
l
n =
+
=
´
+
( )
( )
2 1
4
320
4 1
2 1
 = + ´ = ( ) , , 2 1 80 80 240 n Hz Hz Hz
400 Hz, ...
n
0
0
1
2
320
2
1 =
+
=
´
+
( )
( )
n v
l
n
1.6
= + ´ = ( ) , , n 1 100 200 Hz 100 Hz Hz
300 400 Hz Hz , , ...
\ n n
c o
= = 400 Hz
29. I I
m ax
= 4
0
and I I I
max max
¢ = = 4 16
0
      L¢ = + 10 10 16 dB log
= + = 10 40 2 22 dB dB dB log
30. l
p p
p/
= = =
2 2
2
4
k
m
l = = ´ = 5
4
5
4
4
5
l
m m
32  |  Sound Waves
N
N N
N
C'
B C
341Hz 245 Hz
349Hz
Page 3


¢ Objective Questions (Level 1)
1. Sound cannot travel in vacuum, as it is
mechanical wave.
2. Longitudinal waves can travel through
all mechanical mediums.
3.
g g RT R
32
288
28
=
´
T = ´ = ´ = °
32
28
288
8
7
288 56 K K C
4. Third overtone is 7th harmonic ie, there
4 nodes and 4  antinodes.
5. n µ
v
l
 Þ n µ T so with increase in
temperature, frequency increases.
6. For sound water is rarer medium and air 
is densor medium so, it bends towards
normal while going from water to air.
7. n n
c
c
o
o
v
l
v
l
= = =
4 2
 Þ 
l
l
c
o
= =
2
4
1 2 :
8.
n
n
2
1
2
1
=
F
F
 
Þ  F F
2
2
1
2
1
=
æ
è
ç
ç
ö
ø
÷
÷
n
n
Þ  M M
2
2
1
2
1
2
256
320
10 =
æ
è
ç
ç
ö
ø
÷
÷
=
æ
è
ç
ö
ø
÷
´
n
n
kg
        = 6.4 kg
\    OM M M = - = - = -
2 1
10 6.4 3.6 kg
i.e., Mass has to be decreased by 3.6 kg
9. n n
direct 
=
-
v
v v
s
 and n n
reflected
=
-
v
u v
s
as n n
D R
=  so there will be no beats i e . .,
beat frequency will be zero.
10. Dn n n
l l
l l
l l
= - = - =
-
2 1
2 1
1 2
1 2
v v
v
( )
Þ     v =
l l n
l
1 2
D
D
\      v =
´ ´
=
1
10
3
1.01
0.01
337 m/s
11. n n n n ¢ =
+
= =
v
v v
n
1
2
 Þ n = =
1
2
0.5
12. I I I I NI
max
( ) = + = =
2
4 Þ N = 4
13. n =
+
=
+
v
l e
v
l e 4
3
4
1 2
( ) ( )
Þ l e l e
2 1
3 3 + = +
Þ e
l l
=
-
2 1
3
2
\ e =
- ´
=
42 3 17
2
cm 0.5 cm
v l e = + = ´ + ´
-
4 4 500 17 10
1
2
n ( ) ( ) 8.5
= ´ = 20 350 17.5 m/ s
14. At the moment when velocity of source is 
perpendicular to the line joining source
and observer then there is no Doppler
effect i.e.,  n n n + =
1
 Þ n
1
0 =
15. n =
+
= +
´
= +
( )
( ) ( )
n v
l
n n
1
4
2 1
340
4 1
85 2 1
= 85 255 425 595 765 935 , , , , ,
\  6 frequencies below 1 kHz.
16. Dn n n n =
-
-
-
-
+
= -
+
+
æ
è
ç
ç
ö
ø
÷
÷
v v
v v
v v
v v
v v
v v
s s s
0 0 0
1
= ×
+
+
= ´ = n
v v
v v
s
s
0
10
360
180 5 Hz
17. Dn n n
l l
= = - = - n
v v
1 2
1 2
=
- v ( ) l l
l l
2 1
1 2
 Þ v
n
=
-
l l
l l
1 2
1 2
18.
 Sound Waves   | 31
S
A
o
A'
A B
345Hz
ý
ý
Dn>5 Dn<5
250Hz
355Hz
N
N N A A A N
A
As beat frequency between A and B
decreases on loading A.
i e . ., n n
B A
< Þ n
B
= 345 Hz
After loading A,  n
A
¢ = + = 345 2 247 Hz
and n n
A c
¢ - = ± 6  Þ n l
c A
= ¢ m 6
         = 347 6 m
         = 341 or  353 Hz.
As possible frequency of C are 341 Hz
and 249 Hz then only 341 Hz is justified.
19. e
l l
=
-
=
- ´
=
2 1
3
2
122 3 40
2
1 cm cm
So, 
v
l l e
v
l e ( ) ( )
1 1
5
4 +
=
+
Þ l l l
3 1
5 4 = +
= ´ + ´ = 5 40 4 1 204 cm
20.
D D n
n
=
1
2
F
F
  Þ D
D
n n =
1
2
F
F
= ´ ´ =
1
2
200
1
100
1 Hz
21. n =
+ 2 1
4
n
l
v Þ l
n
v =
+ 2 1
4n 
 
= +
´
=
+
( ) 2 1
340
4 340
2 1
4
n
n
m
=
1
4
3
4
5
4
m m m , , .
As, l
max
= 120 cm Þ l = 25 cm 75 cm.
\ Height of water column 
 = - = 120 75 45 cm cm cm
22. 7
4
105
l
= cm Þ l =
´
=
105 4
7
60 cm
Þ 
l
4
60
4
15 = = cm
So, nodes are at, 
l l l
4
3
4
5
4
, , and 7
4
l
 from
closed end ie ..,  they are at, 
15 45 75 cm cm cm , , and 105 cm.
23. n n
c o
v
l
v
l
v
l
= = = =
4
512
2
2
4
Hz, 
= = ´ = 2 2 512 1024 n
c
Hz Hz
24. M
min
=
´ + ´
+
=
1 32 1 2
1 1
17
v
v
M
M
min
min H
2
2
2
17
= =
H
25. n
a
f >  and  n
r
f <   but n
a
= constant and 
n
r
= cosntant.
So, curve in (b) represents correctly.
26. u
n v
l
m v
l
=
+
-
+ ( ) ( ) 2 1
4
1
2
How, 
( ) ( ) 2 1
4 2
1
4 2
n v
l
m v
l
+
´
-
+
´
   = =
4
2
2  beat/s
27. Dn n n n n = - =
-
-
+
a r
v
v v
v
v v
1 1
=
- +
2
1
1 1
vv
v v v v
n
( ) ( )
=
´ ´ ´
´
=
2 320 4 243
316 324
6 Hz
28. n
c
c
n v
l
n =
+
=
´
+
( )
( )
2 1
4
320
4 1
2 1
 = + ´ = ( ) , , 2 1 80 80 240 n Hz Hz Hz
400 Hz, ...
n
0
0
1
2
320
2
1 =
+
=
´
+
( )
( )
n v
l
n
1.6
= + ´ = ( ) , , n 1 100 200 Hz 100 Hz Hz
300 400 Hz Hz , , ...
\ n n
c o
= = 400 Hz
29. I I
m ax
= 4
0
and I I I
max max
¢ = = 4 16
0
      L¢ = + 10 10 16 dB log
= + = 10 40 2 22 dB dB dB log
30. l
p p
p/
= = =
2 2
2
4
k
m
l = = ´ = 5
4
5
4
4
5
l
m m
32  |  Sound Waves
N
N N
N
C'
B C
341Hz 245 Hz
349Hz
31. d n
n
= + =
+
× ( )
( )
2 1
4
2 1
4
330
660
l
m 
= + = + ´
330
24
2 1 2 1 ( ) ( ) n n cm 13.75 cm.
= 13.75 cm 41.25 cm 68.75 cm, 96.25 cm , ,
etc. 
32. Dn
l l
l l
l l
= - =
- v v v
1 2
2 1
1 2
( )
 
=
´ ´
´
=
-
332 1 10
2
0.49 0.5
13.15 Hz
33. Dn n n
B A B
= - =
-
´ -
300
300 30
300 300
    = 33.33 Hz and n n
A B
¢ ¹ ¢
So both (a) and (b) options are wrong.
34. f
v v
v
f
v
v
f
a
=
+
= +
æ
è
ç
ö
ø
÷
0 0
1 and 
f
v
v
f
r
= -
æ
è
ç
ö
ø
÷
1
0
f
f
v v
v v
a
r
=
+
-
0
0
Þ ( ) ( ) f f v f f v
a r a r
- = +
0
Þ  
v
v
f f
f f
a r
a r 0
=
+
-
.
and       
f f
v
v
f
f f
f f
f
a r
a r
a r
- = =
-
+
æ
è
ç
ç
ö
ø
÷
÷
2
2
0
Þ    f
f f
a r
=
+
2
.
JEE Corner
¢ Assertion and Reason 
1. n
c
n
v
l
= + ( ) 2 1
4
 =
v
l
v
l
v
l 4
3
4
5
4
, , ,...
while, n
0
1
2 2
2
2
3
2
4
2
=
+
=
( )
, , , ,...
n v
l
v
l
v
l
v
l
v
l
it can be seen that n n
c o
¹ at all situation 
and n n
c o
=
1
2
 so assertion is true but
reason is false.
2. Apparent frequency is constant for
constant relative velocity so assertion is
false.
3. At a point of minimum displacement
pressure amplitude is maximum ie . .,
pressure difference is maximum not
pressure. So assertion is false.
4. The deriver receiver two sounds one
direct, n n
0
= and other n n
R
v u
v u
=
+
-
 such
that be detects beats. So reason is true
explanation of assertion.
5. With increase in intensity sound level
increases in lograthmic order so
assertion is false.
6. Speed of sound v
p
=
g
r
, with increase in
only pressure density increases such
that p/r remains constant. Again 
v
RT
M
=
g
 so both assertion and reason
are true but reason is not correct
explanation of assertion.
7. n n
A B
= + 4 when A is loaded with little
wax then n
A
 sightly decreases and then
beat frequency decreases, but if it is
heavily loaded with wax then its
frequency goes much below n
B
 such that
beat frequency increases. So, assertion
and reason are both true but reason is
not correct explanation of assertion.
 Sound Waves   | 33
Page 4


¢ Objective Questions (Level 1)
1. Sound cannot travel in vacuum, as it is
mechanical wave.
2. Longitudinal waves can travel through
all mechanical mediums.
3.
g g RT R
32
288
28
=
´
T = ´ = ´ = °
32
28
288
8
7
288 56 K K C
4. Third overtone is 7th harmonic ie, there
4 nodes and 4  antinodes.
5. n µ
v
l
 Þ n µ T so with increase in
temperature, frequency increases.
6. For sound water is rarer medium and air 
is densor medium so, it bends towards
normal while going from water to air.
7. n n
c
c
o
o
v
l
v
l
= = =
4 2
 Þ 
l
l
c
o
= =
2
4
1 2 :
8.
n
n
2
1
2
1
=
F
F
 
Þ  F F
2
2
1
2
1
=
æ
è
ç
ç
ö
ø
÷
÷
n
n
Þ  M M
2
2
1
2
1
2
256
320
10 =
æ
è
ç
ç
ö
ø
÷
÷
=
æ
è
ç
ö
ø
÷
´
n
n
kg
        = 6.4 kg
\    OM M M = - = - = -
2 1
10 6.4 3.6 kg
i.e., Mass has to be decreased by 3.6 kg
9. n n
direct 
=
-
v
v v
s
 and n n
reflected
=
-
v
u v
s
as n n
D R
=  so there will be no beats i e . .,
beat frequency will be zero.
10. Dn n n
l l
l l
l l
= - = - =
-
2 1
2 1
1 2
1 2
v v
v
( )
Þ     v =
l l n
l
1 2
D
D
\      v =
´ ´
=
1
10
3
1.01
0.01
337 m/s
11. n n n n ¢ =
+
= =
v
v v
n
1
2
 Þ n = =
1
2
0.5
12. I I I I NI
max
( ) = + = =
2
4 Þ N = 4
13. n =
+
=
+
v
l e
v
l e 4
3
4
1 2
( ) ( )
Þ l e l e
2 1
3 3 + = +
Þ e
l l
=
-
2 1
3
2
\ e =
- ´
=
42 3 17
2
cm 0.5 cm
v l e = + = ´ + ´
-
4 4 500 17 10
1
2
n ( ) ( ) 8.5
= ´ = 20 350 17.5 m/ s
14. At the moment when velocity of source is 
perpendicular to the line joining source
and observer then there is no Doppler
effect i.e.,  n n n + =
1
 Þ n
1
0 =
15. n =
+
= +
´
= +
( )
( ) ( )
n v
l
n n
1
4
2 1
340
4 1
85 2 1
= 85 255 425 595 765 935 , , , , ,
\  6 frequencies below 1 kHz.
16. Dn n n n =
-
-
-
-
+
= -
+
+
æ
è
ç
ç
ö
ø
÷
÷
v v
v v
v v
v v
v v
v v
s s s
0 0 0
1
= ×
+
+
= ´ = n
v v
v v
s
s
0
10
360
180 5 Hz
17. Dn n n
l l
= = - = - n
v v
1 2
1 2
=
- v ( ) l l
l l
2 1
1 2
 Þ v
n
=
-
l l
l l
1 2
1 2
18.
 Sound Waves   | 31
S
A
o
A'
A B
345Hz
ý
ý
Dn>5 Dn<5
250Hz
355Hz
N
N N A A A N
A
As beat frequency between A and B
decreases on loading A.
i e . ., n n
B A
< Þ n
B
= 345 Hz
After loading A,  n
A
¢ = + = 345 2 247 Hz
and n n
A c
¢ - = ± 6  Þ n l
c A
= ¢ m 6
         = 347 6 m
         = 341 or  353 Hz.
As possible frequency of C are 341 Hz
and 249 Hz then only 341 Hz is justified.
19. e
l l
=
-
=
- ´
=
2 1
3
2
122 3 40
2
1 cm cm
So, 
v
l l e
v
l e ( ) ( )
1 1
5
4 +
=
+
Þ l l l
3 1
5 4 = +
= ´ + ´ = 5 40 4 1 204 cm
20.
D D n
n
=
1
2
F
F
  Þ D
D
n n =
1
2
F
F
= ´ ´ =
1
2
200
1
100
1 Hz
21. n =
+ 2 1
4
n
l
v Þ l
n
v =
+ 2 1
4n 
 
= +
´
=
+
( ) 2 1
340
4 340
2 1
4
n
n
m
=
1
4
3
4
5
4
m m m , , .
As, l
max
= 120 cm Þ l = 25 cm 75 cm.
\ Height of water column 
 = - = 120 75 45 cm cm cm
22. 7
4
105
l
= cm Þ l =
´
=
105 4
7
60 cm
Þ 
l
4
60
4
15 = = cm
So, nodes are at, 
l l l
4
3
4
5
4
, , and 7
4
l
 from
closed end ie ..,  they are at, 
15 45 75 cm cm cm , , and 105 cm.
23. n n
c o
v
l
v
l
v
l
= = = =
4
512
2
2
4
Hz, 
= = ´ = 2 2 512 1024 n
c
Hz Hz
24. M
min
=
´ + ´
+
=
1 32 1 2
1 1
17
v
v
M
M
min
min H
2
2
2
17
= =
H
25. n
a
f >  and  n
r
f <   but n
a
= constant and 
n
r
= cosntant.
So, curve in (b) represents correctly.
26. u
n v
l
m v
l
=
+
-
+ ( ) ( ) 2 1
4
1
2
How, 
( ) ( ) 2 1
4 2
1
4 2
n v
l
m v
l
+
´
-
+
´
   = =
4
2
2  beat/s
27. Dn n n n n = - =
-
-
+
a r
v
v v
v
v v
1 1
=
- +
2
1
1 1
vv
v v v v
n
( ) ( )
=
´ ´ ´
´
=
2 320 4 243
316 324
6 Hz
28. n
c
c
n v
l
n =
+
=
´
+
( )
( )
2 1
4
320
4 1
2 1
 = + ´ = ( ) , , 2 1 80 80 240 n Hz Hz Hz
400 Hz, ...
n
0
0
1
2
320
2
1 =
+
=
´
+
( )
( )
n v
l
n
1.6
= + ´ = ( ) , , n 1 100 200 Hz 100 Hz Hz
300 400 Hz Hz , , ...
\ n n
c o
= = 400 Hz
29. I I
m ax
= 4
0
and I I I
max max
¢ = = 4 16
0
      L¢ = + 10 10 16 dB log
= + = 10 40 2 22 dB dB dB log
30. l
p p
p/
= = =
2 2
2
4
k
m
l = = ´ = 5
4
5
4
4
5
l
m m
32  |  Sound Waves
N
N N
N
C'
B C
341Hz 245 Hz
349Hz
31. d n
n
= + =
+
× ( )
( )
2 1
4
2 1
4
330
660
l
m 
= + = + ´
330
24
2 1 2 1 ( ) ( ) n n cm 13.75 cm.
= 13.75 cm 41.25 cm 68.75 cm, 96.25 cm , ,
etc. 
32. Dn
l l
l l
l l
= - =
- v v v
1 2
2 1
1 2
( )
 
=
´ ´
´
=
-
332 1 10
2
0.49 0.5
13.15 Hz
33. Dn n n
B A B
= - =
-
´ -
300
300 30
300 300
    = 33.33 Hz and n n
A B
¢ ¹ ¢
So both (a) and (b) options are wrong.
34. f
v v
v
f
v
v
f
a
=
+
= +
æ
è
ç
ö
ø
÷
0 0
1 and 
f
v
v
f
r
= -
æ
è
ç
ö
ø
÷
1
0
f
f
v v
v v
a
r
=
+
-
0
0
Þ ( ) ( ) f f v f f v
a r a r
- = +
0
Þ  
v
v
f f
f f
a r
a r 0
=
+
-
.
and       
f f
v
v
f
f f
f f
f
a r
a r
a r
- = =
-
+
æ
è
ç
ç
ö
ø
÷
÷
2
2
0
Þ    f
f f
a r
=
+
2
.
JEE Corner
¢ Assertion and Reason 
1. n
c
n
v
l
= + ( ) 2 1
4
 =
v
l
v
l
v
l 4
3
4
5
4
, , ,...
while, n
0
1
2 2
2
2
3
2
4
2
=
+
=
( )
, , , ,...
n v
l
v
l
v
l
v
l
v
l
it can be seen that n n
c o
¹ at all situation 
and n n
c o
=
1
2
 so assertion is true but
reason is false.
2. Apparent frequency is constant for
constant relative velocity so assertion is
false.
3. At a point of minimum displacement
pressure amplitude is maximum ie . .,
pressure difference is maximum not
pressure. So assertion is false.
4. The deriver receiver two sounds one
direct, n n
0
= and other n n
R
v u
v u
=
+
-
 such
that be detects beats. So reason is true
explanation of assertion.
5. With increase in intensity sound level
increases in lograthmic order so
assertion is false.
6. Speed of sound v
p
=
g
r
, with increase in
only pressure density increases such
that p/r remains constant. Again 
v
RT
M
=
g
 so both assertion and reason
are true but reason is not correct
explanation of assertion.
7. n n
A B
= + 4 when A is loaded with little
wax then n
A
 sightly decreases and then
beat frequency decreases, but if it is
heavily loaded with wax then its
frequency goes much below n
B
 such that
beat frequency increases. So, assertion
and reason are both true but reason is
not correct explanation of assertion.
 Sound Waves   | 33
8. 150  450, 7 50, 
          3,     5
The frequencies are odd harmonics then
the pipe is closed and fundamental
frequency is also 150 Hz. So assertion
and reason are both true but reason is
not correct explanation of assertion.
9. n µ
+
1
l e
 with increase in diameter end
correction, e, increases and n decreases.
So reason is correct explanation of
assertion.
10. With increasing length of air column,
number of overtone increases and not
the wavelength so assertion is false.
¢ Objective Questions (Level 2)
1. At the boundary between two mediums,
one part of incident wave gets reflected
and other part gets transmitted or
refracted.
2.
3
2
l
p =3.9 Þ l
p
=
3.9
1.5
Þ k= = =
2 2 3 p
l
p
p 3.9
1.5
3.9
 r
0 0
=S kB
Þ   S
kB
0
0
2 5
2
10 10
3
200
= =
´
´ ´
-
r
3.9
1.3 ( )
=
´
´
=
-
3.9
1.3
.025 m
10
12
0
1
  = 2.5 cm
3.
A
A
v
v v
t
i
=
+
=
´
+
=
2 2 100
200 100
2
3
2
1 2
 
4. Dn n n =
+
-
+
v
v v
v
v v
2 1
=
-
+ +
-
- v v v
v v v v
v v
v
n n ( )
( ) ( )
~
( )
1 2
2 1
1 2
\   v v
v
1 2
340 10
1700
- = =
´
=
Dn
n 
2 m/s
5. v gt
s
= = 10 m/s 
Dn n n =
+
-
-
-
+
v v
v v
v v
v v
s s
0 0
=
+
-
-
-
+
æ
è
ç
ç
ö
ø
÷
÷
´
300 2
300 10
300 2
300 10
150 Hz
= -
æ
è
ç
ö
ø
÷
´ =
302
290
298
310
150 12 Hz
6. 7
2
l
= L  Þ l =
2
7
L
A a kx a
L
L
a a = = × = = - cos cos cos
2
2
7
7
p
p
7. For maxima, nl = 3
Þ l n
l
= = = =
3
3
110
n
v nv
n ; .
\ n = 110 220 330 , , , .. Hz etc. maxima will
be formed so maximum will not be
formed at 120 Hz and 100 Hz.
8.
n
w
¢ =
+ °
+ ° -
v
v w v
s
cos
cos
60
60
=
+
+ -
´
300 10
300 10 20
500 Hz
= ´ =
310
290
500 534 Hz
9. Dn n n = - =
+
-
-
+
+
é
ë
ê
ù
û
ú
´
R
v
v
v
v
0
20
10
20
10
500
= -
æ
è
ç
ö
ø
÷
´
360
300
360
350
500 Hz = 31 Hz
10. Dn
p
p
p
p
= - = - =
404
2
400
2
202 200 2 Hz
I
I
max
min
: =
+
-
æ
è
ç
ç
ö
ø
÷
÷
=
2 1
2 1
9 1
2
34  |  Sound Waves
60° W
20 m/s
20 m/s
S
Page 5


¢ Objective Questions (Level 1)
1. Sound cannot travel in vacuum, as it is
mechanical wave.
2. Longitudinal waves can travel through
all mechanical mediums.
3.
g g RT R
32
288
28
=
´
T = ´ = ´ = °
32
28
288
8
7
288 56 K K C
4. Third overtone is 7th harmonic ie, there
4 nodes and 4  antinodes.
5. n µ
v
l
 Þ n µ T so with increase in
temperature, frequency increases.
6. For sound water is rarer medium and air 
is densor medium so, it bends towards
normal while going from water to air.
7. n n
c
c
o
o
v
l
v
l
= = =
4 2
 Þ 
l
l
c
o
= =
2
4
1 2 :
8.
n
n
2
1
2
1
=
F
F
 
Þ  F F
2
2
1
2
1
=
æ
è
ç
ç
ö
ø
÷
÷
n
n
Þ  M M
2
2
1
2
1
2
256
320
10 =
æ
è
ç
ç
ö
ø
÷
÷
=
æ
è
ç
ö
ø
÷
´
n
n
kg
        = 6.4 kg
\    OM M M = - = - = -
2 1
10 6.4 3.6 kg
i.e., Mass has to be decreased by 3.6 kg
9. n n
direct 
=
-
v
v v
s
 and n n
reflected
=
-
v
u v
s
as n n
D R
=  so there will be no beats i e . .,
beat frequency will be zero.
10. Dn n n
l l
l l
l l
= - = - =
-
2 1
2 1
1 2
1 2
v v
v
( )
Þ     v =
l l n
l
1 2
D
D
\      v =
´ ´
=
1
10
3
1.01
0.01
337 m/s
11. n n n n ¢ =
+
= =
v
v v
n
1
2
 Þ n = =
1
2
0.5
12. I I I I NI
max
( ) = + = =
2
4 Þ N = 4
13. n =
+
=
+
v
l e
v
l e 4
3
4
1 2
( ) ( )
Þ l e l e
2 1
3 3 + = +
Þ e
l l
=
-
2 1
3
2
\ e =
- ´
=
42 3 17
2
cm 0.5 cm
v l e = + = ´ + ´
-
4 4 500 17 10
1
2
n ( ) ( ) 8.5
= ´ = 20 350 17.5 m/ s
14. At the moment when velocity of source is 
perpendicular to the line joining source
and observer then there is no Doppler
effect i.e.,  n n n + =
1
 Þ n
1
0 =
15. n =
+
= +
´
= +
( )
( ) ( )
n v
l
n n
1
4
2 1
340
4 1
85 2 1
= 85 255 425 595 765 935 , , , , ,
\  6 frequencies below 1 kHz.
16. Dn n n n =
-
-
-
-
+
= -
+
+
æ
è
ç
ç
ö
ø
÷
÷
v v
v v
v v
v v
v v
v v
s s s
0 0 0
1
= ×
+
+
= ´ = n
v v
v v
s
s
0
10
360
180 5 Hz
17. Dn n n
l l
= = - = - n
v v
1 2
1 2
=
- v ( ) l l
l l
2 1
1 2
 Þ v
n
=
-
l l
l l
1 2
1 2
18.
 Sound Waves   | 31
S
A
o
A'
A B
345Hz
ý
ý
Dn>5 Dn<5
250Hz
355Hz
N
N N A A A N
A
As beat frequency between A and B
decreases on loading A.
i e . ., n n
B A
< Þ n
B
= 345 Hz
After loading A,  n
A
¢ = + = 345 2 247 Hz
and n n
A c
¢ - = ± 6  Þ n l
c A
= ¢ m 6
         = 347 6 m
         = 341 or  353 Hz.
As possible frequency of C are 341 Hz
and 249 Hz then only 341 Hz is justified.
19. e
l l
=
-
=
- ´
=
2 1
3
2
122 3 40
2
1 cm cm
So, 
v
l l e
v
l e ( ) ( )
1 1
5
4 +
=
+
Þ l l l
3 1
5 4 = +
= ´ + ´ = 5 40 4 1 204 cm
20.
D D n
n
=
1
2
F
F
  Þ D
D
n n =
1
2
F
F
= ´ ´ =
1
2
200
1
100
1 Hz
21. n =
+ 2 1
4
n
l
v Þ l
n
v =
+ 2 1
4n 
 
= +
´
=
+
( ) 2 1
340
4 340
2 1
4
n
n
m
=
1
4
3
4
5
4
m m m , , .
As, l
max
= 120 cm Þ l = 25 cm 75 cm.
\ Height of water column 
 = - = 120 75 45 cm cm cm
22. 7
4
105
l
= cm Þ l =
´
=
105 4
7
60 cm
Þ 
l
4
60
4
15 = = cm
So, nodes are at, 
l l l
4
3
4
5
4
, , and 7
4
l
 from
closed end ie ..,  they are at, 
15 45 75 cm cm cm , , and 105 cm.
23. n n
c o
v
l
v
l
v
l
= = = =
4
512
2
2
4
Hz, 
= = ´ = 2 2 512 1024 n
c
Hz Hz
24. M
min
=
´ + ´
+
=
1 32 1 2
1 1
17
v
v
M
M
min
min H
2
2
2
17
= =
H
25. n
a
f >  and  n
r
f <   but n
a
= constant and 
n
r
= cosntant.
So, curve in (b) represents correctly.
26. u
n v
l
m v
l
=
+
-
+ ( ) ( ) 2 1
4
1
2
How, 
( ) ( ) 2 1
4 2
1
4 2
n v
l
m v
l
+
´
-
+
´
   = =
4
2
2  beat/s
27. Dn n n n n = - =
-
-
+
a r
v
v v
v
v v
1 1
=
- +
2
1
1 1
vv
v v v v
n
( ) ( )
=
´ ´ ´
´
=
2 320 4 243
316 324
6 Hz
28. n
c
c
n v
l
n =
+
=
´
+
( )
( )
2 1
4
320
4 1
2 1
 = + ´ = ( ) , , 2 1 80 80 240 n Hz Hz Hz
400 Hz, ...
n
0
0
1
2
320
2
1 =
+
=
´
+
( )
( )
n v
l
n
1.6
= + ´ = ( ) , , n 1 100 200 Hz 100 Hz Hz
300 400 Hz Hz , , ...
\ n n
c o
= = 400 Hz
29. I I
m ax
= 4
0
and I I I
max max
¢ = = 4 16
0
      L¢ = + 10 10 16 dB log
= + = 10 40 2 22 dB dB dB log
30. l
p p
p/
= = =
2 2
2
4
k
m
l = = ´ = 5
4
5
4
4
5
l
m m
32  |  Sound Waves
N
N N
N
C'
B C
341Hz 245 Hz
349Hz
31. d n
n
= + =
+
× ( )
( )
2 1
4
2 1
4
330
660
l
m 
= + = + ´
330
24
2 1 2 1 ( ) ( ) n n cm 13.75 cm.
= 13.75 cm 41.25 cm 68.75 cm, 96.25 cm , ,
etc. 
32. Dn
l l
l l
l l
= - =
- v v v
1 2
2 1
1 2
( )
 
=
´ ´
´
=
-
332 1 10
2
0.49 0.5
13.15 Hz
33. Dn n n
B A B
= - =
-
´ -
300
300 30
300 300
    = 33.33 Hz and n n
A B
¢ ¹ ¢
So both (a) and (b) options are wrong.
34. f
v v
v
f
v
v
f
a
=
+
= +
æ
è
ç
ö
ø
÷
0 0
1 and 
f
v
v
f
r
= -
æ
è
ç
ö
ø
÷
1
0
f
f
v v
v v
a
r
=
+
-
0
0
Þ ( ) ( ) f f v f f v
a r a r
- = +
0
Þ  
v
v
f f
f f
a r
a r 0
=
+
-
.
and       
f f
v
v
f
f f
f f
f
a r
a r
a r
- = =
-
+
æ
è
ç
ç
ö
ø
÷
÷
2
2
0
Þ    f
f f
a r
=
+
2
.
JEE Corner
¢ Assertion and Reason 
1. n
c
n
v
l
= + ( ) 2 1
4
 =
v
l
v
l
v
l 4
3
4
5
4
, , ,...
while, n
0
1
2 2
2
2
3
2
4
2
=
+
=
( )
, , , ,...
n v
l
v
l
v
l
v
l
v
l
it can be seen that n n
c o
¹ at all situation 
and n n
c o
=
1
2
 so assertion is true but
reason is false.
2. Apparent frequency is constant for
constant relative velocity so assertion is
false.
3. At a point of minimum displacement
pressure amplitude is maximum ie . .,
pressure difference is maximum not
pressure. So assertion is false.
4. The deriver receiver two sounds one
direct, n n
0
= and other n n
R
v u
v u
=
+
-
 such
that be detects beats. So reason is true
explanation of assertion.
5. With increase in intensity sound level
increases in lograthmic order so
assertion is false.
6. Speed of sound v
p
=
g
r
, with increase in
only pressure density increases such
that p/r remains constant. Again 
v
RT
M
=
g
 so both assertion and reason
are true but reason is not correct
explanation of assertion.
7. n n
A B
= + 4 when A is loaded with little
wax then n
A
 sightly decreases and then
beat frequency decreases, but if it is
heavily loaded with wax then its
frequency goes much below n
B
 such that
beat frequency increases. So, assertion
and reason are both true but reason is
not correct explanation of assertion.
 Sound Waves   | 33
8. 150  450, 7 50, 
          3,     5
The frequencies are odd harmonics then
the pipe is closed and fundamental
frequency is also 150 Hz. So assertion
and reason are both true but reason is
not correct explanation of assertion.
9. n µ
+
1
l e
 with increase in diameter end
correction, e, increases and n decreases.
So reason is correct explanation of
assertion.
10. With increasing length of air column,
number of overtone increases and not
the wavelength so assertion is false.
¢ Objective Questions (Level 2)
1. At the boundary between two mediums,
one part of incident wave gets reflected
and other part gets transmitted or
refracted.
2.
3
2
l
p =3.9 Þ l
p
=
3.9
1.5
Þ k= = =
2 2 3 p
l
p
p 3.9
1.5
3.9
 r
0 0
=S kB
Þ   S
kB
0
0
2 5
2
10 10
3
200
= =
´
´ ´
-
r
3.9
1.3 ( )
=
´
´
=
-
3.9
1.3
.025 m
10
12
0
1
  = 2.5 cm
3.
A
A
v
v v
t
i
=
+
=
´
+
=
2 2 100
200 100
2
3
2
1 2
 
4. Dn n n =
+
-
+
v
v v
v
v v
2 1
=
-
+ +
-
- v v v
v v v v
v v
v
n n ( )
( ) ( )
~
( )
1 2
2 1
1 2
\   v v
v
1 2
340 10
1700
- = =
´
=
Dn
n 
2 m/s
5. v gt
s
= = 10 m/s 
Dn n n =
+
-
-
-
+
v v
v v
v v
v v
s s
0 0
=
+
-
-
-
+
æ
è
ç
ç
ö
ø
÷
÷
´
300 2
300 10
300 2
300 10
150 Hz
= -
æ
è
ç
ö
ø
÷
´ =
302
290
298
310
150 12 Hz
6. 7
2
l
= L  Þ l =
2
7
L
A a kx a
L
L
a a = = × = = - cos cos cos
2
2
7
7
p
p
7. For maxima, nl = 3
Þ l n
l
= = = =
3
3
110
n
v nv
n ; .
\ n = 110 220 330 , , , .. Hz etc. maxima will
be formed so maximum will not be
formed at 120 Hz and 100 Hz.
8.
n
w
¢ =
+ °
+ ° -
v
v w v
s
cos
cos
60
60
=
+
+ -
´
300 10
300 10 20
500 Hz
= ´ =
310
290
500 534 Hz
9. Dn n n = - =
+
-
-
+
+
é
ë
ê
ù
û
ú
´
R
v
v
v
v
0
20
10
20
10
500
= -
æ
è
ç
ö
ø
÷
´
360
300
360
350
500 Hz = 31 Hz
10. Dn
p
p
p
p
= - = - =
404
2
400
2
202 200 2 Hz
I
I
max
min
: =
+
-
æ
è
ç
ç
ö
ø
÷
÷
=
2 1
2 1
9 1
2
34  |  Sound Waves
60° W
20 m/s
20 m/s
S
11. 3
4
34
l
= cm  Þ l = ´
4
3
34 cm
Þ  n
l
=
v
 Þ v
51
136
3
= = nl n
v
v
16
51
273 16
273 51
289
324
=
+
+
= = =
1 1
1.121 1.1
Þ   nl
nl
16
51
=
1.1
\    l
16
136
3
=
´
=
1.1
41.21 cm
12. 176
22
165 ´
-
-
= ´
+ v v
v
v v
v
Þ          176 v v v v v v ( ) ( ) ( ) - = + - 165 22
\    
176 330 330 165 330 330 22 ´ - = + - ( ) ( ) ( ) v v
or    1.143 ( ) 330 330 - = + v v
or         0.143 2.143 ´ = 330 v Þ v = 22 m/s
13. M
min
=
´ + ´
+
=
2 32 3 48
2 3
41.6 
n
n
m
m
2
1
2
1
1
2
32
= = = =
v
v 41.6
0.77
     = 0.875 Hz = 175
14.       v gt
0
30 = = m/s
    1100
30
1000 =
+
´
v
v
, 1.1v v = + 30
     0.1 v = 30   Þ v = 300 m/s
Passage (Q 5 to 17)
v v
m p
+ = 8 m/s,  50 150 v v
m p
= 
Þ      v v
m p
= 3 , 4 8 v
p
= m/s
     v
p
= 2 m/s and v
m
= 6 m/s
15. n¢ =
+
-
=
v
v
f f
2
6
332
324
0 0
 = constant
16. n¢ ¢ =
-
+
= =
v
v
f f
2
6
328
336
0 0
constant
17. n¢ ¢ < < ¢ f v
0
 and graph is (a)
18.
Both  (a)  and b are correct.
More Than One
19. n =
+ ( ) 2 1
4
n v
l
Þ l
v
n n = + =
´
+
4
2 1
330
4 264
2 1
n
( ) ( ) m
  = + ´ ( ) 2 1 n 31.25 cm
  = 31.25 cm 93.75 cm 156.25 cm , ,
20. (a) v p µ
0
, (b) v T µ Þ v T
2
µ ,
where T is absolute temperature.
(c) v F µ (d) n µ
1
l
\ (c) and (d) are correct.
21. P BA k
0
= ;  B
p
V
V
p
p
p
= - -
D
D
D
D
\ D
D
r
r r
r = = =
p
B
BA k
B
A k
  r ; µ p
Pressure and density equations are in
opposite phase i e . ., Df
p
=
2
 and not p.
So, (a), (b) and (c) are correct.
22.
5
4
3
2
v
l
v
l
c o
= Þ 
125 2 .
l l
c o
= Þ 
l
l
o
c
= =
2 8
5 1.25
.
(a) n
c
c
o
o
v
l
v
l
v
l
= =
´
=
4
4
5
8
2
5
     = × =
4
5 2
4
5
v
l
o
o
n  Þ   n n
c o
<
(b) n
c
c
v
l
v
l
v
l
= =
´
= ×
3
4
3
4
5
8
12
5 2
0
0
     = × = ×
6
5
2
2
6
5
v
l
o
o
n Þ n n
c o
> .
 Sound Waves   | 35
x
y
D E
C B A