DAILY PRACTICE PHYSICS

PROBLEMS SOLUTIONS DPP/CP13

1. (a) For an SHM, the acceleration a = –w2x where w2 is a
M
constant. Therefore,
a
is a constant. The time period 5. (a) We know that T = 2p
x k
aT
T is also constant. Therefore,
x
is a constant. From first case, 2 = 2p M ........ (1)
k
m 4p m 4p m 2 2
2. (b) t1 = 2p or t12 = or k1 = 2 M+2
k1 k1 t1 In second case, 4 = 2p ........ (2)
k
4 p2m 4p 2 m From eq. (1) and eq. (2)
Similarly, k 2 = and (k1 + k 2 ) =
t 22 t 02 4 M+2 2
= Þ 4 = 1+
2 M M
4 p2m 4 p 2m 4 p2m 1 1 1
\ = + or = + 2 2
t 02 t12 t 22 t 02 t12 t 22 = 3 Þ M = kg
M 3
3. (b) y 6. (b) Amplitude of a damped oscillator at any instant t is given
by
A = A0e–bt/2m
B a
where A0 is the original amplitude
P From question,
q
A0
l When t = 2 s, A =
3
A0
\ = A0e–2b/2m
A x 3
O x M 1
or, = e–b/m … (i)
dq 3
= 2 \ q = 2t
dt A0
When t = 6 s, A =
Let BP = a, \ x = OM = a sin q = a sin (2t) n
Hence M executes SHM within the given time period A0
and its acceleration is opposite to x that means towards \ = A0e–6b/2m
left. n
4. (b) The kinetic energy of a particle executing S.H.M. is 1
given by or, = e–3b/m = (e–b/m)3
n
1 2 2 2
K= ma w sin wt 1 æ 1ö
3
2 or, =ç ÷ (Using eq. (i))
where, m = mass of particle n è 3ø
a = amplitude \ n = 33
w = angular frequency 7. (c) Acceleration due to gravity at a depth ‘d’ is given by,
t = time æ dö æ R -d ö g
1 g ' = g ç1 - ÷ = g ç ÷= y
Now, average K.E. = < K > = < mw2 a2 sin2 wt > è Rø è R ø R
2
1 d
= mw2a2 <sin2 wt >
2
1 æ 1ö æ 2 1ö
= mw2a2 çè ÷ø çè Q < sin q > = ÷ø R
2 2 2
1 1 R–d=y
= mw 2 a 2 = ma2 (2pn) 2 (Q w = 2pn)
4 4 or acceleration µ displacement
2 2 2 which is the condition for SHM. So, body will oscillate
or, < K > = p ma n
simple harmonically in tunnel.
 EBD_7156
S-58 DPP/ CP13
13. (a) Here,
T x = x0 cos (wt – p / 4 )
8. (c) Time to complete 1/4th oscillation is s. Time to
4 dx æ pö
\ Velocity, v = = - x0 w sin ç wt - ÷
1 dt è 4ø
complete th vibration from extreme position is Acceleration,
8
obtained from dv 2 æ pö
a= = - x0 w cos ç wt - ÷
a 2p T dt è 4ø
y= = a cos w t = a cos t or t = s
2 T 6 é æ pö ù
= x0 w 2 cos ê p + ç wt - ÷ ú
So time to complete 3/8th oscillation ë è 4øû
T T 5T
+ = 2 æ 3p ö
= = x0 w cos ç wt + ÷ ...(1)
4 6 12 è 4ø
1 Acceleration, a = A cos (wt + d) ...(2)
9. (a) As we know, kinetic energy = mw2 (A 2 – x 2 )
2 Comparing the two equations, we get
1 3p
Potential energy = mw x
2 2 A = x0w2 and d = .
2 4
1 M
mw2 (A 2 – x 2 ) T = 2p
2 1 A2 – x 2 1 14. (c)
\ = Þ = k
1
mw2 x 2
4 x2 4
M + m 5T
2 T ' = 2p =
4 2 k 3
4A2 – 4x2 = x2 Þ x 2 = A 2 \x= A.
5 5 M +m 5 M
\ 2p = ´ 2p
k 3 k
l 1.21l
10. (d) T = 2p and T ' = 2p 25
g g M +m = ´M
9
(Q l ' = l + 21% of l) m 25 m 25 16
1+ = Þ = -1 =
T '- T M 9 M 9 9
% increase = ´ 100
T
15. (d) A = 0.05 m, y = 0.01 m
=
1.21l - l
l
´ 100 = ( )
1.21 - 1 ´ 100 Acceleration, a = 1.0 m/s2
We have, a = -w 2 y or | a |= w 2 y
= (1.1 - 1) ´ 100 = 10% Þ 1.0 = w 2 ´ 0.01
m 1.0
11. (d) T = 2p \ w2 = = 100 Þ w = 10
k 0.01
When a spring is cut into n parts 2 p 2p p
Spring constant for each part = nk Now, time period, T = = = sec.
w 10 5
Here, n = 4 16. (c) We have, U + K = E
m T where, U = potential energy, K = Kinetic energy, E =
T1 = 2p = Total energy.
4k 2
Also, we know that, in S.H.M., when potential energy is
12. (d) x = A cos (wt + d ) maximum, K.E. is zero and vice-versa.
y = A cos(wt + a ) ....(1) \U max + 0 = E Þ U max = E
p Further,
When d = a + 1
2 K .E . = mw 2 a 2 cos 2 w t
2
æp ö
x = A cos ç + wt + a ÷ But by question, K .E. = K0 cos 2 wt
è2 ø
1
x = - A sin (wt + a ) ....(2) \ K 0 = mw 2 a 2
2
Squaring (1) and (2) and then adding 1
x2 + y2 = A2 [cos2 (wt + a) + sin 2 (wt + a)] 2 2
Hence, total energy, E = mw a = K 0
or x2 + y2 = A2, which is the equation of a circle. The 2
present motion is anticlockwise. \U max = K 0 & E = K0 .
DPP/ CP13 S-59
17. (b) Distance covered by lift is given by 22. (c) Here all the three springs are connected in parallel to
y = t2 mass m. Hence equivalent spring constant
\ Acceleration of lift upwards k = K + K + 2 K = 4 K.
d2 y 1 2 2 2
= =
d
(2t) = 2 m / s 2 =
g 23. (a) K.E. = mw (a - x )
2
dt 2 dt 5
1 2 2
l When x = 0, K.E is maximum and is equal to mw a .
Now, T = 2p 2
g
24. (d) y = 5sin(p t + 4p ), comparing it with standard equation
l l 5 æ 2p t ö
T ' = 2p = 2p = T. y = a sin(wt + f ) = a sin ç +f ÷
g 6 6
g+ g è T ø
5 5
2p t
18. (d) In simple harmonic motion, starting from rest, a = 5m and = p t Þ T = 2 sec .
At t = 0 , x = A T
x = Acoswt ..... (i) x
displacement
When t = t , x = A – a 25. (b) T =2π = 2p = 2p / z
When t = 2 t , x = A –3a acceleration zx
From equation (i) 26. (b) Here, x = 2 × 10–2 cos p t
A – a = Acosw t ......(ii) Speed is given by
A – 3a = A cos2w t ....(iii) dx
As cos2w t = 2 cos2 w t – 1 ...(iv) v= = 2 × 10–2 p sin p t
From equation (ii), (iii) and (iv) dt
For the first time, the speed to be maximum,
2
A - 3a æ A-a ö p
= 2ç ÷ -1 sin p t = 1 or, sin p t = sin
A è A ø 2
p 1
A - 3a 2 A2 + 2a 2 - 4 Aa - A2 Þ pt = or,, t = = 0.5 sec.
Þ = 2 2
A A2
Þ 2 2 2 R 64 ´ 106 22 8 ´ 10 3
= 2´ ´
A – 3aA = A + 2a – 4Aa T = 2p = 2p
27. (d)
Þ 2a2 = aA g 9.8 7 7´ 2
Þ A = 2a
2 ´ 22 ´ 8 ´ 1000
a 1 = min = 84.6 min
Þ = 49 ´ 60
A 2
Now, A – a = A coswt 28. (a) x = 3sin 2t + 4cos2t . From given equation
A- a p
Þ cos wt = a1 = 3, a2 = 4, and f =
A 2
1 2p p
Þ cos wt = or t= \ a = a12 + a22 = 32 + 42 = 5
2 T 3
Þ T= 6t Þ vmax = aw = 5 ´ 2 = 10
dy1 æ pö 29. (d) Slope of F - x curve = – k = -
80
Þ k = 400 N/m,
19. (b) v1 = = 0.1 ´ 100p cos ç100pt + ÷
dt è 3ø 0.2

dy2 æ pö m
v2 = = - 0.1p sin pt = 0.1p cos ç pt + ÷ Time period, T = 2p = 0.0314 sec.
dt è 2ø k
30. (b) Phase change p in 50 oscillations.
p p 2 p - 3p p Phase change 2p in 100 oscillations.
\ Phase diff. = f1 - f 2 = - = =–
3 2 6 6 So frequency different ~ 1 in 100.
l l¢
20. (c) f A =
1 g f
and f B = A =
1 g 31. (a) T = 2p ; 2 = 2p l = 2p
2p LA 2 2p L B g g (g / 6 )
Time period will remain constant if on moon,
fA 1 g L LB L l' = l/6 = 1/6 m
\ = ´ 2p B Þ 2 = Þ 4= B ,
f A / 2 2p L A g LA LA 32. (b) Let k be the force constant of spring of length l2. Since
regardless of mass l1 = n l2, where n is an integer, so the spring is made of
21. (d) (n + 1) equal parts in length each of length l2.
 EBD_7156
S-60 DPP/ CP13
1 (n + 1) v pmax75p 3p
\ = or k = (n + 1) K then = =
K k v 50 2
The spring of length l1 (= n l2) will be equivalent to n 37. (b) The equivalent situation is a series combination of two
springs connected in series where spring constant springs of spring constants k and 2k.
k If k' is the equivalent spring constant, then
k¢ = = (n + 1) K / n & spring constant of length l2 is
n (k )(2k ) 2 k
k' = =
K(n+1). 3k 3
33. (c) Given
y = 0.2 sin (10pt + 1.5p) cos (10pt + 1.5p) 3m
Þ T = 2p
We know that 2 sinA cos A = sin2A, we get 2k
y = 0.1 sin 2 (10pt + 1.5p) = 0.1 sin (20pt + 3p) æl ö
38. (a) Time period of simple pendulum T = 2p çç ÷÷ µ l
On comparing with wave equation g è ø
y = a sin (wt + f) we get
w = 20 p where l is effective length.
[i.e distance between centre of suspension and centre
2p 2p 1 of gravity of bob]
T= = = sec . = 0.1sec.
w 20p 10 Initially, centre of gravity is at the centre of sphere. When
34. (a) The displacement of a particle in S.H.M. is given by water leaks the centre of gravity goes down until it is
y = a sin (wt + f) half filled; then it begins to go up and finally it again
goes at the centre. That is effective length first increases
dy
velocity = = wa cos (wt + f) and then decreases. As T µ l , so time period first
dt
The velocity is maximum when the particle passes increases and then decreases.
through the mean position i.e., A
39. (d) At t = 0, x = 5 =
æ dy ö 2
çè ÷ø
dt max = w a p
Þ Initial phase, f = 30° =
The kinetic energy at this instant is given by 6
2 Þ x = A sin (wt + f)
1 æ dy ö 1
mç ÷ = mw2 a2 = 8 × 10–3 joule æ 2p pö æ pö
2 è dt ø max 2 = 10sin ç t + ÷ = 10sin ç pt + ÷
è T 6ø è 6ø
1 40. (b) For block A to move in S.H.M.
or × (0.1) w2 × (0.1)2 = 8 × 10–3
2 N
Solving we get w = ± 4
Substituting the values of a, w and f in the equation of A
S.H.M., we get
y = 0.1 sin (± 4t + p/4) metre.
mg x
1 2 2 mean
35. (d) K.E = k ( A - d ) position
2
mg – N = mw2x
1 2 where x is the distance from mean position
and P.E. = kd
2 For block to leave contact N = 0
At mean position d = 0. At extrement positions d = A
g
p Þ mg = mw2 x Þ x =
36. (b) y = 3sin (50t - x) w2
2
æ p ö l l
y = 3sin ç 25pt - x ÷ on comparing with the 41. (a) t = 2p ; t 0 = 2p
è 2 ø g eff g
standard wave equation
y = a sin (wt – kx) Buoyant
force 1000 Vg
w 25p
Wave velocity v = = = 50 m/sec.
k p/2
The velocity of particle 4
´ 1000 Vg
¶y æ p ö 3
vp = = 75p cos ç 25pt - x÷
¶t è 2 ø Weight
vp max = 75p
DPP/ CP13 S-61
Under the action of second force,
æ4 ö 1000
Net force = çè - 1÷ø ´ 1000 Vg = Vg
F2 = mw 22 y
3 3
Under the action of resultant force,
1000 Vg g
g eff = = F1 + F2 = mw 2 y
4 4
3 ´ ´ 1000 V
3 Þ mw 2 y = mw12 y + mw 22 y
l
\ t = 2p Þ w 2 = w12 + w 22
g/4
2 2 2
t = 2t0 æ 2p ö æ 2p ö æ 2p ö
Þ ç ÷ = ç ÷ +ç ÷
42. (a) K.E. of a body undergoing SHM is given by, è Tø è T1 ø è T2 ø
1 2 2 2
K .E. = ma w cos wt 2 2
2 æ 4 ö æ 3ö
1 T12 T22 çè ÷ø × çè ÷ø
2 2 5 5 12
T .E. = ma w ÞT= = = .
2 T12 + T22 æ 4ö
2
æ 3ö
2
25
Given K.E. = 0.75 T.E. çè ÷ø + çè ÷ø
5 5
2 p
Þ 0.75 = cos wt Þ wt = 44. (d) F = – bV, b depends on all the three i.e, shape and size
6 of he block and viscosity of the medium.
p p´2 1 45. (c) When the bob moves from maximum angular
Þt= Þt= Þt= s displacement q to mean position, then the loss of
6´w 6 ´ 2p 6
gravitational potential energy is mgh
43. (a) Under the action of first force, F1 = mw12 y where h = l(1 – cos q)