Solutions of Waves & Thermodynamics. Lesson 14th to 19th

75 Pages·2013·1.65 MB·English

Solutions
of
Waves & Thermodynamics
Lesson 14 to 19
th th
By DC Pandey

14. Wave Motion
Introductory Exercise 14.1
1 1
1. A funct ion, f can rep re sent wave and y(x, z) = =
1+(x -2w)2 1+(x -1)2
equat ion, if it sati sfy
1
∂2f ∂2f fi w=
=v2 2
∂t2 ∂x2
w 1/2
For, y =asinwt, \ v = = =0.5 m/s
k 1
∂2y
=-w2asinwt =-w2y 10 a
4. y = =
∂t2 5+(x +2t)2 b+(kx + wt)2
∂2y
but, =0 a 10
∂x2 Amplitude, ymax = b = 5 =2 m
So, y do not represent wave equation.
and k=1; w=2
2. y(x, t) =ae-(bx-et)2 =ae-(kx-wt)2 w
v = =2 m/s and is travelling in (–) x
w c k
fi k=b and w=e fi v = =
direction.
k b
10
1 5. y =
3. y(x, t) = rep re sent the (kx -wt)2 +2
1+(4x + wt)2
10 10
given pulse, where, y(x,0) = = fi k=1
k2x2 +2 x2 +2
1 1
y(x,0) = =
1+ k2x2 1+ x2 w=vk=2 m/s ¥1m-1 =2 rad/s
10
fi y =
fi k=1 (x -2t)2 +2
Introductory Exercise 14.2
Ê x t ˆ Ê0.2 0.3 ˆ
1. y(x, t) =0.02 sinÁ + ˜ m cosÁ + ˜
Ë0.05 0.01¯ Ë0.5 0.01¯
= Asin(kx + wt) m =2cos(4 +30)
1 1
fi A =0.02 m, k= m-1, w= s-1 =2cos34
0.05 0.01
=2(-0.85)
w 0.05
(a) v = = m/s =5 m/s =-1.7 m/s
k 0.01
w
∂y 2. Yes, (v ) = Aw= Ak◊ =(Ak)v
(b) vp = ∂t = Awcos(kx + wt) p max k
1 3. l=4cm, v =40 cm/s (given)
v (0.2,0.3) =0.02¥
p 0.01 v 40cm/s
(a) n = = =10 Hz
l 4cm
2 | Waves & Motion
2p
(b) Df= Dx y = Asin(wt -kx)
l Ê 2p 2p ˆ
2p 5p = AsinÁv◊ t - x˜
= ¥2.5cm = rad Ë l l ¯
4cm 4 Ê 2p 2p ˆ
T 1 =0.05sinÁ12¥ t - x˜
(c) Dt = Df= Df Ë 0.4 0.4 ¯
2p 2pn
=0.05sin(60pt -5px)
1 p
= ¥ (b) y(0.25,0.15)
2p¥10 3
=0.05sin(60p¥0.15-5p¥0.25)
1
= s
=0.05sin(9p-1.25p)
60
=0.05sin(7.75p) =0.05sin(1.75p)
(d) v =(v )
p p max
=-0.0354 m =-3.54 cm
=- Aw=-2pAn
T Df 0.25p
=-2p¥2cm ¥10s-1 (c) Dt = Df= =
2p w 60p
=-40p cm/s
1
= s =4.2 ms
=-1.26 cm/s
240
4. (a) y
x
Introductory Exercise 14.3
T T Tl T T
1. v = = = 2. v = =
m m/l m m r◊A
500¥2 100 5 0.98
= = =129.1 m/s = =10 m/s
0.06 3 9.8¥103 ¥10-6
Introductory Exercise 14.4
1. I = P = 1W = 1 W/m2 fi I µ1 and as I µ A2
4pr2 4p¥(1m)2 4p r
1
r fi A µ
2. For line source, I = r
2prl
Waves & Motion | 3
AIEEE Corner
Sub jec tive Ques tions (Level 1)
¢
7p 7
1. y(x, t)=6.50mmcos2p = = m =0.166m
60p 50
Ê p t ˆ 2p
Á - ˜ (b) Df= Dt =2pnDt =2p¥500¥10-3
Á ˜
Ë28.0cm 0.0360s¯ T
Êx t ˆ =p=180∞
= Acos2pÁ - ˜ 6
Ël T¯ 5. y(x, t) =
(kx + wt)2 +3
fi A =6.50mm, l=28.0cm,
1 1 6 6
n = = s-1 =27.78Hz y(x,0) = =
T 0.036 k2x2 +3 x2 +3
v =nl=28.0cm ¥27.78s-1 =778cm/s fi k=1m-1
=7.78m/s fi w=vk=4.5m/s¥1m-1 =4.5rad/s
The wave is travelling along (+)ve x-axis. 6
fi y(x, t) =
Ê x ˆ (x -4.5t)2 +3
2. y =5sin30pÁt - ˜
Ë 240¯ Ê x t ˆ
=5sinÊÁ30pt - pxˆ˜ = Asin(wt -kx) 6. y =1.0sinpËÁ2.0 -0.01¯˜
Ë 8 ¯ Ê x t ˆ
Ê p ˆ =1.0sin2pÁ - ˜
(a) y(2,0) =5sinÁ3p¥0- ¥2˜ Ë4.0 0.02¯
Ë 8 ¯ Êx t ˆ
p 5 = Asin2pÁ - ˜
=-5sin =- =-3.535cm Ël T¯
4 2
(a) A =1.0mm, l=4.0cm, T =0.02s
2p 2p
(b) l= = =16cm ∂y Êx t ˆ
k p/8 (b) v = =-wAcos2pÁ - ˜
p ∂t Ël T¯
w 30p
(c) v = = =240cm/s 2pA Êx t ˆ
k p/8 =- cos2pÁ - ˜
T Ël T¯
w 30p
(d) n = = =15 Hz 2p¥1.0mm Ê x t ˆ
2p 2p =- cos2pÁÁ - ˜˜
0.02s 4.0 0.02s
Ë ¯
3. y =3cm sin(3.14cm-1x -314s-1t) p Ê x t ˆ
=- m/scospÁ - ˜
=3cmsin(pcm-1x =100ps-1t) 10 ËÁ2.0cm 0.01s¯˜
= Asin(kx -wt)
v (1.0cm,0.01s) =
p
(a) (v ) = Aw=3cm ¥100ps-1 p Ê1 0.01ˆ
p max - m/scospÁ - ˜
=300pcm/s =3pm/s =9.4m/s 10 Ë2 0.01¯
p p
(b) a =-w2y =-(100ps-1)2 ¥3cm =- m/scos =0m/s
10 2
sin(6p-111p)
(c) v (3.0,0.01)
=-300psin(-105p) =0 p
p Ê3 ˆ
l v/n 350 p =- cospÁ -1˜ =0m/s
4. (a) Dx = Df= Df= ¥ 10 Ë2 ¯
2p 2p 500¥2p 3
4 | Waves & Motion
p Ê5 ˆ Ê 2p ˆ
v (5.0cm,0.01s) =- mscospÁ -1˜ sinÁ x -2p¥3750t˜
p 10 Ë2 ¯ Ë0.08 ¯
=0m/s =0.06m sin(78.5m-1x -23561.9s-1t)
p Ê7 ˆ v 8.00m/s
v (7.0cm,0.01s) =- m/scospÁ -1˜ 9. (a) n = = =25 Hz
p 10 Ë2 ¯ l 0.32m
=0m/s 1 1
T = = s =0.043 Hz
(d) v (1.0cm,0.011s) n 15
p
p 2p 2p
=- m/s k= = =19.63 rad/m
10 l 0.32m
cospÊÁ1 -0.011ˆ˜ (b) y = Acos(kx + wt) = Acos2pÊÁx + t ˆ˜
Ë2 0.01 ¯ Ël T¯
p Ê 1 ˆ
=- cospÁ -1.1˜ Ê x t ˆ
10 Ë12 ¯ =0.07m cos2pÁÁ + ˜˜
0.32m 0.04 s
p p 3p Ë ¯
=- cos0.6p =- cos =9.7cm/s
Ê0.36 0.15ˆ
10 10 5 (c) y =0.07m cos2pÁ + ˜
Ë0.32 0.04¯
v (1.0cm,0.012s)
p =- p m/scosÊÁ1 -0.012ˆ˜ =0.07m cos2pÊËÁ98 + 380ˆ¯˜
10 Ë2 0.01 ¯
39
p =0.07m cos p
=- cosp(0.5-1.2)
4
10
=- p cos0.7p=18.5cm/s =0.07m cosÊÁ10p- pˆ˜
10 Ë 4¯
p p
v (1.0cm,0.013 s) =- m/s =0.07m cos =0.0495 m
p 10 4
Ê1 0.013ˆ p T Df p+ p/4
cospÁ - ˜=- cos0.8p (d) Dt = Df= =
Ë2 0.01 ¯ 10 2p 2pn 2p¥25
3
=25.4cm/s = s =0.015s
2p 2p p 200
7. (a) k= = = cm-1
l 40cm 20 T T Mg
10. v = = =
m rA rA
=0.157 rad/cm
1 1 2¥9.8
T = = s =0.125s =
n 8 8920¥3.14¥(1.2¥10-3)2
w=2pn =16prad/s =50.26rad/s
2¥9.8¥104
v =nl=8s-1 ¥ 40cm =320cm/s = =22m/s
89.2¥3.14¥1.44
(b) y(x, t) = Acos(kx -wt)
=15.0cm cos(0.157x -50.3t) 11. lµn µ T µ M
8. A =0.06m and 2.5l=20cm l M
fi 2 = 2
20 l M
fi l= cm =8cm 1 1
2.5 8
= = 4 =2.
v 300m/s 2
n = = =3750Hz
l 8cm
fi l =2l
2 1
y = Asin(kx -wt) =0.06m =0.12 m.
Waves & Motion | 5
T(x) 2
12. T(x) =m(L -x)g,v(x) = fi t = ¥2 l
m g 0
8l
= g(L -x) \ t = 0
dx x g
=dt ;
g(L -x) dm
15. m = =kx
dx
Let, L -x = y
fi M =Údm =Ú2 kxdx =1kL2
dx =-dy
0 2
Ú0 -dy =t 2M
fi k=
L g y L2
1 - y 0 L T T TL2 dx
\ t = =2 v(x) = = = =
g 1/2 g m kx 2Mx dt
1
1
13. (a) dm w2R =2Tsindq \ t =Údt =ÚL 2Mx dx = 2M L2+1
0 TL2 TL2 1
+1
2
T
dq dq 2 2ML3 2 2ML
T = =
3 TL2 3 T
R
T Mg
16. (a) v = =
m m
mR2dq w2R =2Tdq 1.5¥9.8
T = =16.3m/s
fi w2R2 = 0.055
m v 16.3m/s
(b) l= = =0.136m
\ Wave speed, v = T = w2R2 =Rw n 120/s
m
(c) lµv µ T µ M i.e., if M is
(b) Kink remains stationary when rope doubled both speed and wavelength
and kink moves in opposite sence increases by a factor of 2 .
i.e., if rope is rotating anticlockwise
17. E =I At =2p2n2a2rvAt
then kink has to move clockwise.
=2p2n2a2(rA)(v.t)
14. x is bei ng meas ured from lover end of the
=2p2n2a2m.l
string
\ m(x) =Údm =Úxm xdx =1m x2 =2p2n2a2m
0 0 2 0 =2¥(3.14)2 ¥(120)2 ¥(0.16¥10-3)2
T(x) m(x)g ¥80¥10-3
\ v(x) = =
m m =582¥10-6 J =582mJ=0.58mJ
1 E
m x2g 18. P = =IA =2p2n2a2rnA =2p2n2a2mv
2 0 1 t
= = gx
m0x 2 =2p2n2a2 Tm
fi Úl dx =Útdt =2¥(3.14)2 ¥(60)2
0 1 0 ¥(6¥10-2)2 80¥5¥10-2
gx
2 =4(3.14¥60¥0.06)2 =511.6W
6 | Waves & Motion
19. P =IA =2p2n2a2 Tm 20. P =2p2v2a2rvA =2p2v2a2mv; v = T
m
=2¥(3.14)2 ¥(200)2
T
¥10-6 60¥6¥10-3 m =
v2
=8¥(3.14)2 ¥10-2 ¥6¥10-1 W
T T
=2p2n2a2 ◊v =2p2n2a2
=0.474 W
v2 v
l
E = Pt = P◊ 2¥(3.14)2 ¥(100)2(0.5¥10-3)2 ¥100
v =
0.474¥2 0.474¥2 100
= = J=9.48mJ
60 100 =2¥(3.14)2 ¥104 ¥0.25¥10-6
6¥10-3 =4.93¥10-2 W =49mW
Obj ect ive Quest ions (Level 1)
¢
150¥2p p 2p
1. w= =5prad/s, A =0.04m and fi = ¥0.04
60 6 l
p
q = \ l=12¥0.04 =0.48m.
4 v 300m/s
Ê pˆ 5. l= = =12m
\ y = Asin(wt + q) =0.04sinÁ5pt + ˜ n 25Hz
Ë 4¯
2p 2p
w Df= Dx = (16-10)m =p
2. w=600p,v =300 fi k= =2p
l 12m
v
fi y = Asin(wt -kx) 6. y =0.02sin(x +30t) = Asin(kx + wt)
=0.04sin(600pt -2px) fi k=1, w=30
Ê 3ˆ
y(0.75,0.01) =0.04sinÁ600p¥0.01-2p¥ ˜ w T
Ë 4¯ v = =30m/s =
Ê 3pˆ k m
=0.04sinÁ6p- ˜
Ë 2 ¯ fi T =mv2 =1.3¥10-4 ¥900=0.117N
Ê pˆ ∂y ∂y ∂x ∂y
=0.04sinÁ4p+ ˜=0.04m 7. v = = ◊ =v =slope¥v
Ë 2¯ p ∂t ∂x ∂t ∂x
1 u
3. y(x, t) = In transverse wave p they are
2+3(kx -wt)2
1 1 v
y(x,0) = =
2+3k2x2 2+3x2
p
perpendicular i.e., . In longitudinal
fi k=1 2
1 1
y(x,2) = =
2+3(x -2w)2 2+3(x -2)2 wave u p , they are either at 0 or
v
w p
fi w=1 \ v = =1m/s p so, 0, and p are the possible angles
k 2
A
4. y = Asin(wt -kx) = between v and v.
2 p
p
fi wt -kx = 8. w=2pn =200prad/s,
6
2p T p w m 3.5¥10-3
\ ◊ - =kx k= =w =200p
T 6 6 v T 35
Waves & Motion | 7
Dl Dl E
=2prad/m Stress =Y =E =
l l 100
y = Acos(wt -kx) = Acos(200pt -2px)
∂y p p p
=2p Asin(200pt -2px) 14. A =4m, w= , k= , q =
∂x 5 9 6
w p/5 9
When, y =0 \ v = = = m/s
fi sin(200pt -2px) =0 k p/9 5
2p 2p
fi sin(200pt -2px) =1 l= = =18 m
p 1 k p/9
\ 2pA = fi A = =0.025m
20 40 w p/5 1
n = = = Hz
\ y =0.025cos(200pt -2px)
2p 2p 10
2p 2p
9. w= = =8 prad/s;
15. w=10pand k=0.1p
T 0.25
w 8p p 2p 2p
k= = = rad/cm fi l= = =20 m
v 48 6 k 0.1p
2p 2p
y = Asin(wt -kx) \ Df= Dx = ¥10=p
Ê p ˆ l 20
= AsinÁ8p¥1- ¥67˜ 2
Ë 6 ¯ 16. y =
p A (2x -6.2t)2 +20
= Asin = Asin30∞= =3cm
6 2 2
fi A = =0.1 m, k=2 rad/m
fi A =6cm 20
v T rA T pd2 and w=6.2rad/s
10. A = A ◊ B = A B w 6.2
\ v = = =3.1m/s
vB rAA TB TBpd2A k 2
w 6.2
d T d T /2 n = = =1Hz
= B A = B B 2p 2¥3.1
d T d /2 T
A B B B 2p 2p
1 l= = =p m
=2¥ = 2 k 2
2 1
17. I =2p2 n2 A2rv = w2A2rv
11. E µ A2n2 for E to con stant, An = 2
constant E IST 2p2 n2A2rv St
u= = =
V V V
A n = A n fi A 4n = A n
A A B B A B B B 1
=2p2 n2A2r = rw2A2
fi A =4A
B A 2
12. k=1 rad/m; v =4 m/s P = E =I.S =2p2n2A2rv◊S
t
fi w=vk=4 rad/s 1
6 6 = rw2A2v◊S
\ y = = 2
(kx -wt)2 +3 (x -4t)2 +3 E P
E = Pt fi P = =IS fi I =
Dl t S
Y
13. v = Y and v = l =v Dl 18. y = Asin(px + pt)
l r t r l l
y(x,0) = Asin(px) fi y =0 for x=0 and 1
l v Dl 1
fi = l =10 \ = a =-w2 y =-w2Asin(px)
Dl v l 100 1 3
t fi a =± w2A at x = and
2 2
8 | Waves & Motion
1 3 Êx tˆ Êx t ˆ
v =pAcos(px) fi v =0 for x = and 20. y = Asin2pÁ - ˜= Asin2pÁ - ˜
P p 2 2 Ëa b¯ Ël T¯
So all the above options are correct. fi l=a, T =b
2p l a
19. y = Asin (x -bt) = Asin(kx -wt) fi v =nl= =
a T b
2p 2pb
k= , w= 21. y a
a a
w 2pb/a
fi v = = =b x fi x
k 2p/a
2p 2p
l= = =a
k 2p/a
as a =-w2y
JEE Corner
Assertion and Reason
¢
1. For prop a ga tion of trans verse waves 7. Elec tro mag netic wave are non-
me dium re quire ten sion which is mec hani c al, they travel dep endi ng upon
poss ib le due to modulus of rig idi ty. And elec tri c and mag netic prop er ties of
in gases there is no such Young’s
me dium. They can travel in me dium as
modulus or sur face ten sion. So the
well as an vacu um. So rea son is false.
reas on given is corr ect explanation.
T 1
2. Sur face ten sion of wat er plays the role of 8. As speed, v = fi v µ in sec ond
m m
modulus of rig idi ty and that is why
transv erse waves can travel on liqu id string m is more (by looki ng) so v will be
sur face. less. Thus reas on is true exp lan at ion of
as ser tion.
3. Both the waves are trav el ling in same
dir ect ion with a phase dif fere nce of p. So 9. At point A both vp and Dl is zero ie, K.E.
reas on is false. and P.E. are min i mum while at B both v
p
4. v = fl is cons tant for a part icu l ar and Dl are max i mum i.e., both K.E. and
med ium so if freq uency is doub led P.E. are max i mum. Thus both as ser tion
wavel ength bec omes half, and speed and reas on are true but not corr ect
rem ains cons tant. Thus ass ert ion is exp lan at ion.
false. 12. y
5. Sound is me chan i cal wave which
re quires ma te rial me dium for P
prop ag a tion and as on moon there is no x
at mo sphere, sound can not travel.
2p
6. Ang ul ar wave numb er, n = while
l
1 If P is moving downword then it shows
wave num ber, k= which is def ined as
l that the wave is travelling in (-) ve x
the num ber of waves per unit length. direction. So assertion is false.