Waves- 1 Practice Questions - DPP for NEET

 Page 1


1. (d)
330
1.29
256
v
vnm
n
=lÞl= ==
2. (a) Time required for a point to move from maximum
displacement to zero displacement is
1
44
T
t
n
==
11
1.47
4 4 0.170
n Hz
t
Þ===
´
3. (b)
1 2 1 2 1 2 12
2 2 2( ) () d d v tv t d d vtt + =´+´ Þ + =+
12
12
() 340 (1.5 3.5)
850
22
vtt
ddm
+ ´+
+= ==
4. (c) At given temperature and pressure
12
21
14
2 :1
1
v
v
v
r
µ Þ = ==
r r
5. (b) The distance between two points i.e. path difference
between them
2 2 3 66
l l pl
D=´f=´ = =
pp
v
n
() \ =l vn
360
0.12 12
6 500
ÞD= ==
´
m cm
6. (d)
1
sin() y a t kx = w- and
2
cos( ) sin
2
y a t kx a t kx
p æö
= w- = w- +
ç÷
èø
Hence phase difference between these two is .
2
p
7. (c)
2 2
11
2
2
2
0.06 4
0.031
æö
===
ç÷
èø
Ia
I
a
8. (d) Comparing the given equation with sin() y a t kx = w- ,
We get 
0,
2
2, a Y fk
p
= w=p=
l
. Hence maximum
particle velocity 
max0
()2
particle
v a Yf = w= ´p
and
wave velocity
2
()
2/
wave
f
vf
k
wp
= = =l
pl
0
max0
( ) 4 24
2
particle wave
Y
v v Y ff
p
\ = = ´ p = l Þ l=
9. (d)
22
() y f x vt =- doesn’t follows the standard wave
equation.
10. (a) 11
2
sin
x
yat
p æö
= w-
ç÷
l
èø
and
222
22
cos sin
2
xx
ya t at
p pp æ ö æö
= w - +f = w - +f+
ç ÷ ç÷
ll
è ø èø
So phase difference = 
2
p
f+ and 
22
lp æö
D = f+
ç÷
p
èø
11. (d) On comparing the given equation with standard
equation 
2
sin ()
p
=-
l
y a vtx . It is clear that wave
speed ( )
wave
vv = and maximum particle velocity
max0
()
particle
v ay = w=´
co-efficient of 
0
2 v
ty
p
=´
l
max0
2
()2()2
particle wave
av
v vy
´p
\ = w Þ = Þ l =p
l
12. (a) Compare the given equation with sin() y a t kx = w+
We get 
50
2 100 n n Hz w= p = Þ=
p
13. (c) A wave travelling in positive x-direction may be
represented as 
2
sin () y A vtx
p
=-
l
. On putting values
2
0.2 sin (360 ) 0.2 sin 2 6
60 60
x
y txyt
p æö
= - Þ = p-
ç÷
èø
14. (a) Comparing the given equation with sin() y a t kx = w-
We get 3000 1500
2
n Hz
w
w= pÞ ==
p
and 
21
12
6
km
p
= = p Þ l=
l
So, 
1
1500 250
6
vnv = lÞ = ´= m / s
15. (b) Given, 0.5sin(20 400 ) y xt =-
Comparing with sin() y a t kx = w-
Gives velocity of wave 
400
20
20
v
k
w
= ==
m / s.
Page 2


1. (d)
330
1.29
256
v
vnm
n
=lÞl= ==
2. (a) Time required for a point to move from maximum
displacement to zero displacement is
1
44
T
t
n
==
11
1.47
4 4 0.170
n Hz
t
Þ===
´
3. (b)
1 2 1 2 1 2 12
2 2 2( ) () d d v tv t d d vtt + =´+´ Þ + =+
12
12
() 340 (1.5 3.5)
850
22
vtt
ddm
+ ´+
+= ==
4. (c) At given temperature and pressure
12
21
14
2 :1
1
v
v
v
r
µ Þ = ==
r r
5. (b) The distance between two points i.e. path difference
between them
2 2 3 66
l l pl
D=´f=´ = =
pp
v
n
() \ =l vn
360
0.12 12
6 500
ÞD= ==
´
m cm
6. (d)
1
sin() y a t kx = w- and
2
cos( ) sin
2
y a t kx a t kx
p æö
= w- = w- +
ç÷
èø
Hence phase difference between these two is .
2
p
7. (c)
2 2
11
2
2
2
0.06 4
0.031
æö
===
ç÷
èø
Ia
I
a
8. (d) Comparing the given equation with sin() y a t kx = w- ,
We get 
0,
2
2, a Y fk
p
= w=p=
l
. Hence maximum
particle velocity 
max0
()2
particle
v a Yf = w= ´p
and
wave velocity
2
()
2/
wave
f
vf
k
wp
= = =l
pl
0
max0
( ) 4 24
2
particle wave
Y
v v Y ff
p
\ = = ´ p = l Þ l=
9. (d)
22
() y f x vt =- doesn’t follows the standard wave
equation.
10. (a) 11
2
sin
x
yat
p æö
= w-
ç÷
l
èø
and
222
22
cos sin
2
xx
ya t at
p pp æ ö æö
= w - +f = w - +f+
ç ÷ ç÷
ll
è ø èø
So phase difference = 
2
p
f+ and 
22
lp æö
D = f+
ç÷
p
èø
11. (d) On comparing the given equation with standard
equation 
2
sin ()
p
=-
l
y a vtx . It is clear that wave
speed ( )
wave
vv = and maximum particle velocity
max0
()
particle
v ay = w=´
co-efficient of 
0
2 v
ty
p
=´
l
max0
2
()2()2
particle wave
av
v vy
´p
\ = w Þ = Þ l =p
l
12. (a) Compare the given equation with sin() y a t kx = w+
We get 
50
2 100 n n Hz w= p = Þ=
p
13. (c) A wave travelling in positive x-direction may be
represented as 
2
sin () y A vtx
p
=-
l
. On putting values
2
0.2 sin (360 ) 0.2 sin 2 6
60 60
x
y txyt
p æö
= - Þ = p-
ç÷
èø
14. (a) Comparing the given equation with sin() y a t kx = w-
We get 3000 1500
2
n Hz
w
w= pÞ ==
p
and 
21
12
6
km
p
= = p Þ l=
l
So, 
1
1500 250
6
vnv = lÞ = ´= m / s
15. (b) Given, 0.5sin(20 400 ) y xt =-
Comparing with sin() y a t kx = w-
Gives velocity of wave 
400
20
20
v
k
w
= ==
m / s.
82
DPP/ P 29
16. (b) With path difference 
2
l
, waves are out of phase at the
point of observation.
17. (b)
222 2 2
12
2 cos cos
23
A a a aa
p
= = + + q Þ q= - Þ q=
18. (c) For interference, two waves must have a constant phase
relationship. Equation ‘1’ and ‘3’ and ‘2’ and ‘4’ have
a constant phase relationship of 
2
p
out of two choices.
Only one 
2
S emitting ‘2’ and 
4
S emitting ‘4’ is given
so only (c) option is correct.
19. (a) The resultant amplitude is given by
222
2 cos 2 (1 cos)
R
A AAAAA = + + q= +q
2 cos /2 A =q
2
( 1 cos 2 cos / 2) \+ q=q
20. (d)
11
sin sin
2
y tt
ab
p æö
= w± w+
ç÷
èø
Here phase difference 
2
p
=
\
 The resultant amplitude
22
1 1 11 ab
a b ab ab
+ æ ö æö
= + = +=
ç ÷ ç÷
è ø èø
21. (a) In a wave equation, x and t must be related in the form
() x vt -
We rewrite the given equations 
2
1
1()
y
x vt
=
+-
For 
0 t =
, this becomes 
2
1
(1)
y
x
=
+
, as given
For 
2 t =
, this becomes
22
11
[1 ( 2 )] [1 ( 1)]
y
xvx
==
+- +-
2v = 1or v = 0 m/s
22. (c)
l
l/2
Q
P
R
y
x
23. (b)
Standard wave equation which travel in negative x-direc-
tion is ( )
0
y Asin?t kx = + +f
For the given wave 
2p
? 2pn 15p, k 10p
?
= = ==
Now 
Coefficient of t ? 15p
v 1.5m / sec
Coefficient of x k 10p
= = ==
and 
2p 2p
? 0.2 m.
k 10p
= ==
24. (a)
2
m
mm
pv BA1
p ,I pAI
v 2 2B
w
= = w Þ=
25. (d),    26. (c),    27. (a).
(i) 
2
1
I
d
µ
, d
initial
 = R,  d
final
 = 3R,
initial initial
final
final
II 9
I
I 19
=Þ=
(ii) During the first half time, wavelength first increases as
the component of velocity of source increases till it be-
comes equal to the velocity of source itself, then it de-
creases till it becomes zeros.
(iii) 
1
/3 2 /3
t , .......
p p+p
=
ww
,  
2
5 /3 2 5 /3
t,
p p+p
=
ww
t=0
t
1
t
2
30°
30°
60°
60°
when  rad/s
3
p
w= ,  t
1
 = 1, 7, 13, .......   t
2
 = 5, 11, 17,
28. (c) The velocity of every oscillating particle of the medium
is different of its different positions in one oscillation
but the velocity of wave motion is always constant i.e.,
particle velocity vary with respect to time, while the
wave velocity is independent of time.
Also for wave propagation medium must have the
properties of elasticity and inertia.
29. (b) Velocity of wave = 
Distance travelled by wave ( )
Time period ( ) T
l
Wavelength is also defined as the distance between two
nearest points in phase.
30. (b) Transverse waves travel in the form of crests and
through involving change in shape of the medium. As
liquids and gases do not possess the elasticity of shape,
therefore, transverse waves cannot be produced in
liquid and gases. Also light wave is one example of
transverse wave.