Chapter 16 : Sound Waves - HC Verma Solution, Physics Class 11 Notes | EduRev

Physics Class 12

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JEE : Chapter 16 : Sound Waves - HC Verma Solution, Physics Class 11 Notes | EduRev

 Page 1


16.1
SOLUTIONS TO CONCEPTS 
CHAPTER – 16
1. V
air
= 230 m/s. V
s
= 5200 m/s. Here S = 7 m
So, t = t
1
– t
2
= ?
?
?
?
?
?
?
5200
1
330
1
= 2.75 × 10
–3
sec = 2.75 ms.
2. Here given S = 80 m × 2 = 160 m.
v = 320 m/s
So the maximum time interval will be 
t = 5/v = 160/320 = 0.5 seconds.
3. He has to clap 10 times in 3 seconds.
So time interval between two clap = (3/10 second).
So the time taken go the wall = (3/2 × 10) = 3/20 seconds.
= 333 m/s.
4. a) for maximum wavelength n = 20 Hz.
as ?
?
?
?
?
?
?
? ?
1
b) for minimum wavelength, n = 20 kHz
? ? = 360/ (20 × 10
3
) = 18 × 10
–3
m = 18 mm
? x = (v/n) = 360/20 = 18 m. ?
5. a) for minimum wavelength n = 20 KHz
? v = n ? ? ? = ?
?
?
?
?
?
?
3
10 20
1450
= 7.25 cm.
b) for maximum wavelength n should be minium
? v = n ? ? ? = v/n ? 1450 / 20 = 72.5 m. ?
6. According to the question,
a) ? = 20 cm × 10 = 200 cm = 2 m
? v = 340 m/s
so, n = v/ ? = 340/2 = 170 Hz.
N = v/ ? ?
2
10 2
340
?
?
= 17.000 Hz = 17 KH
2
(because ? = 2 cm = 2 × 10
–2
m) ?
7. a) Given V
air
= 340 m/s , n = 4.5 ×10
6
Hz
? ?
air
= (340 / 4.5) × 10
–6
= 7.36 × 10
–5 
m.
b) V
tissue 
= 1500 m/s ? ?
t
= (1500 / 4.5) × 10
–6
= 3.3 × 10
–4
m. ?
8. Here given r
y
= 6.0 × 10
–5
m
a) Given 2 ?/ ? = 1.8 ? ? = (2 ?/1.8)
? So, 
?
? ?
?
?
?
2
s / m 10 ) 8 . 1 ( 0 . 6
r
5
y
= 1.7 × 10
–5
m
b) Let, velocity amplitude = V
y
V = dy/dt = 3600 cos (600 t – 1.8) × 10
–5
m/s
Here V
y
= 3600 × 10
–5
  m/s
Again, ? = 2 ?/1.8 and T = 2 ?/600 ? wave speed = v = ?/T = 600/1.8 = 1000 / 3 m/s.
So the ratio of (V
y
/v) = 
1000
10 3 3600
5 ?
? ?
. ?
9. a) Here given n = 100, v = 350 m/s
? ? = 
100
350
n
v
? = 3.5 m.
In 2.5 ms, the distance travelled by the particle is given by
?x = 350 × 2.5 × 10
–3
Page 2


16.1
SOLUTIONS TO CONCEPTS 
CHAPTER – 16
1. V
air
= 230 m/s. V
s
= 5200 m/s. Here S = 7 m
So, t = t
1
– t
2
= ?
?
?
?
?
?
?
5200
1
330
1
= 2.75 × 10
–3
sec = 2.75 ms.
2. Here given S = 80 m × 2 = 160 m.
v = 320 m/s
So the maximum time interval will be 
t = 5/v = 160/320 = 0.5 seconds.
3. He has to clap 10 times in 3 seconds.
So time interval between two clap = (3/10 second).
So the time taken go the wall = (3/2 × 10) = 3/20 seconds.
= 333 m/s.
4. a) for maximum wavelength n = 20 Hz.
as ?
?
?
?
?
?
?
? ?
1
b) for minimum wavelength, n = 20 kHz
? ? = 360/ (20 × 10
3
) = 18 × 10
–3
m = 18 mm
? x = (v/n) = 360/20 = 18 m. ?
5. a) for minimum wavelength n = 20 KHz
? v = n ? ? ? = ?
?
?
?
?
?
?
3
10 20
1450
= 7.25 cm.
b) for maximum wavelength n should be minium
? v = n ? ? ? = v/n ? 1450 / 20 = 72.5 m. ?
6. According to the question,
a) ? = 20 cm × 10 = 200 cm = 2 m
? v = 340 m/s
so, n = v/ ? = 340/2 = 170 Hz.
N = v/ ? ?
2
10 2
340
?
?
= 17.000 Hz = 17 KH
2
(because ? = 2 cm = 2 × 10
–2
m) ?
7. a) Given V
air
= 340 m/s , n = 4.5 ×10
6
Hz
? ?
air
= (340 / 4.5) × 10
–6
= 7.36 × 10
–5 
m.
b) V
tissue 
= 1500 m/s ? ?
t
= (1500 / 4.5) × 10
–6
= 3.3 × 10
–4
m. ?
8. Here given r
y
= 6.0 × 10
–5
m
a) Given 2 ?/ ? = 1.8 ? ? = (2 ?/1.8)
? So, 
?
? ?
?
?
?
2
s / m 10 ) 8 . 1 ( 0 . 6
r
5
y
= 1.7 × 10
–5
m
b) Let, velocity amplitude = V
y
V = dy/dt = 3600 cos (600 t – 1.8) × 10
–5
m/s
Here V
y
= 3600 × 10
–5
  m/s
Again, ? = 2 ?/1.8 and T = 2 ?/600 ? wave speed = v = ?/T = 600/1.8 = 1000 / 3 m/s.
So the ratio of (V
y
/v) = 
1000
10 3 3600
5 ?
? ?
. ?
9. a) Here given n = 100, v = 350 m/s
? ? = 
100
350
n
v
? = 3.5 m.
In 2.5 ms, the distance travelled by the particle is given by
?x = 350 × 2.5 × 10
–3
Chapter 16
16.2
So, phase difference ? = x
2
? ?
?
?
? ) 2 / ( 10 5 . 2 350
) 100 / 350 (
2
3
? ? ? ? ?
?
?
.
b) In the second case, Given ?? = 10 cm = 10
–1
m
? So, ? = 35 / 2
) 100 / 350 (
10 2
x
x
2
1
? ?
? ?
? ?
?
?
. ?
10. a) Given ?x = 10 cm, ? = 5.0 cm
? ? = ? ? ?
?
? 2
= ? ? ?
?
4 10
5
2
.
So phase difference is zero.
b) Zero, as the particle is in same phase because of having same path.
11. Given that p = 1.0 × 10
5
N/m
2
, T = 273 K, M = 32 g = 32 × 10
–3
kg
V = 22.4 litre = 22.4 × 10
–3
m
3
C/C
v
= r = 3.5 R / 2.5 R = 1.4
? V = 
4 . 22 / 32
10 0 . 1 4 . 1
f
rp
5 ?
? ?
? = 310 m/s (because ? = m/v) ?
12. V
1
= 330 m/s, V
2
= ?
T
1
= 273 + 17 = 290 K, T
2
= 272 + 32 = 305 K
We know v ? T
1
2 1
2
2
1
2
1
T
T V
V
T
T
V
V ?
? ? ?
= 
290
305
340 ? = 349 m/s.
13. T
1
= 273 V
2
= 2V
1
V
1
= v T
2
= ?
We know that V ? T ?
2
1
2
2
1
2
V
V
T
T
? ? T
2
= 273 × 2
2
= 4 × 273 K
So temperature will be (4 × 273) – 273 = 819°c.
14. The variation of temperature is given by
T = T
1
+ 
d
) T T (
2 2
?
x …(1)
We know that V ? T ?
273
T
V
V
T
? ? VT = 
273
T
v
? dt = 
T
273
V
du
V
dx
T
? ?
? t = 
?
? ?
d
0
2 / 1
1 2 1
] x ) d / ) T T ( T [
dx
V
273
= 
d
0
1 2
1
1 2
] x
d
T T
T [
T T
d 2
V
273 ?
?
?
? = 
1 2
1 2
T T
T T
273
V
d 2
? ?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
= T = 
1 2
T T
273
V
d 2
?
Putting the given value we get
= 
310 280
273
330
33 2
?
?
?
= 96 ms.
??
?x
cm
20 cm
10cm
A
B
d
x
T 1
T 2
Page 3


16.1
SOLUTIONS TO CONCEPTS 
CHAPTER – 16
1. V
air
= 230 m/s. V
s
= 5200 m/s. Here S = 7 m
So, t = t
1
– t
2
= ?
?
?
?
?
?
?
5200
1
330
1
= 2.75 × 10
–3
sec = 2.75 ms.
2. Here given S = 80 m × 2 = 160 m.
v = 320 m/s
So the maximum time interval will be 
t = 5/v = 160/320 = 0.5 seconds.
3. He has to clap 10 times in 3 seconds.
So time interval between two clap = (3/10 second).
So the time taken go the wall = (3/2 × 10) = 3/20 seconds.
= 333 m/s.
4. a) for maximum wavelength n = 20 Hz.
as ?
?
?
?
?
?
?
? ?
1
b) for minimum wavelength, n = 20 kHz
? ? = 360/ (20 × 10
3
) = 18 × 10
–3
m = 18 mm
? x = (v/n) = 360/20 = 18 m. ?
5. a) for minimum wavelength n = 20 KHz
? v = n ? ? ? = ?
?
?
?
?
?
?
3
10 20
1450
= 7.25 cm.
b) for maximum wavelength n should be minium
? v = n ? ? ? = v/n ? 1450 / 20 = 72.5 m. ?
6. According to the question,
a) ? = 20 cm × 10 = 200 cm = 2 m
? v = 340 m/s
so, n = v/ ? = 340/2 = 170 Hz.
N = v/ ? ?
2
10 2
340
?
?
= 17.000 Hz = 17 KH
2
(because ? = 2 cm = 2 × 10
–2
m) ?
7. a) Given V
air
= 340 m/s , n = 4.5 ×10
6
Hz
? ?
air
= (340 / 4.5) × 10
–6
= 7.36 × 10
–5 
m.
b) V
tissue 
= 1500 m/s ? ?
t
= (1500 / 4.5) × 10
–6
= 3.3 × 10
–4
m. ?
8. Here given r
y
= 6.0 × 10
–5
m
a) Given 2 ?/ ? = 1.8 ? ? = (2 ?/1.8)
? So, 
?
? ?
?
?
?
2
s / m 10 ) 8 . 1 ( 0 . 6
r
5
y
= 1.7 × 10
–5
m
b) Let, velocity amplitude = V
y
V = dy/dt = 3600 cos (600 t – 1.8) × 10
–5
m/s
Here V
y
= 3600 × 10
–5
  m/s
Again, ? = 2 ?/1.8 and T = 2 ?/600 ? wave speed = v = ?/T = 600/1.8 = 1000 / 3 m/s.
So the ratio of (V
y
/v) = 
1000
10 3 3600
5 ?
? ?
. ?
9. a) Here given n = 100, v = 350 m/s
? ? = 
100
350
n
v
? = 3.5 m.
In 2.5 ms, the distance travelled by the particle is given by
?x = 350 × 2.5 × 10
–3
Chapter 16
16.2
So, phase difference ? = x
2
? ?
?
?
? ) 2 / ( 10 5 . 2 350
) 100 / 350 (
2
3
? ? ? ? ?
?
?
.
b) In the second case, Given ?? = 10 cm = 10
–1
m
? So, ? = 35 / 2
) 100 / 350 (
10 2
x
x
2
1
? ?
? ?
? ?
?
?
. ?
10. a) Given ?x = 10 cm, ? = 5.0 cm
? ? = ? ? ?
?
? 2
= ? ? ?
?
4 10
5
2
.
So phase difference is zero.
b) Zero, as the particle is in same phase because of having same path.
11. Given that p = 1.0 × 10
5
N/m
2
, T = 273 K, M = 32 g = 32 × 10
–3
kg
V = 22.4 litre = 22.4 × 10
–3
m
3
C/C
v
= r = 3.5 R / 2.5 R = 1.4
? V = 
4 . 22 / 32
10 0 . 1 4 . 1
f
rp
5 ?
? ?
? = 310 m/s (because ? = m/v) ?
12. V
1
= 330 m/s, V
2
= ?
T
1
= 273 + 17 = 290 K, T
2
= 272 + 32 = 305 K
We know v ? T
1
2 1
2
2
1
2
1
T
T V
V
T
T
V
V ?
? ? ?
= 
290
305
340 ? = 349 m/s.
13. T
1
= 273 V
2
= 2V
1
V
1
= v T
2
= ?
We know that V ? T ?
2
1
2
2
1
2
V
V
T
T
? ? T
2
= 273 × 2
2
= 4 × 273 K
So temperature will be (4 × 273) – 273 = 819°c.
14. The variation of temperature is given by
T = T
1
+ 
d
) T T (
2 2
?
x …(1)
We know that V ? T ?
273
T
V
V
T
? ? VT = 
273
T
v
? dt = 
T
273
V
du
V
dx
T
? ?
? t = 
?
? ?
d
0
2 / 1
1 2 1
] x ) d / ) T T ( T [
dx
V
273
= 
d
0
1 2
1
1 2
] x
d
T T
T [
T T
d 2
V
273 ?
?
?
? = 
1 2
1 2
T T
T T
273
V
d 2
? ?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
= T = 
1 2
T T
273
V
d 2
?
Putting the given value we get
= 
310 280
273
330
33 2
?
?
?
= 96 ms.
??
?x
cm
20 cm
10cm
A
B
d
x
T 1
T 2
Chapter 16
16.3
15. We know that v = ? / K
Where K = bulk modulus of elasticity
? K = v
2
? = (1330)
2
× 800 N/m
2
We know K = ?
?
?
?
?
?
? V / V
A / F
? ?V = 
800 1330 1330
10 2
K
essures Pr
5
? ?
?
?
So, ?V = 0.15 cm
3 ?
16. We know that,
Bulk modulus B = 
0
0
S 2
P
) V / V (
p
?
?
?
?
?
Where P
0
= pressure amplitude ? P
0
= 1.0 × 10
5
S
0
= displacement amplitude ? S
0
= 5.5 × 10
–6
m
? B = 
m 10 ) 5 . 5 ( 2
m 10 35 14
6
2
?
?
? ?
? ?
= 1.4 × 10
5
N/m
2
.
17. a) Here given V
air
= 340 m/s., Power = E/t = 20 W
f = 2,000 Hz, ? = 1.2 kg/m
3
So, intensity I = E/t.A
= 44
6 4
20
r 4
20
2 2
?
? ? ?
?
?
mw/m
2
(because r = 6m)
b) We know that I = 
air
2
0
V 2
P
?
?
air 0
V 2 1 P ? ? ?
? = 
3
10 44 340 2 . 1 2
?
? ? ? ? = 6.0 N/m
2
.
c) We know that I = V v S 2
2 2
0
2
? ? where S
0
= displacement amplitude
? S
0
= 
air
2 2
V
I
? ? ?
Putting the value we get S
g
= 1.2 × 10
–6
m.
18. Here I
1
= 1.0 × 10
–8
W
1
/m
2
; I
2
= ?
r
1
= 5.0 m, r
2
= 25 m.
We know that I ?
2
r
1
? I
1
r
1
2
= I
2
r
2
2
  ? I
2
= 
2
2
2
1 1
r
r I
= 
625
25 10 0 . 1
8
? ?
?
= 4.0 × 10
–10
W/m
2
.
19. We know that ? = 10 log
10
?
?
?
?
?
?
?
?
0
I
I
?
A
= 
o
A
I
I
log 10 , ?
B
= 
o
B
I
I
log 10
? I
A
/ I
0
= 
) 10 / (
A
10
?
? I
B
/I
o
= 
) 10 / (
B
10
?
?
2
2
A
2
B
B
A
5
50
r
r
I
I
?
?
?
?
?
?
? ? ?
2 ) (
10 10
B A
?
? ?
? 2
10
B A
?
? ? ?
? 20
B A
? ? ? ?
? ?
B
= 40 – 20 = 20 d ?. ?
Page 4


16.1
SOLUTIONS TO CONCEPTS 
CHAPTER – 16
1. V
air
= 230 m/s. V
s
= 5200 m/s. Here S = 7 m
So, t = t
1
– t
2
= ?
?
?
?
?
?
?
5200
1
330
1
= 2.75 × 10
–3
sec = 2.75 ms.
2. Here given S = 80 m × 2 = 160 m.
v = 320 m/s
So the maximum time interval will be 
t = 5/v = 160/320 = 0.5 seconds.
3. He has to clap 10 times in 3 seconds.
So time interval between two clap = (3/10 second).
So the time taken go the wall = (3/2 × 10) = 3/20 seconds.
= 333 m/s.
4. a) for maximum wavelength n = 20 Hz.
as ?
?
?
?
?
?
?
? ?
1
b) for minimum wavelength, n = 20 kHz
? ? = 360/ (20 × 10
3
) = 18 × 10
–3
m = 18 mm
? x = (v/n) = 360/20 = 18 m. ?
5. a) for minimum wavelength n = 20 KHz
? v = n ? ? ? = ?
?
?
?
?
?
?
3
10 20
1450
= 7.25 cm.
b) for maximum wavelength n should be minium
? v = n ? ? ? = v/n ? 1450 / 20 = 72.5 m. ?
6. According to the question,
a) ? = 20 cm × 10 = 200 cm = 2 m
? v = 340 m/s
so, n = v/ ? = 340/2 = 170 Hz.
N = v/ ? ?
2
10 2
340
?
?
= 17.000 Hz = 17 KH
2
(because ? = 2 cm = 2 × 10
–2
m) ?
7. a) Given V
air
= 340 m/s , n = 4.5 ×10
6
Hz
? ?
air
= (340 / 4.5) × 10
–6
= 7.36 × 10
–5 
m.
b) V
tissue 
= 1500 m/s ? ?
t
= (1500 / 4.5) × 10
–6
= 3.3 × 10
–4
m. ?
8. Here given r
y
= 6.0 × 10
–5
m
a) Given 2 ?/ ? = 1.8 ? ? = (2 ?/1.8)
? So, 
?
? ?
?
?
?
2
s / m 10 ) 8 . 1 ( 0 . 6
r
5
y
= 1.7 × 10
–5
m
b) Let, velocity amplitude = V
y
V = dy/dt = 3600 cos (600 t – 1.8) × 10
–5
m/s
Here V
y
= 3600 × 10
–5
  m/s
Again, ? = 2 ?/1.8 and T = 2 ?/600 ? wave speed = v = ?/T = 600/1.8 = 1000 / 3 m/s.
So the ratio of (V
y
/v) = 
1000
10 3 3600
5 ?
? ?
. ?
9. a) Here given n = 100, v = 350 m/s
? ? = 
100
350
n
v
? = 3.5 m.
In 2.5 ms, the distance travelled by the particle is given by
?x = 350 × 2.5 × 10
–3
Chapter 16
16.2
So, phase difference ? = x
2
? ?
?
?
? ) 2 / ( 10 5 . 2 350
) 100 / 350 (
2
3
? ? ? ? ?
?
?
.
b) In the second case, Given ?? = 10 cm = 10
–1
m
? So, ? = 35 / 2
) 100 / 350 (
10 2
x
x
2
1
? ?
? ?
? ?
?
?
. ?
10. a) Given ?x = 10 cm, ? = 5.0 cm
? ? = ? ? ?
?
? 2
= ? ? ?
?
4 10
5
2
.
So phase difference is zero.
b) Zero, as the particle is in same phase because of having same path.
11. Given that p = 1.0 × 10
5
N/m
2
, T = 273 K, M = 32 g = 32 × 10
–3
kg
V = 22.4 litre = 22.4 × 10
–3
m
3
C/C
v
= r = 3.5 R / 2.5 R = 1.4
? V = 
4 . 22 / 32
10 0 . 1 4 . 1
f
rp
5 ?
? ?
? = 310 m/s (because ? = m/v) ?
12. V
1
= 330 m/s, V
2
= ?
T
1
= 273 + 17 = 290 K, T
2
= 272 + 32 = 305 K
We know v ? T
1
2 1
2
2
1
2
1
T
T V
V
T
T
V
V ?
? ? ?
= 
290
305
340 ? = 349 m/s.
13. T
1
= 273 V
2
= 2V
1
V
1
= v T
2
= ?
We know that V ? T ?
2
1
2
2
1
2
V
V
T
T
? ? T
2
= 273 × 2
2
= 4 × 273 K
So temperature will be (4 × 273) – 273 = 819°c.
14. The variation of temperature is given by
T = T
1
+ 
d
) T T (
2 2
?
x …(1)
We know that V ? T ?
273
T
V
V
T
? ? VT = 
273
T
v
? dt = 
T
273
V
du
V
dx
T
? ?
? t = 
?
? ?
d
0
2 / 1
1 2 1
] x ) d / ) T T ( T [
dx
V
273
= 
d
0
1 2
1
1 2
] x
d
T T
T [
T T
d 2
V
273 ?
?
?
? = 
1 2
1 2
T T
T T
273
V
d 2
? ?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
= T = 
1 2
T T
273
V
d 2
?
Putting the given value we get
= 
310 280
273
330
33 2
?
?
?
= 96 ms.
??
?x
cm
20 cm
10cm
A
B
d
x
T 1
T 2
Chapter 16
16.3
15. We know that v = ? / K
Where K = bulk modulus of elasticity
? K = v
2
? = (1330)
2
× 800 N/m
2
We know K = ?
?
?
?
?
?
? V / V
A / F
? ?V = 
800 1330 1330
10 2
K
essures Pr
5
? ?
?
?
So, ?V = 0.15 cm
3 ?
16. We know that,
Bulk modulus B = 
0
0
S 2
P
) V / V (
p
?
?
?
?
?
Where P
0
= pressure amplitude ? P
0
= 1.0 × 10
5
S
0
= displacement amplitude ? S
0
= 5.5 × 10
–6
m
? B = 
m 10 ) 5 . 5 ( 2
m 10 35 14
6
2
?
?
? ?
? ?
= 1.4 × 10
5
N/m
2
.
17. a) Here given V
air
= 340 m/s., Power = E/t = 20 W
f = 2,000 Hz, ? = 1.2 kg/m
3
So, intensity I = E/t.A
= 44
6 4
20
r 4
20
2 2
?
? ? ?
?
?
mw/m
2
(because r = 6m)
b) We know that I = 
air
2
0
V 2
P
?
?
air 0
V 2 1 P ? ? ?
? = 
3
10 44 340 2 . 1 2
?
? ? ? ? = 6.0 N/m
2
.
c) We know that I = V v S 2
2 2
0
2
? ? where S
0
= displacement amplitude
? S
0
= 
air
2 2
V
I
? ? ?
Putting the value we get S
g
= 1.2 × 10
–6
m.
18. Here I
1
= 1.0 × 10
–8
W
1
/m
2
; I
2
= ?
r
1
= 5.0 m, r
2
= 25 m.
We know that I ?
2
r
1
? I
1
r
1
2
= I
2
r
2
2
  ? I
2
= 
2
2
2
1 1
r
r I
= 
625
25 10 0 . 1
8
? ?
?
= 4.0 × 10
–10
W/m
2
.
19. We know that ? = 10 log
10
?
?
?
?
?
?
?
?
0
I
I
?
A
= 
o
A
I
I
log 10 , ?
B
= 
o
B
I
I
log 10
? I
A
/ I
0
= 
) 10 / (
A
10
?
? I
B
/I
o
= 
) 10 / (
B
10
?
?
2
2
A
2
B
B
A
5
50
r
r
I
I
?
?
?
?
?
?
? ? ?
2 ) (
10 10
B A
?
? ?
? 2
10
B A
?
? ? ?
? 20
B A
? ? ? ?
? ?
B
= 40 – 20 = 20 d ?. ?
Chapter 16
16.4
20. We know that, ? = 10 log
10
J/I
0
According to the questions
?
A
= 10 log
10
(2I/I
0
)
? ?
B
– ?
A
= 10 log (2I/I) = 10 × 0.3010 = 3 dB. ?
21. If sound level = 120 dB, then I = intensity = 1 W/m
2
Given that, audio output = 2W
Let the closest distance be x.
So, intensity = (2 / 4 ?x
2
) = 1 ? x
2
= (2/2 ?) ? x = 0.4 m = 40 cm. ?
22. ?
1
= 50 dB, ?
2
= 60 dB
? I
1
= 10
–7
W/m
2
, I
2
= 10
–6
W/m
2
(because ? = 10 log
10
(I/I
0
), where I
0
= 10
–12
W/m
2
)
Again, I
2
/I
1
= (p
2
/p
1
)
2
=(10
–6
/10
–7
) = 10 (where p = pressure amplitude).
? (p
2
/ p
1
) = 10 . ?
23. Let the intensity of each student be I.
According to the question
?
A
= 
0
10
I
I 50
log 10 ; ?
B
= 
?
?
?
?
?
?
?
?
0
10
I
I 100
log 10
? ?
B
– ?
A
= 
0
10
I
I 50
log 10 –
?
?
?
?
?
?
?
?
0
10
I
I 100
log 10
= 3 2 log 10
I 50
I 100
log 10
10
? ?
?
?
?
?
?
?
?
?
So, ?
A
= 50 + 3 = 53 dB. ?
24. Distance between tow maximum to a minimum is given by, ?/4 = 2.50 cm
? ? = 10 cm = 10
–1
m
We know, V = nx
? n = 
1
10
340 V
?
?
?
= 3400 Hz = 3.4 kHz. ?
25. a) According to the data
?/4 = 16.5 mm ? ? = 66 mm = 66 × 10
–6=3
m
? n = 
3
10 66
330 V
?
?
?
?
= 5 kHz.
b) I
minimum 
= K(A
1
– A
2
)
2
= I ? A
1
– A
2
= 11
I
maximum 
= K(A
1
+ A
2
)
2
= 9 ? A
1
+ A
2
= 31
So, 
4
3
A A
A A
2 1
2 1
?
?
?
? A
1
/A
2
= 2/1
So, the ratio amplitudes is 2. ?
26. The path difference of the two sound waves is given by
?L = 6.4 – 6.0 = 0.4 m
The wavelength of either wave = ? = 
?
?
?
320 V
(m/s)
For destructive interference ?L = 
2
) 1 n 2 ( ? ?
where n is an integers.
or 0.4 m = 
?
?
? 320
2
1 n 2
? ? = n = Hz
2
1 n 2
800
4 . 0
320 ?
? = (2n + 1) 400 Hz
Thus the frequency within the specified range which cause destructive interference are 1200 Hz, 
2000 Hz, 2800 Hz, 3600 Hz and 4400 Hz. ?
Page 5


16.1
SOLUTIONS TO CONCEPTS 
CHAPTER – 16
1. V
air
= 230 m/s. V
s
= 5200 m/s. Here S = 7 m
So, t = t
1
– t
2
= ?
?
?
?
?
?
?
5200
1
330
1
= 2.75 × 10
–3
sec = 2.75 ms.
2. Here given S = 80 m × 2 = 160 m.
v = 320 m/s
So the maximum time interval will be 
t = 5/v = 160/320 = 0.5 seconds.
3. He has to clap 10 times in 3 seconds.
So time interval between two clap = (3/10 second).
So the time taken go the wall = (3/2 × 10) = 3/20 seconds.
= 333 m/s.
4. a) for maximum wavelength n = 20 Hz.
as ?
?
?
?
?
?
?
? ?
1
b) for minimum wavelength, n = 20 kHz
? ? = 360/ (20 × 10
3
) = 18 × 10
–3
m = 18 mm
? x = (v/n) = 360/20 = 18 m. ?
5. a) for minimum wavelength n = 20 KHz
? v = n ? ? ? = ?
?
?
?
?
?
?
3
10 20
1450
= 7.25 cm.
b) for maximum wavelength n should be minium
? v = n ? ? ? = v/n ? 1450 / 20 = 72.5 m. ?
6. According to the question,
a) ? = 20 cm × 10 = 200 cm = 2 m
? v = 340 m/s
so, n = v/ ? = 340/2 = 170 Hz.
N = v/ ? ?
2
10 2
340
?
?
= 17.000 Hz = 17 KH
2
(because ? = 2 cm = 2 × 10
–2
m) ?
7. a) Given V
air
= 340 m/s , n = 4.5 ×10
6
Hz
? ?
air
= (340 / 4.5) × 10
–6
= 7.36 × 10
–5 
m.
b) V
tissue 
= 1500 m/s ? ?
t
= (1500 / 4.5) × 10
–6
= 3.3 × 10
–4
m. ?
8. Here given r
y
= 6.0 × 10
–5
m
a) Given 2 ?/ ? = 1.8 ? ? = (2 ?/1.8)
? So, 
?
? ?
?
?
?
2
s / m 10 ) 8 . 1 ( 0 . 6
r
5
y
= 1.7 × 10
–5
m
b) Let, velocity amplitude = V
y
V = dy/dt = 3600 cos (600 t – 1.8) × 10
–5
m/s
Here V
y
= 3600 × 10
–5
  m/s
Again, ? = 2 ?/1.8 and T = 2 ?/600 ? wave speed = v = ?/T = 600/1.8 = 1000 / 3 m/s.
So the ratio of (V
y
/v) = 
1000
10 3 3600
5 ?
? ?
. ?
9. a) Here given n = 100, v = 350 m/s
? ? = 
100
350
n
v
? = 3.5 m.
In 2.5 ms, the distance travelled by the particle is given by
?x = 350 × 2.5 × 10
–3
Chapter 16
16.2
So, phase difference ? = x
2
? ?
?
?
? ) 2 / ( 10 5 . 2 350
) 100 / 350 (
2
3
? ? ? ? ?
?
?
.
b) In the second case, Given ?? = 10 cm = 10
–1
m
? So, ? = 35 / 2
) 100 / 350 (
10 2
x
x
2
1
? ?
? ?
? ?
?
?
. ?
10. a) Given ?x = 10 cm, ? = 5.0 cm
? ? = ? ? ?
?
? 2
= ? ? ?
?
4 10
5
2
.
So phase difference is zero.
b) Zero, as the particle is in same phase because of having same path.
11. Given that p = 1.0 × 10
5
N/m
2
, T = 273 K, M = 32 g = 32 × 10
–3
kg
V = 22.4 litre = 22.4 × 10
–3
m
3
C/C
v
= r = 3.5 R / 2.5 R = 1.4
? V = 
4 . 22 / 32
10 0 . 1 4 . 1
f
rp
5 ?
? ?
? = 310 m/s (because ? = m/v) ?
12. V
1
= 330 m/s, V
2
= ?
T
1
= 273 + 17 = 290 K, T
2
= 272 + 32 = 305 K
We know v ? T
1
2 1
2
2
1
2
1
T
T V
V
T
T
V
V ?
? ? ?
= 
290
305
340 ? = 349 m/s.
13. T
1
= 273 V
2
= 2V
1
V
1
= v T
2
= ?
We know that V ? T ?
2
1
2
2
1
2
V
V
T
T
? ? T
2
= 273 × 2
2
= 4 × 273 K
So temperature will be (4 × 273) – 273 = 819°c.
14. The variation of temperature is given by
T = T
1
+ 
d
) T T (
2 2
?
x …(1)
We know that V ? T ?
273
T
V
V
T
? ? VT = 
273
T
v
? dt = 
T
273
V
du
V
dx
T
? ?
? t = 
?
? ?
d
0
2 / 1
1 2 1
] x ) d / ) T T ( T [
dx
V
273
= 
d
0
1 2
1
1 2
] x
d
T T
T [
T T
d 2
V
273 ?
?
?
? = 
1 2
1 2
T T
T T
273
V
d 2
? ?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
= T = 
1 2
T T
273
V
d 2
?
Putting the given value we get
= 
310 280
273
330
33 2
?
?
?
= 96 ms.
??
?x
cm
20 cm
10cm
A
B
d
x
T 1
T 2
Chapter 16
16.3
15. We know that v = ? / K
Where K = bulk modulus of elasticity
? K = v
2
? = (1330)
2
× 800 N/m
2
We know K = ?
?
?
?
?
?
? V / V
A / F
? ?V = 
800 1330 1330
10 2
K
essures Pr
5
? ?
?
?
So, ?V = 0.15 cm
3 ?
16. We know that,
Bulk modulus B = 
0
0
S 2
P
) V / V (
p
?
?
?
?
?
Where P
0
= pressure amplitude ? P
0
= 1.0 × 10
5
S
0
= displacement amplitude ? S
0
= 5.5 × 10
–6
m
? B = 
m 10 ) 5 . 5 ( 2
m 10 35 14
6
2
?
?
? ?
? ?
= 1.4 × 10
5
N/m
2
.
17. a) Here given V
air
= 340 m/s., Power = E/t = 20 W
f = 2,000 Hz, ? = 1.2 kg/m
3
So, intensity I = E/t.A
= 44
6 4
20
r 4
20
2 2
?
? ? ?
?
?
mw/m
2
(because r = 6m)
b) We know that I = 
air
2
0
V 2
P
?
?
air 0
V 2 1 P ? ? ?
? = 
3
10 44 340 2 . 1 2
?
? ? ? ? = 6.0 N/m
2
.
c) We know that I = V v S 2
2 2
0
2
? ? where S
0
= displacement amplitude
? S
0
= 
air
2 2
V
I
? ? ?
Putting the value we get S
g
= 1.2 × 10
–6
m.
18. Here I
1
= 1.0 × 10
–8
W
1
/m
2
; I
2
= ?
r
1
= 5.0 m, r
2
= 25 m.
We know that I ?
2
r
1
? I
1
r
1
2
= I
2
r
2
2
  ? I
2
= 
2
2
2
1 1
r
r I
= 
625
25 10 0 . 1
8
? ?
?
= 4.0 × 10
–10
W/m
2
.
19. We know that ? = 10 log
10
?
?
?
?
?
?
?
?
0
I
I
?
A
= 
o
A
I
I
log 10 , ?
B
= 
o
B
I
I
log 10
? I
A
/ I
0
= 
) 10 / (
A
10
?
? I
B
/I
o
= 
) 10 / (
B
10
?
?
2
2
A
2
B
B
A
5
50
r
r
I
I
?
?
?
?
?
?
? ? ?
2 ) (
10 10
B A
?
? ?
? 2
10
B A
?
? ? ?
? 20
B A
? ? ? ?
? ?
B
= 40 – 20 = 20 d ?. ?
Chapter 16
16.4
20. We know that, ? = 10 log
10
J/I
0
According to the questions
?
A
= 10 log
10
(2I/I
0
)
? ?
B
– ?
A
= 10 log (2I/I) = 10 × 0.3010 = 3 dB. ?
21. If sound level = 120 dB, then I = intensity = 1 W/m
2
Given that, audio output = 2W
Let the closest distance be x.
So, intensity = (2 / 4 ?x
2
) = 1 ? x
2
= (2/2 ?) ? x = 0.4 m = 40 cm. ?
22. ?
1
= 50 dB, ?
2
= 60 dB
? I
1
= 10
–7
W/m
2
, I
2
= 10
–6
W/m
2
(because ? = 10 log
10
(I/I
0
), where I
0
= 10
–12
W/m
2
)
Again, I
2
/I
1
= (p
2
/p
1
)
2
=(10
–6
/10
–7
) = 10 (where p = pressure amplitude).
? (p
2
/ p
1
) = 10 . ?
23. Let the intensity of each student be I.
According to the question
?
A
= 
0
10
I
I 50
log 10 ; ?
B
= 
?
?
?
?
?
?
?
?
0
10
I
I 100
log 10
? ?
B
– ?
A
= 
0
10
I
I 50
log 10 –
?
?
?
?
?
?
?
?
0
10
I
I 100
log 10
= 3 2 log 10
I 50
I 100
log 10
10
? ?
?
?
?
?
?
?
?
?
So, ?
A
= 50 + 3 = 53 dB. ?
24. Distance between tow maximum to a minimum is given by, ?/4 = 2.50 cm
? ? = 10 cm = 10
–1
m
We know, V = nx
? n = 
1
10
340 V
?
?
?
= 3400 Hz = 3.4 kHz. ?
25. a) According to the data
?/4 = 16.5 mm ? ? = 66 mm = 66 × 10
–6=3
m
? n = 
3
10 66
330 V
?
?
?
?
= 5 kHz.
b) I
minimum 
= K(A
1
– A
2
)
2
= I ? A
1
– A
2
= 11
I
maximum 
= K(A
1
+ A
2
)
2
= 9 ? A
1
+ A
2
= 31
So, 
4
3
A A
A A
2 1
2 1
?
?
?
? A
1
/A
2
= 2/1
So, the ratio amplitudes is 2. ?
26. The path difference of the two sound waves is given by
?L = 6.4 – 6.0 = 0.4 m
The wavelength of either wave = ? = 
?
?
?
320 V
(m/s)
For destructive interference ?L = 
2
) 1 n 2 ( ? ?
where n is an integers.
or 0.4 m = 
?
?
? 320
2
1 n 2
? ? = n = Hz
2
1 n 2
800
4 . 0
320 ?
? = (2n + 1) 400 Hz
Thus the frequency within the specified range which cause destructive interference are 1200 Hz, 
2000 Hz, 2800 Hz, 3600 Hz and 4400 Hz. ?
Chapter 16
16.5
27. According to the given data
V = 336 m/s, 
?/4 = distance between maximum and minimum intensity
= (20 cm) ? ? = 80 cm
? n = frequency = 
2
10 80
336 V
?
?
?
?
= 420 Hz.
28. Here given ? = d/2
Initial path difference is given by = d d 2
2
d
2
2
2
? ? ?
?
?
?
?
?
If it is now shifted a distance x then path difference will be
= ?
?
?
?
?
?
? ? ? ? ? ?
?
?
?
?
?
4
d
d 2
4
d
d ) x d 2 (
2
d
2
2
2
?
64
d 169
) x d 2 (
2
d
2
2
2
? ? ? ?
?
?
?
?
?
?
2
d
64
153
? ? ? x d 2 1.54 d ? x = 1.54 d – 1.414 d = 0.13 d.
29. As shown in the figure the path differences 2.4 = ?x = 2 . 3 ) 4 . 2 ( ) 2 . 3 (
2 2
? ?
Again, the wavelength of the either sound waves = 
?
320
We know, destructive interference will be occur
If ?x = 
2
) 1 n 2 ( ? ?
?
?
?
? ? ?
320
2
) 1 n 2 (
) 2 . 3 ( ) 4 . 2 ( ) 2 . 3 (
2 2
Solving we get
? V = ) 1 n 2 ( 200
2
400 ) 1 n 2 (
? ?
?
where n = 1, 2, 3, …… 49. (audible region) ?
30. According to the data
? = 20 cm, S
1
S
2
= 20 cm, BD = 20 cm
Let the detector is shifted to left for a distance x for hearing the 
minimum sound.
So path difference AI = BC – AB 
=
2 2 2 2
) x 10 ( ) 20 ( ) x 10 ( ) 20 ( ? ? ? ? ?
So the minimum distances hearing for minimum 
= 
2
20
2 2
) 1 n 2 (
?
?
?
? ?
= 10 cm
??
2 2 2 2
) x 10 ( ) 20 ( ) x 10 ( ) 20 ( ? ? ? ? ? = 10 solving we get x = 12.0 cm. ?
31.
Given, F = 600 Hz, and v = 330 m/s ? ? = v/f = 330/600 = 0.55 mm
S
D
=x/4
20cm
2 d
D
S
2 2
d 2 ) 2 / d ( ?
x
d
2 2
) 4 . 2 ( ) 2 . 3 ( ?
A
A A
B
A
x
C
20cm
20cm
X
S
? ? P
R
1m
Q
y
S 1
S 2
O
1m
D
S 1
S 2
O ? ?
P
Q
X
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