Class 11-science H C VERMA Solutions Physics Chapter 15: Waves Motion and Waves on a String
Waves Motion and Waves on a String Exercise 323
Solution 1
Speed = 
X= 
5
X= 2m ( in negative direction)
Solution 2
a)
[y]=[A]=[
] ;
and
= 
= [
]
[a]= [
]
[T]= [ 
]
b)
c)
= constant
As time increases, 
must decrease
So, wave travels in negative direction
d) For maximum pulse location
y= A, i.e.
at
Solution 3
At,
Waves Motion and Waves on a String Exercise 324
Solution 4
At t=0
y= 
y(x=0)=a
y(x=
)=0
At t=1 sec
y= 
The wave is moving in positive direction
with wave speed of v so in 1 sec it moves by distance v in positive
direction.
At t=2 sec wave travels distance of 2v
Solution 5
Since wave is travelling in positive direction
Solution 6
a)
and
= [
] [Since Trignometric ratios are dimensionless]
[a]= [
]
b)wave travels in positive direction
Solution 7
In time 
wave travels
distance of 
Since wave is travelling in positive x direction.
Replace
by 
Solution 8
As time increases, decreases
So, wave travels in negative direction.
b) 
= 
= 10 m/s
ω= 2⊓f= 314
f=50 Hz.
c) maximum displacement = A= 0.1mm
Maximum speed = A
=(0.1 
) (314)
=3.14 × 
m/s
=3.14 cm/sec
Solution 9
20 = f 
f = 1000 Hz
ω = 2π×
Equation of wave travelling in positive direction
b)
y=0
Particle velocity
= ω 
= (2
x 
) 
It depends upon displacement not on fashion of wave equation
Solution 10
Comparing equation with 
a)
and 
T=0.02= 20ms
b)
Differentiate the displacement equation
=
C)
cos 
d)at 
cos 
=9.7 cm/s
At t=0.012
cos 
=18 cm/s
At t=0.013
cos 
=25 cm/s
Solution 11
= 5ms
T = 20ms = 
f= 50 Hz
Wavelength:
= 2cm
λ = 4cm
Wave speed:
= f × λ
= 50 × 4 ×
= 2 m/s
Solution 12
a) Amplitude = 
= 1mm
b) wavelength = 4 cm
c) wave matter K = 
= 
k= 1.57 ≅1.6 
d) frequency
ʋ = f × λ
20 = f × 4
f = 5 Hz
Solution 13
ʋ= f × λ
ʋ=
× λ
λ = 10 × 20 ×
λ= 0.2m = 20 cm
b) Displacement equation of particle be
At any instant 1.5 = A sin (t- Kx)
Phase difference for particle at a distance x= 10 cm
= 
× 10
=
The displacement is given by
y'= A sin (
)
Solution 14
ʋ= 
=
=
ʋ = 32 m/s
Solution 15
a) Velocity of the wave = 
=
/
= 20 m/s
Distance travelled by wave = 20cm + 20 cm= 40 cm
T=0.02 sec
In transverse wave, Phase difference of π occurs after reflection from denser medium.
Solution 16
When transverse wave travels from dense medium to a rarer medium no phase change occurs.
a) Distance to travel by wave = 20 + 20
= 40 cm
b) Phase change occurs at fixed end and not at free end.
Distance travelled to regain shape=
30 + 30 = 60 cm
c) v =
2 
= 
= T = 2 
N
Solution 17
= 
= 
=
= 
:
= 1 : 4
Waves Motion and Waves on a String Exercise 325
Solution 18
ʋ= 30 m/s
ʋ =
30 = 
T = 0.108 N
Solution 19
A= 1 cm;
µ = 0.1 kg/m;
T= 90 N
In one direction, displacement becomes
f = 
= 100 Hz
a) Speed of wave
ʋ
= 
= 30 m/s
wavelength:
λ = 0.3m = 30cm
b)
At
, displacement
is maximum wave equation.
Where x is in cm and t is in sec
c) 
=
=
= -5.4 m/s
a= 2km/
Solution 20
=
= 0.05 sec.
Solution 21
Tension in string AB 
Tension in string CD 
ʋ=
For wire AB
For wire CD
ʋ= 
Solution 22
ʋ=
=
ʋ= 100m/s
t = 
= 0.02sec
Solution 23
Tension in string
T= 48N
ʋ= = 
ʋ= 50 m/s
Solution 24
Diving (2) by (1)
=
a = 3.7 m/
Solution 25
Let mass per unit length be 
Mass of small section on 2dѲ; 
Now,
Solution 28
Since 
Now,
P=2
(100)
= 49m
Solution 30
a) 
= = 
ʋ=70 m/s
wavelength
ʋ = f × λ
70 = 440 × λ
λ =16cm
b) Maximum speed of the particle = A
Maximum acceleration of particle =A
= 3.8 Km/s2
c) 
p =0.67W
Solution 31
= 
=
=4
mm
Solution 32
Distance travelled by pulse in 4ms
= (50 
)(4
)
d = 2mm
* Distance travelled by pulse in 6ms
d = (50 )(6)
d = 3mm
* Distance travelled by pulse in 8ms
d = (50 )(8)
d= 4mm
* Distance travelled by a pulse in 12ms
d = (50 )(12)
d=6mm
Waves Motion and Waves on a String Exercise 326
Solution 33
a) 
=2
b) 
=
.(
)
=
.(4)
c)
)
when 
)
when 
)
Solution 34
Fundamental frequency
f= 
f=
=30 Hz
Solution 35
f= = 
100=
Solution 36
Wave speed
ʋ
Fundamental frequency 
= =
= 62.5 Hz
Fourth harmonic frequency = 4
=4 (62.5)
= 250 Hz
Wavelength of the fourth harmonic
100 = 250 × λ
λ = 0.4m = 40cm
Solution 37
Fundamental frequency
261.63=
T ≃1480N
Solution 38
Frequency in IInd harmonic = 
ʋ= 384 m/s
Solution 39
Wire is in IInd harmonic as two loops are formed
Frequency of IInd harmonic =
≃
.
Solution 40
If string forms 4 loops, it will be in 4th harmonic.
f=
128=
T = 164 N
Solution 41
a) Maximum fundamental frequency will be highest common factor of 240 hz and 320Hz
So, 80 Hz is the fundamental frequency.
b) Fundamental frequency = 
80=
Solution 42
If n loops are formed initially then (n+1) loops will be formed in next resonance frequency
So, length of string
0.4n=1.6
n=4
∴ Length of string = 4 × 2
= 8 cm
Solution 43
f = 660 Hz; ʋ = 220 m/s
ʋ = f × λ
220= 660 × λ
λ = 
m
a) 3 loops are formed, so length of wire = 
=
=
m
=50cm
b) Standing wave equation if Node is at x=0
=0.5cm sin( 
).× cos (
= 0.5cm sin ( 
)x cos (2
Solution 44
F
[As velocity remains constant for given medium]
here, 
Now,
=23.8cm
Similarly
and 
Solution 45
Let highest harmonic be 
So. 
= n
n = 70
Solution 46
a) Maximum fundamental frequency will be highest common factor of 90Hz, 150 Hz and 210 Hz.
f0= 30 Hz
b) and c)
when 
= 90 Hz
= 3 (3rd harmonic and 2nd overtone)
When 
= 150 Hz
= 5 (5th harmonic and 4th overtone)
When 
= 210 Hz
= 7 (7th harmonic and 6th overtone)
d) =
30= 
ʋ= 48 m/s
Solution 47
=
; 
[since l is same for both wires]
= 
= 
Solution 48
Fundamental frequency of 1st wire = first overtone of frequency of 2nd wire
= 
-------(i)
Let mass m=4.8kg be placed at distance x from left end.
Now,
= 1.2g + 4.8g (Translatory equilibrium)
4 = 6g
= 12 N
Apply rotational equilibrium at point (A)
Clockwise torque = anticlockwise torque
48x + 2.4 = 4.8
48x = 2.4
x=5cm
Solution 49
Velocity of wave in wire V=
For steel wire
= 
= 71.6 m/s
For Aluminum wire
=
= 71.6 m/s
Let 
harmonic frequency of steel wire matches with 
harmonic of Aluminum wire
P 
= q 
3p=4q
For minimum, p = 4 loops and q= 3 loops
So, minimum frequency = P
= 4
≃ 180 Hz
Solution 50
a) Distance between two consecutive nodes is 
=L
λ = 2L
wave number k=
=
k= 
b) Equation of standing wave having having node at origin and antinode of mean position
y=A sin (
)sin(2
Waves Motion and Waves on a String Exercise 327
Solution 51
a) wavelength λ = 2m
frequency=
200=
b) Equation of a standing wave having node at origin and antinode at maximum displacement at t=0
y
Here x is in meter and t is in second
Solution 52
Comparing equation with y= A.sinkx.cos
a)
2
f=600Hz
b) K=0.314
= 0.314
λ = 20cm
Nodes are at distance of 0, 
from one end i.e., x=0, 10cm, 20cm, 30cm.
c) Length of string 
d) 
= 60 m/s
Solution 53
Here K=0.314
λ = 20 cm
minimum length of string = 
10cm
Solution 54
Fundamental frequency
220= 
T
248.2 N
Y= 
=
Y= 1.98
Solution 55
Since, 
= 10
Solution 56
a) Fundamental frequency
; ʋ
λ= 8m ʋ = 80 m/s
= 
= 
= 10 Hz.
b) 1st overtone frequency 
ʋ= f × λ
80=30 × λ1
λ1 = 2.67m
2nd overtone frequency 
ʋ= f × λ
80=50 × λ2
λ2= 1.6m
Solution 57
Initially, heavy string is fixed at one end only
So, lowest frequency means fundamental frequency = 120 Hz
= 120
Now, when moved by 10cm, heavy string is fixed at both ends
So, minimum frequency = 
=
= 2(120)
= 240Hz
Waves Motion and Waves on a String Exercise 341
Solution 26
a) Let mass per unit length be µ
Tension at point P; 
b) ʋ = 
= 
= 
T= 
c)
Acceleration of pulse wave
Now
=0
=3g/2
=L
+
L = 0 +
3g/2)
t = 
For pulse,
u=0, a=
, t=
s = ut +
a
= 0 + 
s= 
from bottom
Solution 27
=
= 
= 20 m/s
= = 
= 25 m/s
In 20ms, pulse in string A travels = speed 
time
= 20
= 0.4m
Now, 
= 25-20 = 5 m/s
= 0.4m
t = 
= 0.08 sec = 80 ms
Total time = 20 + 80
= 100ms
In 100ms, string A travels a distance of 
Solution 29
a) 
ʋ = 100 m/s
Power = P= 2
s
P= 2
P=0.47W
b)
Length of the string = 2m
Time to cover the distance t= = 0.02 sec
E= 9.4mJ