Oscillations- 1 Practice Questions - DPP for NEET

 Page 1


1. (c)
10 a =
2. (d)
sin22
sin sin
2 12
A A tT
yAt t At
TT
pp
= w = Þ = Þ=
3. (c)
2 2 12
sin sin sin
2 3 23
a tt
ya ta
T
p pp
= Þ = Þ=
2 21
sin sin sec
3 6 3 64
tt
t
p p pp
Þ = Þ = Þ=
4. (a) sin
6
xat
p æö
= w+
ç÷
èø
and ' cos sin
2
xa tat
p æö
= w = w+
ç÷
èø
6 63
p pp æ öæö
\Df=w+ -w+=
ç ÷ç÷
è øèø
tt
5. (c)
2 2 22
10 (4) =w- Þ =w- vaya and
22
8 (5) a =w-
On solving, 
2
2 2 sec T
T
p
w= Þ w= = Þ = p
6. ( b)
3
max
22
(50 10 ) 0.15
2
-
pp
=w= ´= ´ ´= v aa
T
m / s
7. ( d)
max
=w va and 
2
max
Aa =w
max
max
4
2
2
Þw= ==
A
v
rad/sec
8. ( d) At mean position velocity is maximum
i.e., 
max
max
16
4
4
=wÞw= ==
v
va
a
22 22
8 3 44 ayy \u=w - Þ =-
22
192 16(16 ) 12 16 2 y y y cm Þ = - Þ = - Þ=
9. (a) Maximum acceleration 
2 22
4 a an = w = ´p
2 22
0.01 4 ( ) (60) 144 / sec m = ´´p ´ =p
10. (d)
2 max
max
22
7.5
0.61
(3.5)
A
Aaam =wÞ= ==
w
11. ( b) Comparing given equation with standard equation,
sin( ), yat = w +f we get, 2,
2
a cm
p
= w=
2 2
22
max
2/
22
A A cms
pp æö
\ =w = ´=
ç÷
èø
12. (d)
222
1
2
E m a Ea = w Þµ
13. (b)
2
22
2
2
22
1
1 2
2
1
4
2
a
my
Uy
Ea
a
ma
æö
w ç÷
èø
= = ==
w
14. (c) In S.H.M., frequency of K.E. and P .E.
= 2 × (Frequency of oscillating particle)
15. (c) Kinetic energy 
2 222
11
cos
22
= = ww K mv ma t
22
1
(1 cos2)
2
mat = w +w
hence kinetic energy varies periodically with double
the frequency of S.H.M. i.e. 2 w .
16. (a) At mean position, the kinetic energy is maximum.
Hence 
22
1
16
2
ma w=
On putting the values we get
2
10 sec
5
T
pp
w= Þ ==
w
17. (d) From the given equation, 
2 42 n n Hz w=p =pÞ =
18. (a) Using sin x At =w
For 
11
/2, sin 1/2
6
xA TT
p
= w = Þ=
w
For 
12 12
,sin ( )1
2
xA TT TT
p
= w + =Þ +=
w
21
2 2 63
TT
p p pp
Þ =- = -=
w w ww
i.e., 
12
TT <
19. (a) Let the piston be displaced through distance x towards
left, then volume decreases, pressure increases. If P D is
increased in pressure and V D is decreased in volume,
then considering the process to take place gradually
(i.e isothermal)
Gas
P
A
M
h
x
1 1 22
( )() PV PV PV P PVV = Þ = + D -D
PV PV PV P V P V Þ = +D - D -DD
. .0 PV PV ÞD - D=
  (neglecting .) PV DD
.
( ) ()
Px
P Ah P Ax P
h
D = ÞD=
Page 2


1. (c)
10 a =
2. (d)
sin22
sin sin
2 12
A A tT
yAt t At
TT
pp
= w = Þ = Þ=
3. (c)
2 2 12
sin sin sin
2 3 23
a tt
ya ta
T
p pp
= Þ = Þ=
2 21
sin sin sec
3 6 3 64
tt
t
p p pp
Þ = Þ = Þ=
4. (a) sin
6
xat
p æö
= w+
ç÷
èø
and ' cos sin
2
xa tat
p æö
= w = w+
ç÷
èø
6 63
p pp æ öæö
\Df=w+ -w+=
ç ÷ç÷
è øèø
tt
5. (c)
2 2 22
10 (4) =w- Þ =w- vaya and
22
8 (5) a =w-
On solving, 
2
2 2 sec T
T
p
w= Þ w= = Þ = p
6. ( b)
3
max
22
(50 10 ) 0.15
2
-
pp
=w= ´= ´ ´= v aa
T
m / s
7. ( d)
max
=w va and 
2
max
Aa =w
max
max
4
2
2
Þw= ==
A
v
rad/sec
8. ( d) At mean position velocity is maximum
i.e., 
max
max
16
4
4
=wÞw= ==
v
va
a
22 22
8 3 44 ayy \u=w - Þ =-
22
192 16(16 ) 12 16 2 y y y cm Þ = - Þ = - Þ=
9. (a) Maximum acceleration 
2 22
4 a an = w = ´p
2 22
0.01 4 ( ) (60) 144 / sec m = ´´p ´ =p
10. (d)
2 max
max
22
7.5
0.61
(3.5)
A
Aaam =wÞ= ==
w
11. ( b) Comparing given equation with standard equation,
sin( ), yat = w +f we get, 2,
2
a cm
p
= w=
2 2
22
max
2/
22
A A cms
pp æö
\ =w = ´=
ç÷
èø
12. (d)
222
1
2
E m a Ea = w Þµ
13. (b)
2
22
2
2
22
1
1 2
2
1
4
2
a
my
Uy
Ea
a
ma
æö
w ç÷
èø
= = ==
w
14. (c) In S.H.M., frequency of K.E. and P .E.
= 2 × (Frequency of oscillating particle)
15. (c) Kinetic energy 
2 222
11
cos
22
= = ww K mv ma t
22
1
(1 cos2)
2
mat = w +w
hence kinetic energy varies periodically with double
the frequency of S.H.M. i.e. 2 w .
16. (a) At mean position, the kinetic energy is maximum.
Hence 
22
1
16
2
ma w=
On putting the values we get
2
10 sec
5
T
pp
w= Þ ==
w
17. (d) From the given equation, 
2 42 n n Hz w=p =pÞ =
18. (a) Using sin x At =w
For 
11
/2, sin 1/2
6
xA TT
p
= w = Þ=
w
For 
12 12
,sin ( )1
2
xA TT TT
p
= w + =Þ +=
w
21
2 2 63
TT
p p pp
Þ =- = -=
w w ww
i.e., 
12
TT <
19. (a) Let the piston be displaced through distance x towards
left, then volume decreases, pressure increases. If P D is
increased in pressure and V D is decreased in volume,
then considering the process to take place gradually
(i.e isothermal)
Gas
P
A
M
h
x
1 1 22
( )() PV PV PV P PVV = Þ = + D -D
PV PV PV P V P V Þ = +D - D -DD
. .0 PV PV ÞD - D=
  (neglecting .) PV DD
.
( ) ()
Px
P Ah P Ax P
h
D = ÞD=
DPP/ P 27
77
This excess pressure is responsible for providing the
restoring force (F) to the piston of mass M.
Hence .
PAx
F PA
h
=D=
Comparing it with 
2
PA
F kx kM
h
= Þ = w=
2
PA Mh
T
Mh PA
Þw= Þ =p
20. (b) Time taken by particle to move from 
0 x =
(mean
position) to 
4 x =
(extreme position)
1.2
0.3
44
T
= == s
Let t be the time taken by the particle to move from
0 x =
 to 2 x cm =
2 12
sin 2 4sin sin
2 1.2
yat tt
T
pp
= wÞ = Þ=
2
0.1
6 1.2
t ts
pp
Þ = Þ=
.
Hence time to move from 2 x = to 4 x = will be equal
to 0.3 0.1 0.2 s -=
Hence total time to move from 2 x = to 4 x = and
back again 2 0.2 0.4sec =´=
21. (b)
Force constant 
1
(k)
Length of spring
µ
1
1
1
2
3
3
.
2
Þ = = Þ=
l
l K
KK
K ll
22. (c)     
2
2 22
y
2
dy
y Kt a 2K 2 1 2m/s ( K 1m/s )
dt
=Þ===´==Q
Now, 
( )
12
y
ll
T 2 andT2
g ga
=p =p
+
Dividing, 
2
y
11
2
2
2
ga
TT 66
T g 55
T
+
= Þ Þ=
23. (a) At x = 0, v = 0 and potential energy is minimum so
particle will remain at rest.
24. (d) Let simple harmonic motions be represented by
1
yasint
4
p æö
= w-
ç÷
èø
, 
2
y asint =w
and 3
y asint
4
p æö
= w+
ç÷
èø
On superimposing, resultant SHM will be
y a sin t sin t sin t
44
éù pp æ ö æö
= w- + w+ w+
êú ç ÷ ç÷
è ø èø
ëû
= 
a 2sintcos sint
4
p éù
w +w
êú
ëû
 = a[ 2sintsint] w+w
= a(1 2)sint +w
Resultant amplitude = (1 2)a +
Energy in SHM µ (Amplitude)
2
\ 
2
2 Resultant
Single
E A
( 2 1)
Ea
æö
= =+
ç÷
èø
 = (3 2 2) +
Þ E
resultant
 = (3 2 2) + E
single
OR    
45°
45°
a
a
a
  
a a2 +
º ¾¾¾¾®
25. (a) Acceleration µ- displacement, and direction of
acceleration is always directed towards the equilibrium
position.
26. (d) 27. (b)
      Compare given equation with 
2
2
2
dx
x0
dt
+w=
 ; 
2
b
a
w=
2
max
max
a Ab
v Aa
w
= = w=
w
At t = 0, f = p/2
x = A sin (wt + f) = A cos 
b
t
a
28. (b)
29. (b) sin? = xat and ?cos? ==
dx
v at
dt
It is clear that phase difference between ‘x’ and ‘a’ is 
2
p
.
30. (a) The total energy of S.H.M. = Kinetic energy of particle
+ potential energy of particle.
The variation of total energy of the particle in SHM
with time is shown in a graph.
Energy
A
T/4 2 /4 T 3 /4 T
Zero slope
Total energy
Kinetic energy
Potential energy