OSCILLATION AND WAVES 1

EXERCISE - 1 : BASIC OBJECTIVE QUESTIONS

SHM Basics
1
So, m 2 A 2  8  103
1. Which of the following equations does not represent a simple 2
harmonic motion ?
(a) y = a sin  t 1
 (0.1)  2 (0.1)2  8  103
2
(b) y = b cos  t
(c) y = a sin  t + b cos  t  4
(d) y = a tan  t
So, x  A sin (t  )
Ans. (d)
= 0.1 sin (4t + 45º)
2
d y 3. The displacement of a particle is represented by the equation:
Sol.    2 y [Differential equations of SHM]
2
dt  
y = 3 cos   2t 
4 
d2 y
(a)    2 a sin t The motion of the particle is :
dt 2
(a) simple harmonic with period 2/
(b) simple harmonic with period /
d2 y
(b)    2 b cos t
dt 2 (c) periodic but not simple harmonic
(d) non-periodic
d2 y Ans. (b)
(c)    2 (a sin t  b cos t)
2
dt
 
Sol. y  3 cos   2t 
dy 4
(d)  a  sec2 (t)
dt
  
 3 sin     2t  
d2 y 2 4 
 a  2 (2sec t) (sec t) (tan t)
dt 2
 
2  3 sin  2t  
d y  4
Hence 2
   2 y not the equation of SHM
dt
2 
2. A particle of mass 0.1 kg is executing SHM of amplitude 0.1 m. T 
2 
When the particle passes through the mean position, its
–3
K.E. is 8 × 10 J. The equation of motion of the particle, 4. For the two curves choose the correct option.
o
when the initial phase of oscillation is 45 , is
(a) y = 0.1 cos (3 t + /4) (b) y = 0.1 sin (3 t + /4)
(c) y = 0.1 sin (4 t + /4) (d) y = 0.1 cos (4 t + /4)
Ans. (c)
Sol. A = 0.1, m = 0.1 kg
OSCILLATION AND WAVES 2

(a) (Amplitude)A > (Amplitude)B  t1  0.1 sec and t 2  0.5 sec

(b) (Time period)A > (Time period)B So time taken = (0.5 – 0.1) = 0.4 s
(c) Both (a) and (b) are correct 7. The displacement of an oscillating particle varies with time
(d) Both (a) and (b) are incorrect
 t 1
Ans. (d) (in seconds) according to the equation y = sin   ,
2 2 3
Sol. In the given figure the (Amplitude)A < (Amplitude)B
where y is in cm. The maximum acceleration of the particle is
(Timeperiod)A = (Time period)B
approximately
5. The phase difference between x1 = A sin t and x2 = A cos –2 –2
(a) 0.62 cm s (b) 1.81 cm s
t is :
–2 –2
(c) 3.62 cm s (d) 5.2 cm s
(a) /2 (b) 4
Ans. (a)
(c) /3 (d) /6
Ans. (a)   
Sol. y  sin  t   so   , A  1cm
Sol. x1  sin t 4 6 4

 
x 2  cos t  sin  t    2 
a max  2 A     0.616 cm/s
2
 2
 16 
 

hence phase shift 
2
 0.62 cm / s 2
Phase, Displacement, Velocity, Acceleration, Kinetic, Potential
Energy as a funtion of time and positon 8. The kinetic energy of a particle executing S.H.M. is 16 J
when it is in its mean position. If the amplitude of oscillations
6. A particle is performing S.H.M. along X-axis with amplitude
is 25 cm and the mass of the particle is 5.12 kg, the time
4 cm and time period 1.2 sec. The minimum time taken by the
period of its oscillation in second is
particle to move from x = + 2 cm to x = + 4 cm and back again
is given by (a) /5 (b) 2 

(a) 0.6 s (b) 0.4 s (c) 5  (d) 20 

(c) 0.3 s (d) 0.2 s Ans. (a)
Ans. (b)
1 1
Sol. m 2 A 2  16   5.12  2  (0.25)2  16
Sol. x  A sin (t) 2 2

1 16  20
2  A sin (t)  sin t   w2   10 rad/s
2 (5.12  (0.25) 2

 5 2 2 
 t1  and t 2  T   sec
6 6  10 5

 5 9. Two pendulums of time period 3 s and 8 s respectively starts

 t1  and t 2  oscillating simultaneously from two opposite extreme
6 6
positions. After how much time they will be in the same
phase ?
 5
 t1  and t 2 
 2   2  24 12
6  6  (a) s (b) s
 1.2   1.2  5 5
 OSCILLATION AND WAVES 3

24 12 x  A (sin t)
(c) s (d) s
11 11 Hence at t = 0 particle is at mean position
Ans. (b)
So for particle to be at half the maximum value
Sol. y1  A sin (1t  )
Let’s say time taken is t0 then
y 2  A sin ( 2 t)
A
They are in the same phase  A sin (t 0 )
2
1t     2 t
( 2  1 ) t   1 
 sin t 0   t 0 
2 6
 1 1
   2t  
3 8   T
 t0   
6  2  12
5 6 
 2t  1  T
24
12. A particle is executing simple harmonic motion with
12
t sec frequency f. The frequency at which its kinetic energy
5
changes into potential energy is
10. The force constant of a weightless spring is 16 N/m. A body
(a) f/2 (b) f
of mass 1.0 kg suspended from it is pulled down through
5 cm from it’s mean position and then released. The maximum (c) 2 f (d) 4 f
kinetic energy of the system (spring + body) will be Ans. (c)
–2 –2
(a) 2 × 10 J (b) 4 × 10 J Sol. In one oscillation KE converts into PE twice as particles
(c) 8 × 10 J
–2
(d) 16 × 10 J
–2 crosses mean position twice.

Ans. (a) 13. A particle executes simple harmonic motion between x = –A

and x = + A. The time taken for it to go from 0 to A/2 is t1 and
Sol. When 1.0 kg body is suspended from spring, at its
to go from A/2 to A is t2. Then
equilibrium position Mg = Kx
(a) t1 < t2 (b) t1 > t2
1 g g (c) t1 = t2 (d) t1 = 2 t2
x  m
16 16
Ans. (a)
Since spring is pulled from mean position by 5 cm. So
potential energy stored in it will appears as Max K.E. Sol.
2
1  5 
So, K.E. Max =  16  
2  100 
T
 8  25  104 Time taken by the particle from x = 0 to x = +A is
4

 2  102 J A
Time taken from 0 to
11. The time taken by a particle executing S.H.M. of period T to 2
move from the mean position to half the maximum
A
displacement is  A sin (t1 )
2
(a) T/2 (b) T/4
(c) T/8 (d) T/12 1
sin (t1 ) 
Ans. (d) 2
Sol. Lets say equation of SHM is

t1 
6
OSCILLATION AND WAVES 4

(a) (/10) s (b) (/20) s

2 
t1  (c) (/50) s (d) (/100) s
T 6
Ans. (d)
T
t1  sec Sol. TE = PE + KE
12
At M.P. (PE) is min so K.E is maximum
A Hence 9 J  (KE) max  5J  (KE) max  4 J
Time taken from to A
2
(KE)max = 4 J
T
t2   t1 1
4  m 2 A 2  4 J
2
T T
t2   1
4 12  (2)  2 (0.01)2  4
2
T
t2  4 2
6     200 rad / s
2 0.01
(0.01)
t 2  t1
14. For a particle executing simple harmonic motion, the 2 2 
T   sec
displacement x is given by x = A sin t. Identify the graph,  200 100
which represents the variation of potential energy (PE) as a
16. A mass m is performing linear simple harmonic motion, then
function of time t and displacement x
which of the following graph represents correctly the variation
II of acceleration a corresponding to linear velocity v ?
Energy

v2 v2

t
(a) (b)

a2 a2
Energy
v2 v2
III IV

(c) (d)
x
–A +A a2 a2
(a) I, III (b) II, III Ans. (d)
(c) I, IV (d) II, IV Sol. x  A sin t
Ans. (b) dx
v  A cos t
1 1 dt
Sol. PE  Kx 2  K A 2 sin t
2 2  v 2   2 A 2 cos 2 t ........ (1)
15. The potential energy of a harmonic oscillation of mass 2 kg dv
in its mean position is 5 J. If its total energy is 9 J and its a   2 A sin t  a 2   4 A 2 sin 2 t ......(2)
at
amplitude is 0.01 m, its time period of oscillation will be
 2 v2  a 2   4 A 2 cos 2 t   4 A 2 sin 2 t

  2 v2  a 2   4 A 2
 OSCILLATION AND WAVES 5

 ky  x  C Sol. x = A sin t = A sin 2t

17. The relation between acceleration and displacement of four v = A cos t = 2A cos 2t
particles are given below : A 1
 A sin 2t  sin 2t 
which one of the particles is executing simple harmonic 2 2
motion ?
(a) ax = + 2x (b) ax = + 2x
2 3
 cos 2t 
2 2
(c) ax = – 2x (d) ax = – 2x
Ans. (d) 3
v  2A  3A
Sol. For SHM 2
20. For a particle in SHM, if the amplitude of the displacement
a   2 x is a and the amplitude of velocity is v the amplitude of
18. A particle executing SHM has a maximum speed of 30 cm/s acceleration is :
2
and a maximum acceleration of 60 cm/s . The period of
oscillation is : v2
(a) va (b)
a

(a)  s (b) s
2 v2 v
(c) (d)
 2a a
(c) 2 s (d) s
4 Ans. (b)
Ans. (a)
Sol. Given amplitude = a
Sol. max  A
v
30  A ........(1) Maximum velocity  a  v  a   
a
a max  A 2
2
2 v v2
60  A 2 ......... (2) Maximum acceleration  a  a   
a a
divide (2) by (1)
21. The displacement of an object attached to a spring and
2 –2
executing simple harmonic motion is given by x = 2×10
2 cos t m. The time at which the maximum speed first occurs
2
T is :
T   sec (a) 0.5 s (b) 0.75 s
A (c) 0.125 s (d) 0.25 s
19. A block is left from x = + A, its speed at x  is
2 Ans. (a)
( = 2 rad/s). –2
Sol. Given, displacement x = 2 × 10 cos t

dx
The magnitude of velocity, v   2102 sin t
dt

So, the velocity will be maximum when sin t  1 where

(a) (3A) m/s (b) ( 3A) m/ s
 3
t  , , ......
(c) (2A) m/s (d) ( 2A) m/ s 2 2

Ans. (b)
OSCILLATION AND WAVES 6

 1 2
So for first maximum, t  or t  0.5s potential energy   kx  0
2 2
22. What is the ratio of maximum acceleration to the maximum 25. A body executes simple harmonic motion. The potential
velocity of a simple harmonic oscillator ? energy (PE), the kinetic energy (KE) and total energy (TE)
(a)  (b) /2 are measured as function of displacement x. Which of the
(c) 1/ (d) 2 following statements is true ?

Ans. (a) (a) KE is maximum when x = 0

Sol. maximum velocity  a (b) TE is zero when x = 0
(c) KE is maximum when x is maximum
maximum acceleration  a2
(d) PE is maximum when x = 0
2
a Ans. (a)
ratio  
a
Sol. x  A sin t
23. If a conservative force is acting on a system in SHM. The
total mechanical energy is : 1
KE  mA 2 2 cos 2 t
2
(a) time dependent
(b) position dependent 1
PE  mA 2 2 sin 2 t
(c) amplitude dependent 2

(d) Both (a) and (c) are correct 1

TE  mA 2 2
Ans. (c) 2

1 KE is max at x = 0 min at extremes

Sol. E mA 2 2
2 PE is min at x = 0 max at extremes
A – amplitude TE is constant
m – mass Spring Block Oscillation
26. A force of 6.4 N stretches a vertical spring by 0.1 m. The
 – angular frequency
mass that must be suspended from the spring so that it
24. The expression for displacement of an object in SHM is x
oscillates with a period of (/4) second is
T
= A cos (t). The potential energy at t  is : (a) (/4) kg (b) 1 kg
4
(c) () kg (d) 10 kg
1 2 1 2
(a) kA (b) kA Ans. (b)
2 8
Sol. F = kx  6.4 = k (0.1)  k = 64 N/m
1 2
(c) kA (d) zero m
4 T  2
k
Ans. (d)
Sol. x  A cos t  m
 2
4 64
T  2 T 
at t  , x  A cos     0
4  T 4 m 1 m 1
     m  1kg
64 8 64 64

27. A mass of 1 kg attached to the bottom of a spring has a

certain frequency of vibration. The following mass has to
 OSCILLATION AND WAVES 7

be added to it in order to reduce the frequency by half.

A1  2
(a) 1 kg (b) 2 kg 
A 2 1
(c) 3 kg (d) 4 kg
Ans. (c) K

m
Sol. f  

 K
k k
f1 m1 f 1 A1 k2
   
f2 k f k A2 k1
m2 2 m 1
30. The period of oscillation of a mass m suspended from a
spring is 2 seconds. If along with it another mass of 2 kg is
 m 1  2 also suspended, the period of oscillation increases by one
second. The mass m will be
 m  3kg
(a) 2 kg (b) 1 kg
28. If a spring has time period T, and is cut into n equal parts, (c) 1.6 kg (d) 2.6 kg
then the time period of each part will be
Ans. (c)
T
(a) T n (b) m
n Sol. T  2
K
(c) nT (d) T
Ans. (b) m
So 2  2
K
m
Sol. For a spring, T  2
k
m2
3  2
For each piece, spring constant = nk K

m m2 3
 T   2  
nk m 2

m 1 T  4 (m  2)  9 m
 T   2   .
k n n
 5m = 8
29. Two bodies M and N of equal masses are suspended from  m = 1.6 kg
two separate massless springs of spring constant k1 and k2
31. In the figure shown below, the block is moved sideways
respectively. If the two bodies oscillate vertically such that
by a distance A. The |net force| on the block is :
their maximum velocities are equal, the ratio of the amplitude
of M to that of N is

(a) k1/ k2 (b) k1 / k 2

(c) k2 / k1 (d) k 2 / k1 (a) (k1 – k2) A (b) (k2 – k1) A

Ans. (d) (c) (k1 + k2) A (d) None of the above
Ans. (c)
Sol. ( max )1  ( max )2
Sol. The spring on the left of the block is elongated and that
A11  A 2 2 on right is compressed both exert a force on the block
OSCILLATION AND WAVES 8

towards the mean position and they add up. time periods of two systems are T1 and T2. The relation
between T1 and T2 is :
Feff  A (k1  k 2 )

32. A block is in SHM on a frictionless surface as shown in

the figure. The position x = 0 shows the unstretched
position of the spring.

T2
(a) T1 
2
Choose the correct option with reference to the above
system. T1
(b) T2  T
(a) +A and –A are maximum displacements where, A is 2

amplitude (c) T1 = T2
(b) x = 0 indicate the equilibrium position (d) Cannot establish the relation between them
(c) The block executes to and fro motion about the mean Ans. (c)
position, when pulled aside and released
Sol. For both systems both mass and spring constant are the
(d) All of the above same.
Ans. (d)
Sol. A - amplitude and +A, -A are maximum displacements. x = m
T  2 , T1  T2
0 is mean position and block exectues SHM k

33. The time period of a spring mass system shown below is 35. If the block is pulled by a distance x and left, the block will
equal to : start oscillating. The value of x, so that at a moment when
speed of the block become zero the spring become
unstretched.

2m m
(a) 2 (b) 2
k 4k

2 2 m
(c) (d) None of these (a) mg/2k (b) 2k/mg
k
(c) mg/k (d) 2mg/k
Ans. (c) Ans. (d)
Sol. K eff  k  k  2k
1 2
Sol. mgx  kx
2
m m
T  2  2
K eff 2k 2mg
x
k
m 22 m 36. Two identical springs are connected in series and parallel
T  42  
2k k as shown in the figure. If fs and fp are frequencies of series

34. There are two springs mass systems as shown in the fs

and parallel arrangements, what is ?
figures. Resistive forces are absent everywhere. Let the fp
 OSCILLATION AND WAVES 9

pendulum will be

2T 5T
(a) (b)
5 2

5 2
(c) (d)
2T 5T
Ans. (a)
(a) 1 : 2 (b) 2 : 1

(c) 1 : 3 (d) 3 : 1 Sol. T  2 .... (1)
g
Ans. (a)
When lift starts accelerating
1 k
Sol. f  f  K eff 
2 m T   2
g eff

k g 5g
Ks  , K p  2k g eff  g  
2 4 4

k 4
T   2 ....... (2)
fs Ks 2  1 1 5g
  
fp Kp 2k 4 2 divide 1 by 2

37. A coin is placed on a horizontal platform which undergoes T 2


vertical simple harmonic motion of angular frequency . The T 5
amplitude of oscillation is gradually increased. The coin will 39. A simple pendulum is oscillating without damping, as showm
leave contact with the platform for the first time (2006) in figure. When the displacement of the bob is less than

(a) at the highest position of the platform maximum, its acceleration vector a is correctly shown in
(b) at the mean position of the platform

g
(c) at an amplitude of
2

g2 (a) (b)
(d) for an amplitude of a
2
Ans. (c) a
Sol. In vertical simple harmonic motion, maximum acceleration
(a 2 ) and so the maximum force (ma 2 ) will be at extreme
positions. At highest position, force will be towards mean
position and so it will be downwards. At lowest position,
force will be towards mean position and so it will be upwards. (c) (d)
This is opposite to weight direction of the coin. The coin
will leave contact with the platform for the first time when m a
a
2
(a )  mg at the lowest position of the platform.
Ans. (c)
Angular SHM
38. A man measures time period of a simple pendulum inside a
stationary lift and find it to be T. If the lift starts accelerating
upwards with an acceleration g/4, then the time period of
OSCILLATION AND WAVES 10

(T1/T2) is :
(a) 1/9 (b) 1/3

(c) 3 (d) 9

Ans. (b)
Sol.
l
Sol. T  2
g

Hence resultant acceleration should be given by vectorial g g

sum of centripetal acceleration and tangential g at height 2R  2

 2R  9
(acceleration) (at). l  
 R 
40. If we do an experiment by swinging a small ball by a thread
of length 100 cm, what will be the approximate time for
1
complete to and fro periodic motion ? As T 
g
(a) 4s (b) 2s
(c) 6s (d) 1s
g
Ans. (b) T1 9  T1  1

T2 g T2 3
l 1m
Sol. T  2  2  2s
g 9.8 m / s 2 43. A hollow sphere is filled with water through the small hole
in it. It is then hung by a long thread and made to oscillate.
41. A simple pendulum is suspended from the ceiling of a lift. As the water slowly flow out of the hole at the bottom, the
When the lift is at rest its time period is T. With what period of oscillation will :
acceleration should the lift be accelerated upwards in order
(a) continuously decrease
to reduce its period to T/2? (g is acceleration due to
gravity). (b) continuously increase

(a) 2g (b) 3g (c) first decrease then increase

(c) 4g (d) g (d) first increase then decrease

Ans. (b) Ans. (d)

Sol. The time period of the pendulum
l
Sol. T  2
g l
T  2
g
as the lift moves up g becomes g + a

1  T l
As T 
g Initially the centre of mass of the sphere is at the centre of
the sphere. As the water slowly flows out of the hole at
T ga the bottom, the CM of the liquid (hollow sphere) first goes
  g  a  4g  3g  a on downward and then upward.
T g
2 hence, the effective length of the pendulum first increases
and then decreases.
42. A simple pendulum has a time period T1 when on the Earth’s
44. What is the velocity of the bob of a simple pendulum at its
surface and T2 when taken to a height 2R above the Earth’s
mean position, if it is able to rise to vertical height of 10
surface, where R is the radius of the Earth. The value of –2
cm? (g = 9.8 ms ).
 OSCILLATION AND WAVES 11

The amplitude and maximum velocity will be respectively

(a) 5, 10 (b) 3, 2
(c) 3, 4 (d) 4, 2
Ans. (a)
Sol. x = 3 sin 2t + 4 cos 2t

 
–1 –1
 3 sin (2t)  4 sin  2t    A net (sin 2t  d)
(a) 2.2 ms (b) 1.8 ms  2
–1 –1
(c) 1.4 ms (d) 0.6 ms

Ans. (c) A net  A12  A 22  2A1 A 2 cos  
2
Sol. From conservation of energy

1  32  42  2 (3) (4) 0  5
m 2  mgh    2gh
2
Vmax  A  2 (5)  10
   2  9.8  0.1  1.96 47. The displacement of a particle varies with time according to
the relation : y = asin t + bcos t.
   1.4 m / s
Choose the correct statement.
Miscellaneous SHM problems (a) The motion is oscillatory but not SHM.
45. Four simple harmonic motions ; x1 = 8 sin t ; x2 = 6 sin (b) The motion is SHM with amplitude (a + b).
(t + /2) ; x3 = 4 sin (t + ) and x4 = 2 sin (t + 3 /2) are 2 2
(c) The motion is SHM with amplitude (a + b )
superimposed on each other. The resulting amplitude and
its phase difference with x1 are respectively (d) The motion is SHM with amplitude a 2  b2 .
–1
(a) 20, tan (1/2) (b) 4 2 , /2
Ans. (d)
–1
(c) 20, tan (2) (d) 4 2 , /4 Sol. y  a sin t  b cos t
Ans. (d)
 
 a sin t  b sin  t  
Sol. x1  8 sin t  2

 
x 2  6 sin  t    6 cos t  a 2  b 2 sin (t  d)
 2

b
x 3  4 sin  t     4 sin t where tan d =
a

 3  48. A block of mass 1 kg hangs without vibrating at the end of

x 4  2 sin  t     2 cos t
 2  a spring whose force constant is 200 N/m and which is
attached to the ceiling of an elevator. The elevator is rising
x1  x 2  x 3  x 4  4 sin t  4 cos t with an upward acceleration of g/3 when the acceleration
suddenly ceases. The angular frequency of the block after
 4 2 sin (t  45º ) the acceleration ceases is

 (a) 13 rad/s (b) 14 rad/s

Hence A  4 2, phase diff  rad
(c) 15 rad/s (d) None of these
4
46. The displacement equation of a particle is Ans. (b)

x = 3 sin 2 t + 4 cos 2 t
 OSCILLATION AND WAVES 12

As there is positive sign between x and t terms, hence wave

K travel in -x direction.
Sol.  
m
Coefficient of t b
Wave speed  
200 Coefficient of x a
  14.1 rad / s
1
51. A particle of mass m moving along the x-axis has a potential
2
49. A body is moving in a room with a velocity of 20 m/s energy U(x) = a + bx where a and b are positive constants.
perpendicular to the two walls separated by 5 m. There is no It will execute simple harmonic motion with a frequency
frictrion and the collisions with the walls are elastic. The detemined by the value of
motion of the body is (a) b alone (b) b and a alone
(a) not periodic (c) b and m alone (d) b, a and m alone
(b) periodic but not simple harmonic Ans. (c)
(c) periodic and simple harmonic
Sol. U  a  bx 2
(d) periodic with variable time period
Ans. (b) dU
F
Sol. Motion will be periodic because ball repeats its motion after dx
regular interval. The periodic motion will not be SHM as
force should be directly proportional to negative of F   [0  2bx]
displacement. F = – 2bx
 
Fx K = 2b

50. The transverse displacement y (x, t) of a wave on a string is K 2b

given by  
m m

y  x, t   e

 ax 2  bt 2  2 ab xt  52. A metre stick swinging in vertical plane about a fixed
horizontal axis passing through its one end undergoes small
This represents a (AIEEE 2011) oscillation of frequency f0. If the bottom half of the stick
were cut off, then its new frequency of small oscillation
b would become.
(a) wave moving in –x direction with speed
a

(b) standing wave of frequency b

1
(c) standing wave of frequency
b

a (a) f0 (b) 2 f0
(d) wave moving in +x direction with speed
b
(c) 2f0 (d) 2 2 f 0
Ans. (a)
2 2
Ans. (b)
Sol. y (x, t)  e  (ax  bt  2 abxt)
g 1
 ( ax  bt )2
Sol. f    
y (x, t)  e ...... (i)  

Comparing equation (i) with standard equation

y (x, t) = f (ax + bt)
 OSCILLATION AND WAVES 13

Applying work energy theorem

f 
  f  2 f0 w g  KE
f
So 0 
2
1 2
 mg (2l )  I  0
53. Which of the following expressions corresponds to simple 2
harmonic motion along a straight line, where x is the
displacement and a, b, c are positive constant?
4 mgl
(a) a + bs – cx2 (b) a – bx + cx2  
(c) bx2
(d) – bx I
Ans. (d) Where I is moment of inertia of body about point of
Sol. In SHM the restoring force  discplacement from mean suspension.
position (x = 0).
For small angular displacement of body about stable
If x  0 at mean position than force is a linear function of equation position we can write
x and direction of force always toward mean position.
2
In Option (a), (b) and (c) function depends upon x
 it can not be SHM
In option (d) it is (–bx)
 F = –bx (it is SHM)
54. A physical pendulum is positioned so that its centre of
gravity is above the suspension point. When the pendulum
is released it passes the point of stable equilibrium with an
angular velocity . The period of small oscillations of the
pendulum is

4 2
(a) (b)
 

     mgl sin 
(c) (d)
 2
 I    mgl 
Ans. (a)
Sol. Let’s assume the centre of gravity is at height l above point  mgl 
  
of suspension (0) (as shown)  I 
Comparing with standard SHM equation.

   2 
We can write angular frequency of angular SHM motion

mgl 
0  
I 2

 2 4
So T   
0  / 2 

55. A wave y = a sin (t – kx) on a string meets with another

wave producing a node at x = 0. Then the equation of the
unknown wave is :
(a) y = a sin (t + kx) (b) y = – a sin (t + kx)
OSCILLATION AND WAVES 14

(c) y = a sin (t – kx) (d) y = – a sin (t – kx) (2n  1) 

 x
Ans. (b) 2 2
Sol. For standing wave, wave will reflect from denser end with
Waves
phase difference .
Wave Parameter
y   a sin (t  kx)
58. A string of 5.5 m length has a mass 0.035 kg. If the tension in
56. Length of a string tied to two rigid supports is 40 cm. the string is 77 N, then the speed of wave on the string is :
Maximum length (wavelength in cm) of a stationary wave (a) 77 m/s (b) 102 m/s
produced on it is :
(c) 110 m/s (d) 164 m/s
(a) 20 (b) 80 Ans. (c)
(c) 40 (d) 120 Sol. Speed of wave
Ans. (b)
Sol. Maximum wavelength is when string has only 1 antinode T 77 5.5  77
    110 m / s
and at that time m 0.035 / 5.5 0.035
l

L
2 59. The equation which represents a sinusoidal (harmonic)
= 2 × L wave travelling along the positive direction of the X-axis
= 2 × 40 is :
= 80 cm (a) y (x, t) = a sin (kx – t + )
57. A travelling wave represented by y = A sin (t – kx) is (b) y (x, t) = a sin (kx + t + )
superimposed on another wave represented by
(c) y (x, t) = a sin (kx + t)
y = A sin (t + kx). The resultant is
(d) y (x, t) = a sin (t + kx)
 1 Ans. (a)
(a) A standing wave having nodes at x   n   ,
 22
n = 0, 1, 2 Sol. y  a sin (kx  t  ) travels along + ve X - axis.
(b) A wave travelling along + x direction 60. A sinusoidal travelling wave is described by y (x, t) = a sin
(c) A wave travelling along –x direction (kx – t + ), where y(x, t) is the displacement as a function
of position x and time t.
n
(d) A standing wave having nodes at x  ; n = 0, 1, 2 With reference to the above equation, match the items in
2
Column-I with terms in Column-II and choose the correct
Ans. (a) option from the codes given below.
Sol. y net  y1  y 2 Column-I Column-II
(A) a denotes 1. angular frequency of the wave
 2A sin (t) cos (kx)
(B)  denotes 2. angular wave number
This is equation of a stationery waves for nodes, ynet = 0
(C) k denotes 3. amplitude of the wave
 cos (kx)  0
(D)  denotes 4. Initial phase angle at x = 0, t = 0
 Codes
 kx  (2n  1)
2 A B C D

2  (a) 3 2 4 2
 x  (2n  1)
 2 (b) 3 2 1 4
(c) 3 1 2 4
 OSCILLATION AND WAVES 15

(d) 4 1 2 3 64. The equation of progressive wave is

Ans. (c)  t x 
y  0.2sin 2    , where x and y are in metres
Sol. Here  0.01 0.3 
A - amplitude and t is in seconds. The velocity of propagation of the
 - angular frequency wave is :
–1 –1
K - angular wavenumber (a) 30 ms (b) 40 ms
–1 –1
(c) 300 ms (d) 400 ms
 - initial phase
Ans. (a)
61. The minimum distance between the two points having the
same phase is : 1
(a) wavelength of the wave  0.01
   0.3  100  30 m / s
Sol. k 1
(b) amplitude of the wave
0.3
(c) wave number
65. A cylinderical tube, open at both ends, has a fundamental
(d) frequency of the wave
frequency, f, in air. The tube is dipped vertically in water so
Ans. (a) that half of it is in water. The fundamental frequency of the
Sol. The wavelength in a longitudinal wave is the distance air-column is now
between two consecutive points that are in phase.
f
62. A wave equation is given by (a) f (b)
2
  t x 1  3f
y  4 sin        (c) (d) 2f
  5 9 6  4

where, x is in cm and t is in seconds. The wavelength of Ans. (a)

the wave is: Sol. When tube is opened at both ends

(a) 18 cm (b) 9 cm 1
f Vs ........(1)
(c) 36 cm (d) 6 cm 2L
When tube is cut into two equal part and closed at one end
Ans. (a)
1
2 2 f  Vs
   18 cm l ....... (2)
k  4
Sol. 2
9
From (1) and (2)
63. The distance travelled by the wave pattern in the time
f  f
required for one full oscillation by any constituent of the
medium is equal to : 66. A long string having mass density as 0.01 kg/m is
subjected to a tension of 64 N. The speed of the transverse
(a) wavelength of the wave
wave on the string is:
(b) amplitude of the wave (a) 100 ms
–1
(b) 120 ms
–1

(c) wave number of the wave (c) 80 ms

–1
(d) 90 ms
–1

(d) both (a) and (b) Ans. (c)

Ans. (a)
T 64
Sol. The distance travelled by the wave pattern in the time Sol.  
 0.01
required for one full oscillation by any constituent of the
medium is equal to wavelength.
  6400  80m / s
OSCILLATION AND WAVES 16
–4
Standing Wave on Stretched String 69. A stretched string of length 1 m and mass 5 × 10 kg, fixed
67. A standing wave consisting of 3 nodes and 2 antinodes is at both ends, is under a tension of 20 N. If it is plucked at
formed between the two atoms having a distance of 1.21 Å points situated at 25 cm from one end, it would vibrate with
between them. The wavelength of the standing wave is : a frequency :

(a) 1.21 Å (b) 2.42 Å (a) 400 Hz (b) 200 Hz

(c) 6.05 Å (d) 3.63 Å (c) 100 Hz (d) 256 Hz

Ans. (a) Ans. (b)

V
Sol. f 


Sol.
T 20
Where V = velocity of wave = 
 5  104
For three nodes and 2 Anti-nodes situation will be similar as
shown.  4  104  200 m / s
Hence  = 1.21 Å.
Since it’s plucked at 25 in from node hence antinode will be
68. A string is stretched between fixed points separated by 75.0 cm.
formed at 25 cm from node as such
It is observed to have resonant frequencies of 420 Hz and
315 Hz. There are no other resonant frequencies between 
 25 cm    100 cm
these two. Then the lowest resonance frequency for this 4
string is f = 200 Hz
(a) 1.05 Hz (b) 1050 Hz Direction for questions 79 to 81
(c) 10.5 Hz (d) 105 Hz Answer to these questions are based on the given
Ans. (d) paragraph. Choose the correct option from those given
Sol. For wire with fixed ends it’s ends should behave as nodes below for each question.
as such A standing wave is formed on a string fixed at both the
ends. The individual waves i.e., incident wave and reflected
 2l wave are y1 (x, t) = a sin (kx – t) and y2 (x, t) = a sin (kx +
n l 
2 n t), respectively. The two waves have same wavelength
‘’.
V
f 70. The position of nodes is given as

n
(a) x  ; n = 0, 1, 2, 3, ...
 V 2
n 
 2l 
(2n  1)
(b) x  ; n = 0, 1, 2, 3, ...
Hence frequencies will be in the ratio of 1 : 2 : 5.7 .... and so 2
th
on. Let’s assume 315 Hz as n harmonic then 420 Hz will be (c) x = n; n = 0, 1, 2, 3, ...
th
(n + 1) harmonic (d) x = (2n + 1) ; n = 0, 1, 2, 3, ...
i.e. nf0 = 315 Ans. (a)
(n +1) f0 = 420 Sol. a sin (kx  t)  a sin (kx  t)
 f0 = 105 Hz
 OSCILLATION AND WAVES 17

 2a sin (kx) cos (t) Hence frequency decreases.

74. Column-I has figures showing different modes of
n n n oscillation of the system (a string tied at both the ends)
nodes at kx  n, x   
k 2 2 and Column-II has name of the corresponding modes.
 Match the items in Column I with terms in Column II and
71. The position of anti-nodes is given by the equation, choose the correct option from the codes given below.

 1
(a) x   n   ; n = 0, 1, 2, 3, ...
 2 2

n
(b) x  ; n = 0, 1, 2, 3, ...
2

(2n  1)
(c) x  ; n = 0, 1, 2, 3, ...
2

(d) x = (2n + 1); n = 0, 1, 2, 3, ...

Ans. (a)

Sol. a sin (kx  t)  a sin (kx  t)

 2a sin (kx) cos ( t)

(2n  1) 
anti nodes at kx  .
2

1 Codes :

n 
(2n  1)   (2n  1)  2 A B C D
x  
2k 2 2 (a) 4 2 3 1
2

(b) 4 3 1 2
72. The distance between any two consecutive anti-nodes is (c) 3 2 1 4

(a)  (b) /2 (d) 2 3 1 4

Ans. (b)
(c) 3/2 (d) 2
th
Sol. N harmonic has n loops
Ans. (a)
75. A uniform wire of length L, diameter D and density S is
Sol. Distance between two successive anti nodes or nodes is
stretched under a tension T. The correct relation between
wavelength.
its fundamental fequency f, the length L and the diameter
73. When temperature increase, the frequency of a tuning D is :
fork (2002)
(a) increases 1 1
(a) f  (b) f 
LD L D
(b) decreases
(c) remains same 1 1
(c) f  2 (d) f 
(d) increases of decreases depending on the material D LD 2
Ans. (b) Ans. (a)
Sol. When temperature increases, l increases. Sol. Given that,
Frequency = f
Diameter = D
OSCILLATION AND WAVES 18

Length = L 77. When sound propagates through air, the region of high
Density =  density of air molecules is called:
(a) compression (b) rarefaction
Tension = T
(c) denser (d) None of the above
Now, we know that
Ans. (a)
c Sol. When air is in compression more number of air molecules
f 
 are located in that volume hence more dense.
78. The relation for Bulk modulus of a medium is given by
1 T
f 
2L  p V
(a) B   (b) B  
V / V p / p
 is mass per unit length

p p
D2 (c) B  (d) B  
   V / V V / V
4
Ans. (a)
Now, the frequency is
p
Sol. B
1 T V
f 
2L D2 V

4 79. Three sound waves of equal amplitudes have frequencies
(f – 1), f, (f + 1). They superpose to give beats. The number
1 of beats produced per second will be
f
LD (a) 4 (b) 3
(c) 2 (d) 1
1 Ans. (c)
Hence, the correct relation is f 
LD
Sol. Beat produced between (f–1) and f is 1. Beat produced
76. A wire under tension vibrates with a fundamental between f and (f + 1) is 1. Beat produced between (f–1)
frequency of 600 Hz. If the length of the wire is doubled, and (f + 1) is between (f–1) and (f + 1) is 2.
the radius is halved and the wire is made to vibrate under  No. of beats produced per second will be 2.
one-ninth the tension. Then, the fundamental frequency 80. The speed of a longitudinal wave in air is given by:
will become:
(a) 400 Hz (b) 600 Hz  B
(a) v  (b) v 
(c) 300 Hz (d) 200 Hz B 

Ans. (d)
(V / V)
(c) v  (d) both (a) and (c)
T p
Sol. f 
LR
Ans. (b)

f1 T L R f
  1 2 2  1 3 
B
f2 T2 L1R1 f2 Sol.


f1 600 p
 f2    200 Hz B
3 3 V
Sound Wave V
 OSCILLATION AND WAVES 19

1 32
p (c) (d)
 8 17
V
 V   pV Ans. (b)
 V
Sol. Let one mole of each gas has same volume as V. When
they are mixed, then density of mixture is
81. Match the items in Column I with terms in Column-II and
choose the correct option from the codes given below. mass of O 2  mass of H 2
mixture 
Column-I Column-II volume of O 2  volume of H 2
(A) Bulk modulus (isothermal) 1. p
32  2 34 17
  
(B) Bulk modulus (adiabatic) 2.  V  V 2V V
(C) Laplace correction/Netwon’s 3. p
2
formula Also, H 2 
V
Codes :
A B C 1
 P  2 1
(a) 1 2 3 Now, velocity     or  

  p
(b) 2 3 1
(c) 3 1 2
(d) 3 2 1 mixture  H 2 
  
H 2  mixture 
Ans. (c)
Sol. pV  nRT  pV  Vp  0 (isothermal)
 2/V   2
    
V p  17 / V   17 
 
V p
Standing Wave in organ Pipe
83. A pipe closed at one end produces a fundamental note of
p
Biso   p 412 Hz. It is cut into two equal length, the fundamental notes
V
produced by the two pieces are
V
(a) 206 Hz, 412 Hz (b) 206 Hz, 824 Hz
 1 
pV  const  pV V  pV  0 (c) 412 Hz, 824 Hz (d) 824 Hz, 1648 Hz
Ans. (d)
V p
 
V p

p
Badi    p Sol.
V
V

82. Oxygen is 16 times heavier than hydrogen. Equal volumes

1
of hydrogen and oxygen are mixed. The ratio of speed of 412  us ....... (1)
sound in the mixture to that in hydrogen is : 4L

2
(a) 8 (b)
17
OSCILLATION AND WAVES 20

Sol. For Doppler effect in sound component of velocity

perpendicular to line joining observer and source are not
considered hence no effect will be observed or frequency
observed by observer.
86. An open pipe is in resonance in 2nd harmonic with frequency
f1. Now one end of the tube is closed and frequency is
1 increased to f2 such that the resonance again occurs in nth
f1  us ..... (2) harmonic. Choose the correct option.
4L / 2
3 5
1 (a) n = 3, f2 = f (b) n = 3, f2 = f
f2  us ....... (3) 4 1 4 1
2L / 2
5 3
From (1) & (2) (c) n = 5, f2 = f (d) n = 5, f2 = f
4 1 4 1
f1  824 Hz Ans. (c)
From (1) & (3) Sol. For open pipe

f 2  1648 Hz 2V V
2 harmonic frequency f1  
84. An open pipe is suddenly closed at one end with the result 2l l
that the frequency of third harmonic of the closed pipe is
found to be higher by 100 Hz than fundamental frequency nV
For closed pipe nth harmonic f 2 
of the open pipe. The fundamental frequency of the open 4l
pipe is
where n can have only odd value
(a) 200 Hz (b) 300 Hz
i.e. n  [1, 3, 5, 7, 9, 11 and so on]
(c) 240 Hz (d) 480 Hz
Ans. (a) also f 2  f1

Sol. Assume the length of pipe to be l. Haves next value could be 5, 7 .... Thus n = 5

V 5V 5
for open pipe f  f2   f1
2l 4l 4
If pipe is closed at one end then third harmonic, 87. If the length of a closed organ pipe is 1 m and velocity of
sound is 330 m/s, then the frequency of 1st overtone is :
 V 3
f  3   f
 4l  2 (a) 4 (330/4) Hz (b) 3 (330/4) Hz
(c) 2 (330/4) Hz (d) none of these
3
f   f  100 Hz  f  f  100  f  200 Hz Ans. (b)
2

85. A vehicle with a horn of frequency n is moving with a (2n  1) V

Sol. f cop 
velocity of 30 m/s in a direction perpendicular to the straight 4l
line joining the observer and the vehicle. The observer
perceives the sound to have a frequency (n + n1). If velocity Where n is the overtone and l is length of closed organ pipe
of sound in air is 300 m/s, n1 would be
 V  3  330  330 
(a) n1 = 10 n (b) n1 = 0 1st overtone  3    3 
4
 l 4  1  4 
(c) n1 = 0.1 n (d) n1 = – 0.1 n
88. An open organ pipe of length l vibrates in its fundamental
Ans. (b)
 OSCILLATION AND WAVES 21
–1 –1
mode. The pressure variation is maximum : (a) 5 s (b) 10 s
–1 –1
(a) at the two ends (c) 15 s (d) 20 s

(b) at the distance l/2 inside the ends Ans. (b)

(c) at the distance l/4 inside the ends V 1 T

Sol. f    
(d) at the distance l/8 inside the ends   R2
Ans. (b)
Sol. Pressure is maximum at nodes and in open organ pipe 2
f1 l2 T1  e2   R 2 
vibrates in its fundamental mode, node will be toured midway 
f 2 l1 T2  e1   R1 
of pipe.
89. A pipe closed at one end and open at the other end resonates
2
with sound waves of frequencies 135 Hz and also 165 Hz 36 8  2  1 f1 36
      
but not with any wave of frequency intermediate between 35 1  1 4 f 2 35
these two. The frequency of the fundamental note is :
(a) 30 Hz (b) 15 Hz Given f1  360 Hz

(c) 60 Hz (d) 7.5 Hz So f 2  350 Hz

Ans. (b)
Beat frequency = 360 – 350 = 10 Hz
Sol. First resonance frequencies be to resonant frequency in
92. Two sources A and B are sounding notes of frequency
closed organ pipe are in ratio of odd natural no.
680 Hz. A listener moves from A to B with a constant velocity
–1
So n f0  135 u. If speed of sound is 340 ms , what should be the value of
u so that he hears 10 beats/s ?
(n  2) f0  165 (a) 2.0 ms
–1
(b) 3.0 ms
–1

–1 –1
 2f 0  30  f 0  15 Hz (c) 2.5 ms (d) 3.5 ms
Ans. (c)
90. A pipe closed at one end and open at the other end resonates
with sound waves of frequency 135 Hz and also 165 Hz but
not with any wave of frequency intermediate between these
two. Then the frequency of the fundamental note is
Sol.
(a) 30 Hz (b) 15 Hz
(c) 60 Hz (d) 7.5 Hz
Ans. (b) 340  4
f2  f
Sol. First resonance frequencies be to resonant frequency in 340
closed organ pipe are in ratio of odd natural no.
340  4
f1  f
So n f0  135 340

(n  2) f0  165
 24 
f 2  f1   f
 340 
 2f 0  30  f 0  15 Hz
Beats 24
10   680
91. Two wires are fixed on a sonometer. Their tensions are in the 340
ratio 8 : 1, their lengths are in the ratio 36 : 35, the diameters
are in the ratio 4 :1 and densities are in the ratio 1 : 2. If the
–1
note of the higher pitch has a frequency 360 s , the frequency
of beats produced is
OSCILLATION AND WAVES 22

420 Hz. The value of f is :

10
   2.5 m / s (a) 200 Hz (b) 210 Hz
4
(c) 205 Hz (d) 195 Hz
93. Two sound waves with wavelength 5.0 m and 5.5 m
respectively, each propagate in a gas with velocity 330 m/s. Ans. (c)
We expect the following number of beats/sec. Sol. First case : Frequency = f; No. of beats/sec = 5 and
(a) 6 (b) 12 frequency sounded with = 200 Hz.
(c) 0 (d) 1 Second case : Frequency = 2 f; No. of beats/sec = 10 and
Ans. (a) frequency sounded with = 420 Hz.

Sol. f  f In the first case frequency f  200  5  205 or 195 Hz

and
330
f1   66 Hz In the second case frequency 2f  420  10
5
or f  210  5  205 or 215
330
f2   60 Hz So common value of f in both the cases is 205 Hz.
5.5
Dopplers Effect
no. of beats  f1  f 2  66  60  6 97. When a source is going away from a stationary observer
94. Three sound waves of equal amplitudes have frequencies (f with a velocity equal to velocity of sound in air, then the
– 1), f and (f + 1). They superpose to give beats. The number frequency heard by the observer will be
of beats per second will be : (a) same (b) double
(a) 4 (b) 3 (c) half (d) one third
(c) 2 (d) 1 Ans. (c)
Ans. (c)
 V 
Sol. f    f
Sol. Number of beats/s  f1  f 2  f  1  (f  1)  2 V  u

95. A tuning fork vibrating with a sonometer having 20 cm wire Where V is speed of sound
produces 5 beats/s. The beat frequency does not change if
U  speed of receding source
the length of the wire is changed to 21 cm. The frequency of
the tuning fork must be :
V  U 1
(a) 200 Hz (b) 210 Hz So f     f  f
u  v  2v  2
(c) 205 Hz (d) 215 Hz
98. A car sounding its horn at 480 Hz moves towards a high wall
Ans. (c) –1
at a speed of 20 ms , the frequency of the reflected sound
Sol. The frequency of sonometer decreases on increasing the heard by the man sitting in the car will be nearest to ; (speed
length of sound 330 m/s)
v (a) 480 Hz (b) 510 Hz
Thus f1 – f0 = 5 = 2 (0.2)  f 0
(c) 540 Hz (d) 570 Hz
Ans. (c)
v
and f 0  f 2  5  f 0  2(0.21) Sol. The wall will reflects the sound with frequency as observed
by the wall
 v = 84 m/s
Thus f0 = 205 Hz  330 
So f R   f
96. A source of frequency f gives 5 beats/s when sounded
330  20 
with a frequency 200 Hz. The second harmonic of source
gives 10 beats/s when sounded with a source of frequency
 OSCILLATION AND WAVES 23

VA V
  0.1 and B  0.2
V V

VB 0.2
  2
VA 0.1

100. A car is moving towards a high cliff. The driver sounds a

horn of frequency f. The reflected sound heard by the driver
Sound frequency as observed by person sitting in car
has a frequency 2 f. If v be the velocity of sound, then the
 330  20  velocity of the car, in the same velocity units would be
fD   f
 330  R (a) v/4 (b) v/2

(c) v/ 2 (d) v/3

350 330
  f
330 (330  20) Ans. (d)
Sol. Let f’ be the reflected frequency as heard by observer and
35
 (480)  541  540
31  V  VC 
Vc be the speed of car then, f '   f
99. A siren placed at a railway platform is emitting sound of  V  VC 
frequency 5 k Hz. A passenger sitting in a moving train A
records a frequency of 5.5 k Hz, while the train approaches
 V  VC 
the siren. During his return journey in a different train B, he  2f   f
records a frequency of 6.0 k Hz, while approaching the same  V  VC 
siren. The ratio of the velocity of train B to that of train A is
(a) 242/252 (b) 2 V
 VC 
(c) 5/6 (d) 11/6 3

Ans. (b) 101. A police car with a siren of frequency 8 kHz is moving with
Sol. Let’s assume speed of train A and B are V A and V B uniform velocity of 36 km/h towards a tall building which
respectively then reflects the sound waves. The speed of sound in air is 320 m/s.
The frequency of the siren heard by the car driver is :
 V  VA 
fA  f
 V  (a) 8.50 kHz (b) 8.25 kHz
(c) 7.75 kHz (d) 7.50 kHz
 V  VB  Ans. (a)
fB   f
 V 

Where fA and fB are apparent frequency as observed by  V  VC 

Sol. f    f
person in train A and B respectively and V  speed of  V  VC 
sound in air
Where f’ is frequency as heard by observer in car
 V  VA 
5.5   5
 V  5
VC  36 km / h   36 m / s  10 m / s
18
 V  VB 
6 5
 V   320  10   33 
f    8 KHZ    8  8.5 KHZ
 320  10   31 
OSCILLATION AND WAVES 24

102. Two trains, each moving with a velocity of 30 m/s, cross (a) f1 = 10 f (b) f1 = 0
each other. One of the trains gives a whistle whose frequency (c) f1 = 0.1f (d) f1 = –0.1 f
is 600 Hz. If the speed of sound is 330 m/s, the apparent
Ans. (b)
frequency for passengers sitting in the other train before
crossing would be : Sol. Doppler effect in sound is not valid for component of velocity
perpendicular to line joining object and source hence no
(a) 600 Hz (b) 630 Hz
effect will be observed hence f1  0
(c) 920 Hz (d) 720 Hz
Miscellaneous
Ans. (d)
105. The path difference between the two waves
 330  30  360
Sol. f     600   600  720 Hz  2x 
330  30 300 y1  a1 sin  t  
  
103. A whistle producing sound waves of frequencies 9500 Hz
 2x 
and above is approaching a stationary person with speed v and y 2  a 2 cos  t     is
  
m/s. The velocity of sound in air is 300 m/s. If the person
can hear frequencies upto a maximum of 10,000 Hz, the    
maximum value of v upto which he can hear the whistle is : (a)  (b)  
2 2  2

(a) 15 / 2 m/s (b) 15 m/s 2   2

(c)   (d)  
  2 
(c) 30 m/s (d) 15 2 m/s Ans. (b)

Ans. (b)  2x 

Sol. y1  a1 sin  t  
  
 Vs 
Sol. f    V  V  f  2x 
 s  y2  a 2 cos  t   
  

As f   10000 Hz   2x  
 a 2 sin    t    
 2    
 Vs 
hence  V  V  f  10000  2x   
 s   a 2 sin  t     
   2 

300 10000  

300  V 9500
So phase difference, d     
2

 95  3  300  V d x 
Now   x  (d)
2  2
 V  300  95  3  15 m / s
  
    
104. A vehicle with a horn of frequency f is moving with a velocity 2 2
of 30 m/s in a direction perpendicular to the straight line 106. The equation of a wave travelling on a string is
joining the observer and the vehicle. The observer perceives
the sound to have a frequency f + f1. If the velocity of  x
y  4sin  8t  
sound in air is 300 m/s, f1 would be : 2 8

if x and y are in centimetres, then velocity of wave is

 OSCILLATION AND WAVES 25

(a) 64 cm/sec in –ve x-direction

(b) 32 cm/sec in –ve x-direction
(c) 32 cm/sec in +ve x-direction
(d) 64 cm/sec in +ve x-direction
Ans. (d)

coeff of t
Sol. Velocity of wave  coeff of x (a) positive y-axis and positive y-axis respectively
(b) negative y-axis and positive y-axis respectively

 (c) positive y-axis and negative y-axis respectively

(8)
 2   64 cm / s (d) negative y-axis and negative y-axis respectively
  1
   Ans. (b)
2  8

 dy 
Here -ve sign suggest wave is travelling in +x direction. Sol. VP   V   
dx
107. A transverse wave is described by the equation
Where VP is velocity of particle and wave respectively,
 x
y  y0 sin 2  ft  
  Also as wave is moving in +x-direction hence V  0

The maximum particle velocity is equal to four times the

wave velocity if  dy 
as  dx  0
at x  2.5
y0 y0
(a)   (b)  
4 2
 dy 
and  dx  0
(c)   y0 (d)   2y0 at x  1.5
Ans. (b)
hance VP (at x  1.5)  0

Sol. Velocity of wave V   f 
T VP (at x  2.5)  0

dy  x 109. A progressive wave is given by

Velocity of particle VP   (2y0 f ) cos 2  ft 
 
dt  y = 3 sin 2[(t/0.04) – (x/0.01)]
Where x, y are in cm and t in s. The frequency of wave and
So (VP )max  2y0 f
maximum acceleration will be
If (VP )max  4 V (a) 100 Hz, 4.7 × 10 cm/s
3 2

3 2
 2y0 f  4f (b) 50 Hz, 7.5 × 10 cm/s
4 2
(c) 25 Hz, 4.7 × 10 cm/s
y 0 4 2
  (d) 25 Hz, 7.5 × 10 cm/s
2
Ans. (d)
108. Wave pulse on a string shown in figure is moving to the
right without changing shape. Consider two particles at Sol. Equation given is in the form
positions x1 = 1.5 m and x2 = 2.5 m. Their transverse velocities
 t x
at the moment shown in figure are along directions y  A sin 2   
 T 

Where symbols have their usual meaning

OSCILLATION AND WAVES 26

1 100 If
here, T   0.04  f   25 Hz   1.2589
f 4 IL

2 So percentage change in intensity

  2 
a max  2 A    (3)
 0.04  I I  IL
  100%  f  100%  (1.259  1)  100%
I IL
 7.40  104 cm / s 2
= 25.89%
110. Which of the following is not true for the progressive wave
112. The intensity level of two sounds are 100 dB and 50 dB.
 t x  What is the ratio of their intensities?
y  4sin 2    1 3
 0.02 100  (a) 10 (b) 10
5 10
(c) 10 (d) 10
Where x and y are in cm and t in seconds.
Ans. (c)
(a) The amplitude is 4 cm
(b) The wavelength is 100 cm  I
(c) The frequency is 50 Hz Sol.  (dB)  10 log10  I 
 0
(d) The velocity of propagation is 50 cm/s
Ans. (d) I0  1012 ω/m 2
Sol. Equation of standard plane progressive wave is
I1
100  10 log10
 t x  Symbols have their  I0
y  A sin 2     
 T   usual meaning 
I1
On camparing with given equation we get 10  log10 ......(1)
I0
A = 4 cm
I2
1 1 50  10 log10
f   50 Hz I0
T 0.02

  100 cm I2
5  log10 .......(2)
I0
V  f  50  100 cm/s  50 m/s
From equation (1) and (2)
111. An increase in intensity level of 1 dB implies an increase in
Intensity of (given antilog10 0.1 = 1.2589) I1 I
10  5  log10  log10 2
(a) 1 % (b) 3.01 % I0 I0
(c) 26 % (d) 0.1 %
Ans. (c) I  I I
5  log10  1   1  antilog10 5  1  105
I
 2 I 2 I 2
I 
Sol.   10 log  f 
 IL  113. A source of sound emits 200 W power which is uniformly
distributed over a sphere of radius 10 m. What is the
Where  is change in intensity (in dB) loudness of sound on the surface of the sphere?
(a) 70 dB (b) 74 dB
I
So for 1 dB = 10 log f (c) 80 dB (d) 117 dB
IL
Ans. (d)
 OSCILLATION AND WAVES 27

Power 200  1
Sol. I   
Area 2 2
4 (10)

I
Loudness, B (in double)  10 log
I0

 1 
 
 10 log  2 
12
 10 
 

 10 [log (1012 )  log 2 ]

= 10 (12 – 0.30)
= 10 (11.7)
= 117 dB