8. STATIONARY WAVES
1. Given :
Equation of stationary wave,
y = 10 sin 2ππππ (100 t – 0.02 x)
+ 10 sin 2ππππ (100 t + 0.02 x)
To Find :
Loop length = ?
Frequency = f = ?
Maximum-Amplitude = A = ?
Solution :
If two waves y1 and y
2 of same frequency (n)
and same amplitude (a) are approching towards
each other then the Resultant wave equation is,
Y = y1 + y
2
y1
= a sin 2ππππ (nt – x )
λ λ λ λ
y2
= a sin 2ππππ (nt + x )
λ λ λ λ
∴∴∴∴ y = a sin 2ππππ (nt – x ) λ λ λ λ
+ a sin 2ππππ (nt + x ) λ λ λ λ
...(i)
∴∴∴∴ y = [ 2 a cos (2ππππx )] sin (2.ππππ.n.t.) λ λ λ λ
...(ii)
∴∴∴∴ y = A sin (2 ππππ.n.t)
Where A = Resultant Amplitude
A = 2 a cos 2ππππx
...(iii) λ λ λ λ
y = 10 sin 2ππππ (100 t – 0.02 x)
+ 10 sin 2ππππ (100 t + 0.02 x)
We can modify above equation,
y = 10 sin 2ππππ 100 – x
( 100) 2
+ 10 sin 2ππππ 100 + x
( 100) 2
Comparing with equation (i), We get
∴∴∴∴ λλλλ =100
= 50 units 2
n = 100 units
a = 10 units
Its Resultant equation will be,
y = [ 2 ×××× 10 cos (2ππππx) ] sin (2ππππ100.t)
50
...from equation (ii)
i) loop length :
loop length =λλλλ
=50
2 2
= 25 units
ii) Frquency :
n = 100 units
iii) Maximum Amplitude :
Resultant Amplitude, A= 2 ×××× 10 cos (2ππππx) 50
...From (iii)
For maximum Amplitude, cos (2ππππx) = 1 50
∴∴∴∴ Maximum amplitude
= 2 ×××× 10 (1) = 20 units
Stationary Waves
HOMEWORK SOLUTIONS
2. Given :
y = 0.01 cos (4ππππx) sin (200 ππππt)To Find :
amplitude (a) = ?
Velocity (v) = ?
Solution :
Given equaiton compare with standard
equation
y = 2a cos (2ππππx) sin (2ππππnt) λ λ λ λ
∴∴∴∴ 2a = 0.01
a = 0.005 m
2ππππx= 4ππππx
λ λ λ λ∴∴∴∴ λλλλ = 0.5 m
∴ ∴ ∴ ∴ 2ππππn = 200ππππn = 100 Hz
∴∴∴∴ V = n.λλλλ∴∴∴∴ V = 100 × × × × 0.5 m = 50 m/s
3. Given :
l = 1.6 m
To Find :
Distance= ?
Solution :
For 1st overtone, P =2,
λλλλ =2l
P
=2 ×××× 1.6
2
∴∴∴∴ λλλλ = 1.6 m.
(Distance between node ) =λλλλ
and very next antinode 4
=1.6
4
= 0.4 m
4. Given :
Mass of wire (M) = 0.5 gm
= 0.5 x 10-3 kg
Length of wire (L) = 0.5 m
Tension in wire (T) = 2 kg.wt
= 19.6 N
To Find :
Fundamental frequency (n) = ?
Formula :
n =1 T
2L √ µSolution :
Mass per unit length of wire (m)
m =M
= 0.5
x 10-3 = 10-3 kg/m L 0.5
∴∴∴∴ n =1 T
2L √ µ∴∴∴∴ n =
1 19.6
2(0.5) √ 10-3∴∴∴∴ n = √√√√19600
∴∴∴∴ n = 140 Hz
5. Given :
Length of the wire (L) = 60 cm
= 0.6m
Fundamental frequency (n) = 120 Hz
Tension (T) = 3kg.wt.
= 3 x 9.8 N
To Find :
Linear density (m) = ?
Formula :
n = 1 T
2L √m
Solution :
n = 1 T
2L √ m
∴∴∴∴ 120 = 1 3 x 9.8
2(0.6) √ m
MAHESH TUTORIALS SCIENCE .. 93
Stationary Waves
.. 94 MAHESH TUTORIALS SCIENCE
Stationary Waves
∴∴∴∴ 240 x 0.6 = 3 x 9.8√ m
∴∴∴∴ 144 = 3 x 9.8√ m
∴∴∴∴ m = 3 x 9.8
144 x 144
∴∴∴∴ = 4.9
48 x 72
∴∴∴∴ m = 1.42 x 10–3 kg/m
6. Given :
For two wires of same material &
cross-section
Tension in 1st wire (T1) = 8 kg.wt.
Tension in 2nd wire (T2) = 2 kg.wt.
Length of 1st wire (L1) = 80 cm.
= 0.8 m.
To Find :
Length of 2nd wire = ?
Formula :
n = 1 T
2L √ m
Solution :
For 1st wire.
n1
= 1 T
1
2L1 √m
For 2nd wire.
n2
= 1 T
2
2L2 √ m
∴∴∴∴n1 =
L2
T1
n2
L1 √ T
2
But, n1
= n2
∴∴∴∴L1 =
T1
L2
√T2
∴∴∴∴80
= 8 x g
L2
√2 x g
∴∴∴∴80
= 2L2
∴∴∴∴ L2
=80
= 40 cm 2
7. Given :
For 1st string :
Tension (T1) = 196 N
Length (L1) = 1m
Density (ρρρρ1) = 8 x 103 kg/m3
For 2nd string :
Tension (T2) = 49 N
Length (L2) = 1m.
Density (ρρρρ2) = 2 x 103 kg/m3
To Find :
The ratio of their frequencies (n1 : n
2) = ?
Formula :
m=
1 T
2L √ m
Solution :
Linear density = Density x Area
= ρρρρ x ππππr2
Since both wires have same diameter
∴∴∴∴ radii are also same.
∴∴∴∴ Linear density of 1st wire
(m1) = 8 x 103 x ππππr
12
Linear density of 2nd wire
(m2) = 2 x 103 x ππππr
22
∴∴∴∴m
1 = 4 { ... r
1 = r
2}
m2
Now,
n1
= 1 T
1
2L1 √ m
1
MAHESH TUTORIALS SCIENCE .. 95
Stationary Waves
n2
= 1 T
2
2L2 √m
2
∴∴∴∴n1 =
L2
T1 m
2
n2
L1 √ T
2 m
1
∴∴∴∴n1 =
1 196 x 1
n2
1 √ 49 4
∴∴∴∴n1 =
14 x 1
n2
7 2
∴∴∴∴ n1/n
2= 1/1
8. Given :
Area of cross section (A) = 0.2 mm2
= 0.2 x 10-6m2
Density of wire (ρρρρ) = 8000 kg/m3
Tension (T) = 5 kg wt
= 5 x 9.8 N
To Find :
Velocity of the transverse wave (v)
Formula :
v = √ T
m
m = ρρρρ x ππππr2
Solution :
The mass per unit length of wire is given
by
m = ρρρρ x A= 8000 x 0.2 x 10-6
m = 16 x 10-4 kg/m
Now the velocity of transverse wave is
given by
v = √ T
m
= √ 5 x 9.8
16 x 10-4
= √ 49 16 x 10-4
=7 x 102
4
= 1.75 x 102
v = 175 m/s
9. Given :
For 1st fork :
Frequency is n1
The length of the wire with which it is
in unison (l1) = 90 cm.
For 2nd fork :
Frequency is n2
The length of the wire with which it is
in unison (l2) = 91 cm.
|n1 – n
2| = 5
To Find :
The frequencies n1 & n
2 = ?
Formula :
For same wire under constant tension.
n1
∝∝∝∝1
(for 1st wire)l1
n2
∝∝∝∝1
(for 2nd wire)l2
Solution :
n1 =
l2 =
91
n2
l1
90
n1
> n2
∴∴∴∴ n1 – n
2= 5
∴∴∴∴91
n2 – n
2= 5
90
∴∴∴∴ n2
= 450 Hz
∴∴∴∴ n1
= 455 Hz
10. Given :
Initial tension of wire (T1) = 100 N
New tension of wire (T2) = 102 N
Initial frequency of wire = n1
New frequency = n2
|n1 - n
2| = 3
To Find :
n1
= ?
.. 96 MAHESH TUTORIALS SCIENCE
Stationary Waves
11. Given :
Length of wire (l1) = 60 cm
New length of wire (l2) = 55 cm
|n1 - n
2| = 10 beats/s
To Find :
The frequency of the fork (n1) = ?
Formula :
n ∝∝∝∝1
l
Solution :
n ∝∝∝∝1
l
n1/n
2 =l2/l
1
l2 =
55
l1
60
∴∴∴∴n1 =
55
n2
60
∴∴∴∴ n2
> n1
∴∴∴∴ n2 - n
1= 10
∴∴∴∴60
n1 - n
1= 10
55
∴∴∴∴ 5n1
= 550
∴∴∴∴ n1
= 110 Hz
12. No. of beats in 5 sec = 12
No. of beats in 1 sec =12
5
= 2.4
l1
= 84 cm
l2
= 85 cm
n =v ...(i)
2l1
n ± 2.4 = v
2l2
... l2 > l
1 , we use n – 2.4 =
v ...(ii)
2l2
n=
l2 =
85
n – 2.4 l1
84
...(From (i) and (ii))
n = 204 Hz
n =v
2l1
v = 204 ×××× 2 ×××× 0.84
= 342.72 m/s
Formula :
n =1 T
2l √m
Solution :
For a given wire
n ∝∝∝∝ √√√√T
∴∴∴∴n1 =
T1
n2
√T2
∴∴∴∴n2 =
102 =
√√√√102n1
√100 10∴∴∴∴
n2 =
10.1
n1
10
∴∴∴∴ n2
> n1
∴∴∴∴ n2 - n
1= 3
∴∴∴∴10.1n
1 - n
1= 3
10
∴∴∴∴ 0.1 n1
= 30
∴∴∴∴ n1
= 300 Hz
13. Given :
l1
= 32 cm
l2
= 32.4 cm
No. of beats per second = 5
To Find :
Frequency of tuning fork = N = ?
Solution :
Let n1 be frequency of air column having length
32 cm and n2 be frequency of air column having
length 32.4 cm
∴∴∴∴ N = n1 ∝ ∝ ∝ ∝
1
l1
Similary, n2 ∝ ∝ ∝ ∝
1
l2
MAHESH TUTORIALS SCIENCE .. 97
Stationary Waves
15. Given :
Initial length of wire (l1) = 34cm
Final length of wire (l2) = 35 cm
Let n1 be the frequency of the wire
when it is of length l1 & n
2 be the
frequency when it is of length l2.
Let f be the frequency of the fork.
∴∴∴∴ |n - n1| = 5 &
|n - n2| = 5
To Find :
The frequency of the fork.
Solution :
For a given wire.
n αααα1
l
∴∴∴∴n1 =
l2
n2
l1
∴∴∴∴ n1
= 35 ∴∴∴∴ n1 > n
2
Also,
n1 - n = 5
& n - n2
= 5
∴∴∴∴ n1 - n
2= 10
35 n
2 - n
2= 10
34
∴∴∴∴ n2
= 340 Hz
n = 340 + 5
∴∴∴∴ n = 345 Hz
∴∴∴∴n1 =
l2
n2
l1
∴∴∴∴n1 =
32.4 = 1.0125
n2
32
∴∴∴∴ n1
= 1.0125 n2
As l1
< l2
∴∴∴∴ n1
> n2
∴∴∴∴ n1 – n
2= 5
∴∴∴∴ 1.0125 n2 – n
2= 5
∴∴∴∴ 0.0125 n2
= 5
∴∴∴∴ n2
= 5
=400 Hz
0.0125
∴∴∴∴ n1
= n2 + 5 = 400 + 5
= 405 Hz
∴∴∴∴ Frequency of tuning = N
= n1
= 405 Hz
14. Given :
T = 1000 g. wt
= 1000 ×××× 10–3 kg wt
= 1 ×××× 9.8 N
= 9.8 N
V = 68 m/s
ρρρρ = 7900 kg/m3
To Find :
A = ?
Formula :
V = T√m
Calculation :
Since mass of the wire,
M = Vρρρρ = A.l.ρρρρ
Also, m =M
=A.l.ρρρρ
l l
∴∴∴∴ m = A.ρρρρ.
From Formula
V = T√ A.ρρρρ
∴∴∴∴ V2 = T
A.ρρρρ
∴∴∴∴ A = T
= 9.8
V2.ρρρρ (68)2 ×××× 7900
∴∴∴∴ A = 2.683 ×××× 10–7 m2.
= 0.2683 mm2
.. 98 MAHESH TUTORIALS SCIENCE
Stationary Waves
16. Given :
Initial Tension (T1) = 12.1 N
Final Tension (T2) = 10 N
|n - n1| = 5
& |n - n2| = 5
To Find :
Frequency (f) of fork
Solution :
n αααα √√√√T
∴∴∴∴n1 =
T1
n2
√ T2
n1 = √√√√1.21
n2
n1 = 1.1
n2
n1
> n2
n1 - n = 5
n - n2
= 5
n1 - n
2= 10
∴∴∴∴ 1.1 n2 - n
2= 10
∴∴∴∴ n2
= 100 Hz
But n = n2 + 5
n = 105 Hz
=60
x 4
x 9
x 159 3 √16
=60
x4 x
3
59 3 4
n1 =
60
n2
59
∴∴∴∴ n1
> n2
n1 – n
2= 4
∴∴∴∴60
n1 – n
2= 4
59
∴∴∴∴ n2
= 236 Hz
∴∴∴∴ n1
= 240 Hz
17. Given :
No. of beats per sec = 4
Ratio of the radii = 3/4
Ratio of the tensions = 9/16
Ratio of the lengths = 59/60
To Find :
The frequencies of the two wires.
Formula :
n = 1 T
2lr √ πρπρπρπρSolution :
For 1st wire,
n1
= 1
x T
1
2l1r1
√ πρπρπρπρ1
For 2nd wire,
n2
= 1
x T
2
2l2r2
√ πρπρπρπρ2
∴∴∴∴n1 =
l2 x
r2
T1 x ρ ρ ρ ρ
2
n2
l1
r1 √ T
2 ρ ρ ρ ρ
1
18. Data :
L1= 100 cm,
L2= 90 cm,
ρρρρw= 1 gm/cc,
To Find :
ρρρρ = ?
Formula :
T
2= Constant
Solution :
By law of tension, T
L = constant
L1 > L
2
T1 = ρρρρ V
g and T
w < V
g
T1 > T
2, But T
2 = T
1 – T
w
T2 = ρρρρV
g – V
g = (ρρρρ – 1) V
g
T1 =
T2
L12 L
22
T2 =
( L2 )T1
L1
(ρρρρ –1) Vg= ( 90 ) = ( 81 )
ρρρρ Vg 100 100
ρρρρ – 1 =81 ρρρρ
= 0.80 ρρρρ100
0.19 ρρρρ = 1
ρρρρ = 5.263 g/cm3
MAHESH TUTORIALS SCIENCE .. 99
Stationary Waves
19. Given :
Initial no. of loops formed (P1) = 4
Corresponding tension (T1) =8 x 980 dyne
No. of loops formed (P2) = 8
To Find :
The tension in the string (T2) = ?
Formula :
TP2 = constant
Solution :
TP2 = constant
∴∴∴∴ T1P12 = T
2P22
∴∴∴∴ (8 x 980) (4)2 = (T2) (8)2
T2
=(8 x 980) (42)
(82)
T2
= 2 x 980 dynes
= 2 gm
20. Given :
No. of loops (P) = 6
Length (l) = 1.5
Tension in string (T) = 10gm.wt.
T = 10–2 x 9.8 N
Mass of string (m) = 9 x 10–5kg
To Find :
Frequency of the tuning fork (n) = ?
Formula :
n = P T
2l √ m
Solution :
Linear density of the wire is given by
m = M
=9 x 10–5
l 1.5
m = 6 x 10–5 kg/m
n = P
x T
2l √ m
∴∴∴∴ n = 6 10–2 x 9.8
2 (1.5) √ 6 x 10–5
∴∴∴∴ n=
2 98
x 10 √ 6
∴∴∴∴ n = 2√ √ √ √ 49 x 2
x 10 3 x 2
∴∴∴∴ n =2 x 7 x 10
√√√√3
∴∴∴∴ n = 140/1.732
∴∴∴∴ n = 80.83 Hz.
21.Given :
Let mo be mass of pan
Case I :
Mass in pan (m1) = 6 gm
Tension (T1) = (m
0 + m
1) gm wt
= (m0 + 6) gm wt
No. of loops (P1) = 5
Case II :
Mass in pan (m2) = 10.5 gm
Tension (T2) = (m
0 + m
2) gm wt
= (m0 +10.5) gm wt
No. of loops (P2) = 4
Case III
Tension in string (T3) = m
0 gm.wt.
To Find :
No. of loops formed (P3) = ?
Formula :
TP2 = constant
Solution :
T1P12 = T
2P22
∴∴∴∴ (m0 + 6) (5)2 = (m
0 + 10.5) (4)2
∴∴∴∴ 25 m0 + 150 = 16 m
0 + 168
∴∴∴∴ 9m0
= 18
∴∴∴∴ m0
= 2gm
Now,
T3P32 = T
1P12
∴∴∴∴ P32 =
(m0 + 6) (5)2
= (2 + 6) (5)2
m0 2
.. 100 MAHESH TUTORIALS SCIENCE
Stationary Waves
22. Given :
P1
= 5 loops
P2
= 4 loops
T2
= (T1 + 0.018) kgwt
To Find :
T2
= ?
Solution :
T1P12 = T
2P22 ;
T1 (5)2 = (T
1 + 0.018) (4)2
25 T1
= 16 (T1 + 0.018)
25T1
= 16T1 + (16 ×××× 0.018)
9T1
= (16 ×××× 0.018)
∴∴∴∴ T1
=16 ×××× 0.018 9
= 16 ×××× 0.002T1
= 0.032 kg.wt.
=8 x 25
2
P32 = 100
∴∴∴∴ P3
= 10
23. Given :
Frequency of sound wave (n) = 1000 Hz
Speed of the wave (v) = 340 m/s
To Find : The distance at which the next
successive node will be formed.
Formula :
n = v
4L
Solution :
Let n = 1000 Hz. be the fundamen-
tal mode of vibration of tube
∴∴∴∴ n = v/4L
∴∴∴∴ L = v/4n
= 340
4 x 1000
L = 0.085 m = 8.5 cm
Now, the distance betwen two nodes is λ λ λ λ/2where λλλλ is the wavelength of the wave.
But, for a closed pipe in its fundamental
mode.
L = λλλλ/4
∴∴∴∴ 2L = λλλλ/2 = 2 x 8.5
∴∴∴∴ λλλλ/2 = 17 cm.
24. Given :
Velocity of sound (v)= 333 m/s
Length of air column (L) = 33.3 cm
To Find :
The frequency of 5th overtone when pipe
i) closed at one end.
ii) open at both ends.
Formula :
For a closed pipe
n = v ×××× (2p + 1)4L
...For pth overtone
For an open pipe
n = v × × × × (p + 1)2L
...For pth overtone
Solution :
i) For a closed pipe :
The frequency of the 5th overtone is
given by
n =11
xv
4 L
∴∴∴∴ n =11
x 333
4 0.333
∴∴∴∴ n = 11 x 250
n = 2750 Hz
ii) For an open pipe :
The frequency of the 5th overtone is
n =6
xv
2 L
MAHESH TUTORIALS SCIENCE .. 101
Stationary Waves
25. Given :
Length of 1st air column (L1) = 16 cm
Length of 2nd air column(L2) = 24 cm
Smaller frequency (n2) = 320 Hz
To Find :
Frequency of other fork (n1) = ?
Formula :
n ∝∝∝∝1
L
Solution :
∴∴∴∴ n ∝∝∝∝1
L
∴∴∴∴ The frequency will be smaller for the
air column of length 24 cm.
n1/n
2= L
2/L
1
∴∴∴∴n1/320
=24/16
∴∴∴∴ n1
=24
x 32016
∴∴∴∴ n1
= 480 Hz
27. Given :
Let n1 & n
2 be the frequencies of the
two pipes closed at one end |n1-n
2|= 3
Length of 1st pipe (L1) = 51 cm
Length of 2nd pipe (L2) = 52 cm
To Find :
Velocity of sound in air = ?
Formula :
n = v
4L
Solution :
For a closed pipe
n ∝∝∝∝1
L
∴∴∴∴n1 =
L2 =
52
n2
L1
51
∴∴∴∴ n1
> f2
∴∴∴∴ n1 - n
2= 3
∴∴∴∴52n
2 - n
2= 3
51
∴∴∴∴ n2
= 153 Hz
Now n = v
4L
∴∴∴∴ v = 4 (L)n
∴∴∴∴ v = 4 (52) 153
= 31824 cm/s
∴∴∴∴ v = 318.24 m/s
28. Given :
The frequency of the 3rd overtone of a
closed pipe is in unison with the
frequency of the 5th overtone of an
open pipe.
To Find :
The ratio of the lengths of the two pipes
(L1/L
2) = ?
∴∴∴∴ n = 3 x 1000
∴∴∴∴ n = 3000 Hz.
26. Given :
Length of air column for
fundamental mode (l1) = 15 cm.
Length of air column for
1st overtone (l2) = 48cm
To Find :
End correction (e) = ?
Formula :
l2 + e
= 3l1 + e
Solution :
∴∴∴∴48 + e
= 315 + e
∴∴∴∴ 48 + e = 45 + 3e
∴∴∴∴ 2e = 3
∴∴∴∴ e = 1.5 cm
.. 102 MAHESH TUTORIALS SCIENCE
Stationary Waves
30. Given :
Frequency of tuning fork (n) = 300Hz.
Length of air column (l1) = 25 cm
Length of air column in next resonance
(l2) = 80 cm
To Find :
i) The velocity of sound in air (v) = ?
ii) The internal diameter (d) = ?
Formula :
i) n = v
4L
ii) l
2 + e
= 3 l
1 + e
Solution :
i) l
2 + e
= 3 l
1 + e
∴∴∴∴ 80 + e
= 3 25 + e
∴∴∴∴ 2e = 5
e = 2.5 cm
e = 0.3 d
∴∴∴∴ 2.5 = 0.3 d
∴∴∴∴ d =2.5
=25
0.3 3
d = 8.333 cm
ii) n = v
4 L
∴∴∴∴ n = v
4 (l1+e)
∴∴∴∴ v = 4n ( l1 + e)
∴∴∴∴ v = 4 x 300 x (25+2.5) x 10–2
∴∴∴∴ v = 1200 x (27.5) x 10–2
∴∴∴∴ v = 12 x 27.5 x 102 x 10–2
∴∴∴∴ v = 330 m/s
Solution :
Let n1 be the fundamental frequency of the
closed pipe.
∴∴∴∴ n1
= v
4 L1
∴∴∴∴ The frequency of 3rd overtone
(n3(c)) =
7 v ...(i)
4 L1
Let n2 be the fundamental frequency of
the open pipe
n2
= v
2 L2
∴∴∴∴ The frequency of the 5th overtone of
this pipe is
From (i) and (ii)
∴∴∴∴ n5(o)
= n3(c)
n5(o)
=
6v ...(ii)
2L2
∴∴∴∴7 v
=6 v
4 L1
2 L2
∴∴∴∴L1 =
7
L2
12
29. Given :
Initial lengths (l1) = 12.5 cm
Final length (l2) = 40 cm
To Find :
End correction (e) = ?
Formula :
l2 + e
= 3l1 + e
Solution :
l2 + e
= 3l1 + e
40 + e= 3
12.5 + e
∴∴∴∴ 3e + 37.5 = 40 + e
∴∴∴∴ 2e = 2.5
∴∴∴∴ e = 1.25 cm
MAHESH TUTORIALS SCIENCE .. 103
Stationary Waves
31. Given :
Frequency of the tuning fork = 550 Hz.
End correction (e) = 0.005 m
Length of the tube (L) = 31 x 10–2m
To Find :
Velocity of the sound in air (V).
Solution :
n = V
2(L + 2e)
∴∴∴∴ V = 2n(L + 2e)
= 2(550) (31 x 10–2 + 2(0.5) x 10–2)
= 1100 (32) x 10–2
V = 352 m/s.