Oscillations- 2 Practice Questions - DPP for NEET

 Page 1


1. (b) When the particle of mass m at O is pushed by y in
the direction of A . The spring A will compressed by
y while spring B and C will be stretched by
' cos 45 yy =
o
. So that the total restoring force on the
mass  m  along OA.
A
B
C
O
m
F
C
F
B
F
A
cos 45 cos 45
netABC
FFFF =++
oo
2 '45 2 ( cos 45 ) cos 45 2 ky ky ky k y ky =+ =+ =
o oo
Also
 ' ' 2 '2
net
F ky ky ky k k = Þ = Þ=
22
'2
mm
T
kk
=p =p
2. (b) When mass 700 gm  is removed, the left out mass (500
+ 400) gm oscillates with a period of 3 sec
(500 400
32 t
k
+
\ = =p
......(i)
When 500 gm mass is also removed, the left out mass
is 400 gm.
400
'2 t
k
\ =p ...(ii)
3 900
' 2sec
' 400
t
t
Þ = Þ=
3. (a) Slope is irrelevant hence
1/2
2
2
æö
=p
ç÷
èø
M
T
k
4. (a) Tension in the string when bob passes through lowest
point 
2
= + = +w
mv
T mg mg mv
r
 () \ =w vr
Putting 2 v gh = and 
22
2 T
pp
w = = =p
we get ( 2) T m g gh = +p
5. (b)
k
k
Force constant 
1
() k
Length of spring
µ
1
1
1
2
3
3
2
Þ = = Þ=
l
l k
kk
k ll
6. (b) Initially time period was
2
l
T
g
=p
When train accelerates, the effective value of g becomes
22
() ga + which is greater than g.
Hence, new time period, becomes less than the initial time
period.
a
g
eff
g
7. (b) In accelerated frame of reference, a fictitious force
(pseudo force) ma acts on the bob of pendulum as
shown in figure.
Page 2


1. (b) When the particle of mass m at O is pushed by y in
the direction of A . The spring A will compressed by
y while spring B and C will be stretched by
' cos 45 yy =
o
. So that the total restoring force on the
mass  m  along OA.
A
B
C
O
m
F
C
F
B
F
A
cos 45 cos 45
netABC
FFFF =++
oo
2 '45 2 ( cos 45 ) cos 45 2 ky ky ky k y ky =+ =+ =
o oo
Also
 ' ' 2 '2
net
F ky ky ky k k = Þ = Þ=
22
'2
mm
T
kk
=p =p
2. (b) When mass 700 gm  is removed, the left out mass (500
+ 400) gm oscillates with a period of 3 sec
(500 400
32 t
k
+
\ = =p
......(i)
When 500 gm mass is also removed, the left out mass
is 400 gm.
400
'2 t
k
\ =p ...(ii)
3 900
' 2sec
' 400
t
t
Þ = Þ=
3. (a) Slope is irrelevant hence
1/2
2
2
æö
=p
ç÷
èø
M
T
k
4. (a) Tension in the string when bob passes through lowest
point 
2
= + = +w
mv
T mg mg mv
r
 () \ =w vr
Putting 2 v gh = and 
22
2 T
pp
w = = =p
we get ( 2) T m g gh = +p
5. (b)
k
k
Force constant 
1
() k
Length of spring
µ
1
1
1
2
3
3
2
Þ = = Þ=
l
l k
kk
k ll
6. (b) Initially time period was
2
l
T
g
=p
When train accelerates, the effective value of g becomes
22
() ga + which is greater than g.
Hence, new time period, becomes less than the initial time
period.
a
g
eff
g
7. (b) In accelerated frame of reference, a fictitious force
(pseudo force) ma acts on the bob of pendulum as
shown in figure.
DPP/ P 28
79
Hence tan
maa
mgg
q==
1
tan
a
g
-
æö
Þq=
ç÷
èø
 in the backward direction.
8. (c) 2
l
T
g
=p (Independent of mass)
9. (b)
11
1% 0.5%
22
Tl
Tl
Tl
DD
µ Þ = = ´=
10. (c) If suppose bob rises up to a height h as shown then
after releasing potential energy at extreme position
becomes kinetic energy of mean position
h = l (1 – cos ) q
l
l
q
2
max max
1
2
2
mgh mv v gh Þ = Þ=
Also, from figure 
cos
lh
l
-
q=
(1 cos) hl Þ = -q
So, 
max
2 (1 cos) v gl = -q
11. (a) If initial length 
1
100 l = then 
2
121 l =
By using 
11
22
2
Tl l
T
g Tl
=p Þ=
Hence, 
1
21
2
100
1.1
121
T
TT
T
= Þ=
% increase 
21
1
100 10%
TT
T
-
= ´=
12. (d) After standing centre of mass of the oscillating body
will shift upward therefore effective length will
decrease and by 
Tl µ
, time period will decrease.
13. (a) No momentum will be transferred because, at extreme
position the velocity of bob is zero.
14. (c) The effective acceleration in a lift descending with
acceleration 
3
g
is
2
33
eff
gg
gg = -=
3
2 22
2/32
eff
L LL
T
g gg
æö
æ ö æö
\ =p =p =p ç÷
ç ÷ ç÷
ç÷
è ø èø
èø
15. (c) In series 
12
12
eq
kk
k
kk
=
+
so time period
12
12
()
2
mkk
T
kk
+
=p
16. (c) Spring constant 
1
()
Length of the spring ( )
k
l
µ
as length becomes half, k becomes twice i.e. 2k
17. (b) Standard equation for given condition
2
cos 0.16 cos( ) xatxt
T
p
= Þ=-p
[As 
0.16 a =-
meter , 2sec] T =
18. (d) 1
1
2 =p
m
t
k
and 2
2
2 =p
m
t
k
Equivalent spring constant for shown combination is
12
+ kk . So time period t is given by
12
2 =p
+
m
t
kk
By solving these equations we get
2 22
12
t tt
- --
=+
19. (a) With mass 
2
m alone, the extension of the spring 
l
is
given as
2
m g kl = ........(i)
With mass 
12
() mm + , the extension 
' l
is given by
12
( ) () m mg kll + = +D .......(ii)
The increase in extension is 
l D
which is the amplitude
of vibration. Subtracting (i) from (ii), we get
1
mg kl =D
or
1
mg
l
k
D=
20. (a) If 
11
sin y at =w and 
2
sin() at w +p
112
21
121
yya
yy
aaa
Þ + Þ=
This is the equation of straight line.
Page 3


1. (b) When the particle of mass m at O is pushed by y in
the direction of A . The spring A will compressed by
y while spring B and C will be stretched by
' cos 45 yy =
o
. So that the total restoring force on the
mass  m  along OA.
A
B
C
O
m
F
C
F
B
F
A
cos 45 cos 45
netABC
FFFF =++
oo
2 '45 2 ( cos 45 ) cos 45 2 ky ky ky k y ky =+ =+ =
o oo
Also
 ' ' 2 '2
net
F ky ky ky k k = Þ = Þ=
22
'2
mm
T
kk
=p =p
2. (b) When mass 700 gm  is removed, the left out mass (500
+ 400) gm oscillates with a period of 3 sec
(500 400
32 t
k
+
\ = =p
......(i)
When 500 gm mass is also removed, the left out mass
is 400 gm.
400
'2 t
k
\ =p ...(ii)
3 900
' 2sec
' 400
t
t
Þ = Þ=
3. (a) Slope is irrelevant hence
1/2
2
2
æö
=p
ç÷
èø
M
T
k
4. (a) Tension in the string when bob passes through lowest
point 
2
= + = +w
mv
T mg mg mv
r
 () \ =w vr
Putting 2 v gh = and 
22
2 T
pp
w = = =p
we get ( 2) T m g gh = +p
5. (b)
k
k
Force constant 
1
() k
Length of spring
µ
1
1
1
2
3
3
2
Þ = = Þ=
l
l k
kk
k ll
6. (b) Initially time period was
2
l
T
g
=p
When train accelerates, the effective value of g becomes
22
() ga + which is greater than g.
Hence, new time period, becomes less than the initial time
period.
a
g
eff
g
7. (b) In accelerated frame of reference, a fictitious force
(pseudo force) ma acts on the bob of pendulum as
shown in figure.
DPP/ P 28
79
Hence tan
maa
mgg
q==
1
tan
a
g
-
æö
Þq=
ç÷
èø
 in the backward direction.
8. (c) 2
l
T
g
=p (Independent of mass)
9. (b)
11
1% 0.5%
22
Tl
Tl
Tl
DD
µ Þ = = ´=
10. (c) If suppose bob rises up to a height h as shown then
after releasing potential energy at extreme position
becomes kinetic energy of mean position
h = l (1 – cos ) q
l
l
q
2
max max
1
2
2
mgh mv v gh Þ = Þ=
Also, from figure 
cos
lh
l
-
q=
(1 cos) hl Þ = -q
So, 
max
2 (1 cos) v gl = -q
11. (a) If initial length 
1
100 l = then 
2
121 l =
By using 
11
22
2
Tl l
T
g Tl
=p Þ=
Hence, 
1
21
2
100
1.1
121
T
TT
T
= Þ=
% increase 
21
1
100 10%
TT
T
-
= ´=
12. (d) After standing centre of mass of the oscillating body
will shift upward therefore effective length will
decrease and by 
Tl µ
, time period will decrease.
13. (a) No momentum will be transferred because, at extreme
position the velocity of bob is zero.
14. (c) The effective acceleration in a lift descending with
acceleration 
3
g
is
2
33
eff
gg
gg = -=
3
2 22
2/32
eff
L LL
T
g gg
æö
æ ö æö
\ =p =p =p ç÷
ç ÷ ç÷
ç÷
è ø èø
èø
15. (c) In series 
12
12
eq
kk
k
kk
=
+
so time period
12
12
()
2
mkk
T
kk
+
=p
16. (c) Spring constant 
1
()
Length of the spring ( )
k
l
µ
as length becomes half, k becomes twice i.e. 2k
17. (b) Standard equation for given condition
2
cos 0.16 cos( ) xatxt
T
p
= Þ=-p
[As 
0.16 a =-
meter , 2sec] T =
18. (d) 1
1
2 =p
m
t
k
and 2
2
2 =p
m
t
k
Equivalent spring constant for shown combination is
12
+ kk . So time period t is given by
12
2 =p
+
m
t
kk
By solving these equations we get
2 22
12
t tt
- --
=+
19. (a) With mass 
2
m alone, the extension of the spring 
l
is
given as
2
m g kl = ........(i)
With mass 
12
() mm + , the extension 
' l
is given by
12
( ) () m mg kll + = +D .......(ii)
The increase in extension is 
l D
which is the amplitude
of vibration. Subtracting (i) from (ii), we get
1
mg kl =D
or
1
mg
l
k
D=
20. (a) If 
11
sin y at =w and 
2
sin() at w +p
112
21
121
yya
yy
aaa
Þ + Þ=
This is the equation of straight line.
80
DPP/ P 28
21. (c) Energy of particle is maximum at resonant frequency
i.e., 
20
w =w . For amplitude resonance (amplitude
maximum) frequency of driver force
w =
2 22
0 10
2 bm w - Þ w ¹w
22. (b)
c
A ; when b 0, a c, amplitude
a bc
= ==
+-
A. ®¥
 This corresponds to resonance.
23. (b) Let the velocity acquired by A and B be V , then
v
mv mV mV V
2
= + Þ=
Also 
2 2 22
1 1 11
mv mV mV kx
2 2 22
= ++
Where x is the maximum compression of the spring.
On solving the above equations, we get 
1/2
m
xv
2k
æö
=
ç÷
èø
At maximum compression, kinetic energy of the
A – B system 
2
2 22
1 1 mv
mV mV mV
224
= + ==
24. (d)
L
L
mg cos q
mg sin q
mg
q
– f
+ f
From following figure it is clear that
T Mg cos? Centripetal force -=
2
Mv
T Mgcos?
L
Þ-=
Also tangential acceleration  
T
a g sin?. =
25. (a) Except (4) all statements are wrong.
26. (b)        27. (b).
For minimum time period w
2
A = mg
2
2
4
A mg
T
p
=
, T = 0.2 sec,
At t = 0.05 sec.
y = A sin wt = 1 sn 
2
0.2
p
 × 0.05cm. = 1cm.
PE = mgy = 1 × 10 × 
1
100
 = 0.1 Joule
28. (c) Statement -1 is False, Statement-2 is True.
29. (a) The time period of a oscillating spring is given by ,
1
2
m
TT
k k
= p Þµ
Since the spring constant is large for hard spring,
therefore hard spring has a less periodic time as
compared to soft spring.
30. (d) Time period of simple pendulum of length l is,
2
l
T Tl
g
= p Þµ
1
2
Tl
Tl
DD
Þ=
\ 
1
3 1.5%
2
D
= ´=
T
T