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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


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

= ?


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


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


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


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


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


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


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


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.

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