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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= 2f= 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=

T1480N

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

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