6. Wave on String

7/29/2019 6. Wave on String

1/13

Rise Academy, 607/608-A, Talwandi , Kota (Raj.) -324005. 121

EXERCISE # 11.4 Clear from the figure fp=k ls Li"V gSA

1.7 = 2f = 4 sec1

K =2

= 2 m1

y = 0.5 cos (2x + 4t)

1.9 =T

2=

28

2

=3

K =v

=

63

=18

y = 0.5 sin

x

18t

3

at t = 2, x = 8 , y = 0

=9

7

y = 0.5 sin

9

7x

18t

3

1.10* Standard equation

2.5 By definition

2.6 VAB

= 31010

g4.6

= 6400 = 80 sec

m

VCD

= 3108

g2.3

= 4000 = 20 10 sec

m

VDE

= 31010

g6.1

= 1600 = 40 m/s

2.7 P A2 2

2

2

4

40.0

P P = 1.6 watt

2.8 As

= 22 f2 A2v put values

90 = 2 10 f2 25 104 4 102 2104

100

2.9 Pav

= 22f2A2v =32

20252 625 w

WAVE ON STRING


2/13


2.10 f = 200 Hz

A = 103 m

= 6 103 Kg/mT = 60 N

V =

T

= 100 sec

m

(A) Pav

= 22 f2 A2V = 0.47 Watt

3.1 Amplitude varies between A1+ A

2to A

1 A

2

vk;keA1+ A

2ls A

1 A

2ds chp ifjofrZr gksxkA

3.4 By difination ifjHkk"kk ls

3.7 Second string is denser so speed will decreases.

nwljh jLlh l?ku gS vr% pky de gksxhA

4.6 (A) V = 3102.7

150

=

3

250

sec

m~ 144 m/s

(B)2

3= 90 cm = 60 cm

(C) V = f f =

v=

603

100250

=

33

1250Hz

5.5 (A)

2 = 4f

0= 400 f0 = 100 Hz

(B)

2

7= =

7

2

f

v=

7

2 f = 7

2

v= 700 Hz

f = 30 Hz

5.71 T 1 Tn

2L m 2 ML MSincem L

= 1 8002 0.01 0.5 = 200 Hz.

5.82

n

T

= 90

2

1n

T= 150

2

2n

T= 210

1n

n

=

150

90 n = 1.5


3/13


2

n

T

= 90 =2

nV

2

V=

n

90=

5.1

90= 60

4

V= 30

(B)L4

V3= 90 and

4

V5= 150

and4

V7= 210

(C) overtones are 2nd, 4th, 6th

(D)8.02

v

= 30 v = 48 m/sec.

5.13

A = 1 106 m, T = 10g = 100N

1

= 2.6 103kg/m2 2

= 1.04 104 kg /m3

361 106.2101

100

v = sec/m26

103

v2

= 461004.1101

100

= 262

103

f1

= f2

266.02

10n 31

=

269.022

10n 32

3

1

n

n

2

1

Lowest f requency =1

11

L2

vn=

6.0226

101 3

=2612

104

= 162 vibration/sec.

2

1

n

n=

3

1

One side 1 loop and other side 3 loop

So excluding the two nodes at ends total nodes is 3


4/13


EXERCISE # 2PART - I

4. m = 104 kg/m, v =30

k 1 = 30 m/s

v =Tm

T = mv2 = 104 (30)2 = 0.09 N.

9. (i)

2T sin2

d= dm.2R

2T 2d =

mRd.2R T =

22

Rm

V = T

=

mT

= R2 V =

2

L

(ii) VP/R

=

2

L

VP./G

VR/G

=

2

L V

P/G

2

L=

2

L

VP/G =

2

L

2 VP/G =

L

(iii) VP/R

=

LV

P/G

2

L =

2

L

VP/G = 0

10. n =1 T

2L m; n =

44T T1 100

M402100

1.44n' 1.2 2.n 0.6 0.6

11. 2nA

= 3nB

2 2

A BA B

1 T 1 T2. 3.2 2r d r d

A B

B A

r2 2 1 1.3 r 3 2 3


5/13


12.2

n

T= 384;

2

1n

T= 288

1n

n=

3

4

n = 4

Now, 4

L2

V= 384

Put L = 75 cm

V = 144 m/sec.

14. As, T =20

YA4

0=

242

1

4.

20

YA.......(1)

and T =20

YA6 =

262

1

6

.20

YA.......(2)

0

=

26

24

4

6

> 0

19. V1

= 3104

10

= 50 m/sec

V2

= 31016

10

=

16

104

=4

100= 25 m/sec

20.

f =

vat lowest point

f =1 mg

f1 = 1

1v

at highest point

f1 = 11

g)mM(

f1 = f : no any change in source frq.

1

mg=

1

g)mM(

m

mM1(D)


6/13


23. n =Mg Vdg1 1

2 m 2 m

; n =

VV dg g1 22 m

d

n' 1 12 1 1 .n d 2 d 2

d

2 1n' n

2

= 3001/ 2

2 1

2

.

24. 256 =1

Mg V g1 12 m 2 m

...(1)

256 =2 2

Mg B V g V g1 1 .2 m 2 m

...(2)

Equation (1) and (2)

2 1

1 8 1 1 .87 8

25.

(A) X = 7.5 = 60 AS = 3.5 = A sin Kx

3.5 = A sin

5.7

60

2 A = 3.5 2 mm

A = 3.5 2 mm(B) 3rd overtone which covers 120 cm length of the string

28.v 1 T 1 Tn n m n Ad

A A B

B B A

T A.

B A

(As frequency n and density d are same for both wires)

= 1 22. .1/ 2 1

29. 420 = p .v2

; 315 = (p 1)v2

p420315 p 1

p = 4. ; Lowest resonant frequency n =

v2

=4204

= 105 Hz.


7/13


34. = 1 m = 10 sin (80 t 4x)For superimposed IInd wave is

2

= 10sin (80 t + 4x)Amplitude of stationary wave = 2A = 2 10 = 20m

K = 4 =

2

= 24

=2

1= 0.5 m

= 1m

( = 2)

PART-II

7.

Steel LVhy Aluminium ,Y;wfefu;eT = 104

1

= 7.8 gm / cm3 2

= 2.6 gm /cm3

A1

= A2

= 1 mm2

V1 =11A

T

= 663

10110

108.7

104

V1

=

3

200m/ sec.

v2

22A

T

= 3106.2

104

= 2 102 m/sec.

for lowest frequency

f1

= f2

1

11

L2

Vn=

2

22

L2

Vn

35032

200n1

=

1602

200n2

2

5

n

n

2

1

So lowest frequency

f2

=L2Vn2 =

6.022002

=

31000 Hz


8/13


8. 2f = 6, f = 3, T =3

1sec, v = 3 m/sec

3 =T

= 1m

x = vt

3 = 3t t = 1

total time = t +4

T3

= 1 +4

3

3

1=

4

5= 1.25 sec.

9. = 90cm = 0.9m

=9.0

1044 3=

2

3

1090

1044

f =L2v4 =

Lv2 60 =

v2

v =2

60= 30l = 30 0.9 = 27

T

= 27

T = (27 27 ) = 27 27 23

1090

1044

(T = 36 N)

10.21 = 1 = 2 2 =

1

2

=

2

1

2

1

= 2

2

1

V

V=

2

1

T

T= 2

/ / / / / / / / / / / / / / / / / / / / / / / /

4.8g 1.2g

T1 T2

x

T1= 4T

21)

T1+ T

2= 60

T1= 48 N

T2= 12 N

No taking moments about right endT

140 = 12 20 + 48 (40 x)

Put T1& T

2

x = 5 cm i. e 5 cm from left.


9/13


12. The frequency of transverse waves in a stretched string is given by

p Tn ,2l m

where the string is vibrating with p loops, T = tension in the string and m = mass per unit length of the string.

As the frequency of the wave in both strings (A vibrating with p loops and B vibrating with q loops must be

the same, so)

A C

0.3 m 0.75 m

A A B B

p qT T2l m 2l m

A A A A

B B B B

l m lp

q i m l

0.3 6.3 30.75 2.8 5

So, p = 3, q = 5

No. of antinodes = p + q

= 3 + 5 = 8

13.

Fy= T sin = T tan

x

yT~

y = A sin (t kx)

= 0.02 sin

x

280.0

2t

xy

= 0.02

80.0 cos

x

80.0t

At x = 0

x

y

=

80

2 cos t

maxx

y

=40

Fy =

x

yT

= 40

10

= 4

Ns


10/13


EXERCISE # 1

8. Let p & q be the no. of loops ;

AL2

P

A

T

= BL2

q

B

T

q

p=

B

A

B

A

L

L

=

5

3

Now for A 2

3= 0.3 = 1/5

VA

=A

T

= s103.6

mg3

= f = f5

1

Where s is the cross section of wire

f =35

s70m Hz

Antinode = 8

9. f =2

1

A

T

2

1

f

f=

2

2

r

T

L4

1

r4

T

L2

1

=

1

1

10. v = x

100

0 = dt

dx

xdx o = 10 dt

t =15

1 2/30

2/3o )a(

11. Vvel.

= 10 + 10 = 20sec

m

when string is flat v = f

20 = t

1

= 20 t = 10 m.

12. By definition

13. f2 VA2

14. Energy A22

2

1

E

E= 22

22

)2(A

A

E2 = 4E1


11/13


15. After 2 sec.D = D

1+ D

2= (2 2) (2 2) = 0

As their amplitude is same P.E. = 0 purely kinetic

16. f =L2

5

g9now f =

L2

3

Mg

as f = f M = 25 Kg

17. Energy tkZ E =

0

2dmv2

1=

0

22xdmA

2

1

=

0

222 kxsinAdxm

2

1

=

0

222 dxkxsinAm

2

1

=22mA

4

1

As = 2f =22

v=

T=

m

T

Energy tkZ =4

1m a2

2

2

m

T=

4

Ta 22

18. Maximum particle velocity vmax

= A = 3 m/s

Maximum particle acceleration amax

= 2 A = 90 m/s2

amax

= vmax

= 3 = 90 m/s2 = 30 s1 A =

3=

30

3= 0.1 m

k =v

=

20

30=

2

3[where v is velocity of wave] [tgk v rjax dh pky gS ]

Equation of wave in string y = 0.1 sin

x

2

3t30 [where is initial phase]

19.2

1= 1 = 2

2

= 2

1

= 2

2f/v

f/v

2

1

2

1

v

v= 2 =

/T

/T

2

1

2

1

T

T= 4 (1)

Now moment about P : T1

x = T2

( x) x = 4x x =

//5


12/13


20. V = 22 yA

VP

= 2f22 yA

VP

= 2

V22

yA =

5.0

2 0.1 22

)05.0()1.0(

VP

=50

3j m/s

21. v =

T=

2.0/10

5.03 = 10 m/sec.

v = f

10 = (100) = 0.1 m = 10 cm

distance between two successive nodes =2 = 5 cm

22. Aeq

= cosAA2AA 2122

21

Aeq

=2

cos)3)(4(234 22

Aeq

= 5.

PART - II

1. y1+ y

2= a sin (w t kx) a sin (w t +s kx)

= 2a cos t sin kx

y1

+ y2

= 0 at x = 0 ij

2.2

= l l = 80 cm

3. V =

2

600

k

= 300 m/sec

4. = T

2

2

= 100 Hz

5. V =20

100

k

= 5 m/sec


13/13


6. (n + 1)2

v= 420 ......(1)

2

nv= 315 ......(2)

(1) (2) V

= 105 Hz f min

= = 105 Hz

7.2

= a a =05.0

2= 25p

T

2= b = 2 T = p

8. By equation lehdj.k ls

f =04.0

1and = 0.5

V =04.0

1 0.5 =

2

25

by V = T

2

2

25

=

04.0

T =

4

625 0.04

T = 6.25 N

6. Wave on String

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