8. STATIONARY WAVES
A sonometer wire of length 0.5 m is Formula :
stretched by a weight of 5 kg. The
fundamental frequency of vibration is
n
=
100 Hz. Determine the linear density of
Solution :
material of wire.
Given :
l
=
0.5 m
n
=
T
=
Mg = 5 × 9.8 N
n
=
100 Hz
1
To Find :
Since, n1 ∝
l1
m
=
?
Formula :
n1
∴
=
n2
1 T
n
=
∴
n1l1
=
2l m
∴
100
n
=
1
Solution :
1.
n
∴
=
1 T
2l m
∴
1
5 × 9.8
100 =
=
2 × 0.5
m
Squaring both sides, we get
(100)2
=
∴
m
=
∴
m
=
49
m
49
m
49
( 100 )2
0.0049 kg/m
0.0049 kg/m
=
1 T
2l m
1
and n2 ∝
l2
l2
l1
n2l2
90 n2
100
n
90 1
10
n
9 1
then n2 > n1
=
8
=
∵
∴
∴
=
n2
1 T
2l m
l1 > l2
n2 –n1
10
n – n1 =
9 1
10
n1  – 1 
 9

∴
n1
∴
n1
8
=
8
=
=
9×8 =
n
=
72 Hz
72 Hz
2.
A sonometer wire 100 cm long produces
a resonance with a tuning fork. When 3.
Two wires of the same material and
its length is decreased by 10 cm, 8 beats
having the same radius have their
per second are heard. Find the frequency
fundamental frequencies in the ratio
of tuning fork.
1 : 2, and tension in the ratio 1 : 8
Given :
compare ratio of their lengths.
l1
=
100 cm
Given :
l2
=
100 – 10
=
90 cm
n1
1
N
=
n1 ~ n2
=
8s–1
=
n2
2
To Find :
n1
=
?
T1
1
=
T2
8
Stationary Waves
MAHESH TUTORIALS SCIENCE
.. 40
To Find :
l1
l2
4.
=
?
=
1 T
2l m
=
1 T
2l m
A wire, 1 m long and weighing 2 g, will
be in resonance with a frequency of 300
Hz. Find tension on stretching the wire.
Given :
l
=
1m
M
=
2 gm
= 2 × 10–3 kg
n
=
300 Hz
To Find :
T
=
?
Formula :
Formula :
n
Solution :
n
∴
∴
n1
l1
=
=
1
2 l1
1
2n 1
n
T1
m
T1
m
=
Solution :
M
l
... (i)
=
=
For second case,
n2
∴
l2
=
=
1
2 l2
1
2n 2
T2
m
T2
m
∴
l1
l2
=
2n 2
2n 1
=
n2
n1
=
2 1
 
 1 8
=
0.707 : 1
T1 m
×
m T2
T1
T2
Now, n
=
1 T
2l m
∴
300
=
1
2×l
600
=
∴
∴
∴
T
2 × 10–3
(600)2 =
T
T
=
=
T
2 × 10–3
T
2 × 10–3
36 × 104 × 2 × 10–3
720 N
5.
A stretched sonometer wire is in unison
with a tuning fork, when the length is
increased by 4 %, the number of beats
heard per second is 6. find the frequency
of the fork.
Given :
l2
=
1.04 l1
∴
l2
l1
N
To Find :
n1
Stationary Waves
2 × 10 –3
1
2 × 10–3 kg/m
... (ii)
[m is constant because both wires are
made of same material]
Divide (i) by (ii)
l1
l2
1 T
2l m
=
1.04,
=
6s–1
=
?
MAHESH TUTORIALS SCIENCE
Formula :
n
=
1 T
2l m
=
1 T
2l m
Solution :
Since mass of the wire,
M
=
Vρ =
Alρ
Also,
Solution :
n
∴
Since l2
>
l1
then n1
>
n2
∴ n1 – n2 =
6 Hz
∴
n2
=
n1 – 6
For two wires of same material
n1l1 =
n2l2
But , n2
=
n1 – 6
=
1.04 l1
and l2
∴
n1l1 =
(n1 – 6) (1.04 l1)
n1
=
(n1 – 6) (1.04)
∴
∴ 1.04 n1 – n1 =
6.24
∴
0.04 n1 =
6.24
∴
n1
=
6.24
0.04
∴
n1
=
156 Hz
6.
.. 41
=
T
m
=
m
=
M
l
Aρ
=
Now, v
=
T
m
∴
v
=
T
Aρ
∴
v2
=
∴
A
=
A
=
A
=
∴
Alρ
l
T
Aρ
T
v2 ρ
9.8
( 68 )
2
× 7900
2.683 × 10–7 m2
7.
A transverse wave is produced on a
streched string 0.7 m long and fixed at
its ends. Find speed of transverse wave,
when it vibrates, emitting the second
overtone of frequency 300 Hz.
Given :
l
=
0.7 m
n
=
300 Hz
To Find :
v
=
?
Formula :
v
=
nλ
λ
Solution :
v
=
nλ
λ
In 2nd overtone, 3 loops are formed
The speed of a transverse wave along a
uniform metal wire, when it is under a
tension of 1000 g wt. is 68 m/s. If the
density of metal is 7900 kg/m3. Find the
area of cross section of the wire.
Given :
T
=
1000 g wt.
=
1000 × 10–3 kg wt.
=
1 × 9.8 N = 9.8 N
v
=
68 m/s
=
7900 kg/m3
ρ
To Find :
A
=
?
∴
Formula :
v
m
∴
l
=
3
λ
2
λ
=
2l
3
Stationary Waves
MAHESH TUTORIALS SCIENCE
.. 42
Now, v
=
nλ
v
=
 2l 
n 
3
v
=
v
=
∴
9.
2

300  × 0.7 
3


140 m/s
8.
A uniform wire under tension, is fixed
at its ends. If the ratio of tension in the
wire to the square of its length is
360 dyne/cm2 and fundamental frequency
of vibration of wire is 300 Hz. Find its
linear density.
Given :
T
=
l2
n
=
To Find :
m
=
Formula :
360 dyne/cm2
300 Hz
n1
n2
?
=
ρ
ρ–1
=
ρ
ρ–1
250
=
240
ρ
ρ–1
25
24
ρ
ρ–1
Solution :
=
1 T
2l m
n
=
1 T
2l m
n2
=
n
A wire is in unison with a fork of
frequency 250 Hz, when streched by a
weight hanging vertically. On immersing
the weight in water, the wire produces
ten beats per second with the same fork.
Calculate density of material of weight,
Given :
When wire is stretched by a weight
hanging vertically, n1 = 250 Hz,
Frequency of wire when the weight is
immersed in water producing 10 beats
per second = n2
n2 = n1 – 10 = 250 – 10 = 240 Hz
∴
ρ w = 1 g/cc
To Find :
=
?
ρ
Formula :
n1
n2
Solution :
∴
∴
m
=
m
=
m
=
m
m
m
=
=
=
Stationary Waves
1
4l
.
2
T
m
∴
Squaring both sides,
T
.
2  2 
4n  l 
1
1
4 × ( 300 )
2
90
90000
10–3
10–3 g/cm
10–4 kg/m
=
×
360
1
ρ
625
=
ρ–1
576
625 (ρ – 1)
=
∴
∴ 625 ρ – 576 ρ
∴
49 ρ
576 ρ
625
625
=
=
∴
ρ
=
625
49
∴
ρ
=
12.76 g/cm3
MAHESH TUTORIALS SCIENCE
10.
.. 43
Two simple harmonic progressive waves
∴
are represented by
x 

y1 = 2 sin 2π  100t –  cm and
60


 2 πx 
4 cos 
 sin 2π (100 t)
 60 
...(i)
Comparing above equation with,
y
=
R sin (2πt)
We get,
y
=
x 

y2 = 2 sin 2π  100t +  cm.
60 

 2 πx 
=
R
4 cos 

The waves combine to from a stationary
 60 
wave.
Find :
 2 πR 
But, R
=
2π cos 

i)
amplitude at antinode
 λ 
ii)
distance between adjacent node
∴
=
60 cm
λ
and antinode
Amplitude at antinode is maximum value
iii) loop length
of R.
iv) wave velocity
 2 πx 
Given :
i.e, R is maximum when cos 
 = 1
λ


x 

y1
=
2 sin 2π  100t –  cm
∴
R
=
4
×
1
=
4
cm
60 

y2
To Find :
i)
ii)
iii)
iv)
Formula :
y
x 

2 sin 2π  100t +  cm
60


=
R
λ
4
l
v
where, R
=
?
=
?
=
=
?
?
=
R sin 2πnt
=
 2 πx 
2A cos 

 λ 
Solution :
i)
Resultantant equation of wave is given
by
y
=
y1 + y2
=
x 

2 sin 2π  100t – 
60 

x 

+ 2 sin 2π  100t + 
60 

∴
y
ii)
∴
∴
iii)
λ
=
60 cm
λ
60
=
=
15 cm
4
4
Distance between successive node and
antinode = 15 cm
l
=
=
length of loop
λ
=
2
30 cm
60
2
∴
l
=
iv)
∴
v
v
From
n
λ
v
v
v
v
=
wave velocity
=
nλ
equation (i), we get,
=
100 Hz
=
60 cm
=
nλ
=
100 × 60
=
6000 cm/s
=
60 m/s
∴
∴
=
nλ
 x 
= 2 × 2 sin 2π (100 t)cos 2π  
 60 
Stationary Waves
MAHESH TUTORIALS SCIENCE
.. 44
11.
The equation of a standing wave is
given by y = 0.02 cos (πx) sin (100 π t) m.
Find the amplitude of either wave
interfering, wavelength, time period,
frequency and wave velocity of
interfering waves.
Given :
y
=
0.02 cos (πx) sin (100 πt)
To Find :
A
=
?
λ
=
?
T
=
?
n
=
?
v
=
?
Formula :
y
=
R sin 2πnt
where,
R
=
Calculation :
y
=
2 A cos
2 πx
λ
12.
In Melde’s experiment, find weight
added in the pan when number of loops
on the string changes from 4 to 2. If
initial tension on the string is 1960
dyne and mass of the pan in one gram.
Given :
p1
=
4
p2
=
2
M0 =
1g
=
(M0 + M1)g
T1
=
1960 dyne
T2
=
(M0 + M2)g
To Find :
M2 =
?
Formula :
T1p12 =
T2p22
Solution :
T1p12 =
T2p22
0.02 cos (πx) sin (100 π t)
 2 πx 
0.02 cos 
 sin[2π (50)t]
 2 
Comparing the given equation with,
y
=
R sin 2πnt
where,
∴
y
R
We get,
A
=
=
=
λ
=
n
=
=
T
=
v
∴
T2
=
=
2 A cos
2 πx
λ
amplitudeof interfering waves
0.01 m
Wavelength of interfering waves
2m
Frequency of interfering waves
50 Hz
Time period interfering waves
=
1
n
=
=
v
Velocity of interfering waves
nλ
= 50 × 2
=
100 m/s
Stationary Waves
=
1
50
= 0.025
=
=
∴
T2
=
=
∴
∴
∴
∴
∴
T1p21
p22
( M 0 + M 1 ) g ( 4 )2
( 2 )2
16 ( M 0 + M 1 ) g
4
1960 × 16
4
4 × 1960
T2
=
But (M0 + M2)g
(M0 + M2)g
(1 + M2) 980
(M2 + 1)
M2 + 1
M2
M2
=
=
=
=
=
=
=
T2
4 × 1960
4 × 1960
4×2
8
8–1
7 g wt
MAHESH TUTORIALS SCIENCE
.. 45
13.
In Melde’s experiemnt, fork was
6 × 7 × 10
70
arranged in parallel position and 6 loops
N
=
=
7.2
1.2
were formed along a length of 7.2 m
∴
N
=
58.33
Hz
when stretched by a weight of 10 g. If
mass of the string is 14.4 × 10–2 g, find
14. Find the frequency of fifth overtone of
the freqeuncy of tuning fork.
an air column vibrating in a pipe closed
Given :
at one end, length of pipe is 42.10 cm
p
=
6
and speed of sound in air at room
L
=
7.2 m
temperature is 350 m/s.
M
=
10 g
[Inner diameter of pipe is 3.5 cm]
–3
=
10 × 10
Given
:
=
10–2 kg
l
=
42.10cm
M′ =
14.4 × 10–2 g
=
42.10 × 10–2 m
=
14.4 × 10–5 kg
=
0.4210 m
To Find :
v
=
350 m/s
N
=
?
d
=
3.5 cm = 3.5 × 10–2 m
Formula :
=
0.035 m
To Find :
P T
n
=
Frequecy of fifth overtone n5 = ?
2L m
Formula
:
Solution :
Fundamental frequency in air column
P T
closed at one end is given by
n
=
2L m
V
n
=
For parallel position, frequency of fork
4L
is given by
Solution :
N
=
2n
L
=
l + 0.3 d
=
0.4210 + 0.3 × 0.035
2.P T
∴
N
=
=
0.42 10 + 0.0105
2L m
∴
L
=
0.4315 m
P T
V
N
=
...(i)
Now, n
=
L m
4L
M′
350
Now, m
=
n
=
l
4 ( 0.4315 )
14.4 × 10 –5
7.2
∴
m
=
2 × 10–5 kg/m
Substituting the values in (i), we get,
m
=
N
=
6 10 –2 × 9.8
7.2 2 × 10–5
N
=
6
49 × 10 2
7.2
∴
∴
∴
∴
∴
350
4 × 0.4315
n
=
202.78 Hz
th
P overtone is given by np = (2p + 1)n
Frequencey of fifth overtone is,
n5
=
(2 × 5 + 1)n
n5
=
11 n
=
11 × 202.78
=
2230.59 Hz
n5
=
2230.59 Hz
Stationary Waves
n
=
MAHESH TUTORIALS SCIENCE
.. 46
15.
Two organ pipes, open at both ends, are
sounded together and 5 beats are heard
per second. The length of shorter pipe
is 0.25 m. find the length of the
otherpipe. (Given : velocity of sound in
air = 350 m/s, end orrection at one end =
0.015 m same for both pipes)
Given :
l1
=
0.25 m
v
=
350 m/s
n1 – n2
=
5
e
=
0.015 m same for both pipes.
To Find :
l2
=
?
Formula :
n
=
V
2L
Using (i), we get,
625 – n2 =
5
n2
=
620 Hz
∴
Also, n2
=
v
2L 2
∴
620
=
350
2L 2
∴
2L 2
=
350
620
∴
L2
=
350
2 × 620
L2
=
L2
l2
l2
l2
=
=
=
=
∵
∴
35
=
124
l2 + 2e
L2 – 2e
0.2823 – 0.03
0.2823 m
Solution :
Fundamental frequency of organ pipe
open at both ends
∴
0.2523 m
v
n
=
16. The fundamental frequency of a pipe
2L
closed at one end is unison with the
where
third overtone of an open pipe.
L
=
l+e
Calculate the ratio of their lengths of
< L2 then n1 > n2
air column.
1
Given :
n1 – n2 =
5
... (i)
no
=
nc
Since pipe is open at both ends hence end
where
correction is (2e)
no
=
frequency of third overtone
st
∴
correct length of 1 pipe
of open pipe
=
l1 + 2e
L1
nc
=
fundametal freqeuncy of
=
0.25 + 2 × 0.015
closed pipe
To
Find
:
L1
=
0.25 + 0.03
= 0.28 m
Lc
v
=
?
∴
n1
=
L
o
2L
S i n
c e
L
1
=
=
∴
n1
=
Formula :
350
2 × 0.28
350
=
0.56
625 Hz
n
625 Hz
V
2L
Solution :
3rd overtone of open pipe is given by
no
Stationary Waves
=
=
 v 
4

 2Lo 
MAHESH TUTORIALS SCIENCE
Fundamental frequency of closed pipe at ∴
one end is given by,
2e(n1 – n2)
=
∴
e
=
18.
nc
=
v
4L c
∵
nc
=
no
∴
v
4L c
=
 v 
4

 2Lo 
∴
Lo
Lc
=
8
∴
Lc
Lo
=
1
8
∴
Lc
Lo
=
1:8
17.
.. 47
ii)
∴
Solution :
Let,
l1 and l2
= Vibrating lengths of pipe
n1 and n2 = Resonating freqeuncy
v
=
Velocity of sound in air
e
=
End correction
∴
i)
For the first resonance
∴
ii)
∴
v
=
2n1 (l1 + 2e)
For the second resonance
n2
=
v
2 ( l2 + 2e )
v
=
2n2 (l2 + 2e)
iii)
From (i) and (ii) we get,
∴
∴
∴
n1(l1 + 2e)
n1l1 + 2e n1
2e n1 – 2e n2
n2(l2 + 2e)
n2l2 + 2e n2
n2l2 – n1l1
L1
=
n 2 l2 – n 1l1
2 (n1 – n2 )
=
l2 – 3l1
2
l+e
l1 + e =
λ
4
...(i)
3λ
...(ii)
4
Subtract equation (i) from equation (ii)
L2
=
l2 +e =
(l2 + e) – (l1 + e)
=
3λ
λ
–
4
4
(l2 – l1)
=
λ
2
∴
v
2 ( l1 + 2e )
=
=
=
e
Solution :
L
=
n 2 l2 – n 1l1
e = 2 n –n .
( 1 2)
=
l – n1l1
2 2
In a resonance tube experiment a tuning
fork resonates wth an air column 10 cm
long and again resonates when, it is 32.2
cm long. Calculate the wavelength of
wave and the end correction.
Given :
l1
=
10 cm
l2
=
32.2 m
To Find :
λ
=
?
e
=
?
Formula :
i)
L
=
l+e
Show that for a pipe open at both ends
the end correction is
n1
n
∴
λ
∴
λ
=
=
=
=
2(l2 – l1)
2(32.2 – 10.0)
2(22.2)
44.4 cm
Now, e
=
l2 – 3l1
2
∴
e
=
∴
e
=
∴
e
=
...(i)
...(ii)
( 32.2 ) – 3 ( 10 )
2
2.2
2
1.1 cm
Stationary Waves