13

Oscillations
Quick Review
Periodic Motion " If initial phase is ¢, then
displacement in SHy
A motion which repeats itself over and over again after y= Asin(ot + )
aregular interval of time is called a periodic motion. and y= Acos(ot + )
e.g., revolution of planets around the sun, rotation of
the earth about its polar axis etc. Velocity in SHM
The function which are used to represent The velocity of particle executing SHM is det
motion are called periodic functions. periodic the time rate of change of its displacement at nan
instant.
One of the simplest periodic function is given by
V=
f(t)=A cos ot
. When the particle is at the mean position ie.:
Oscillatory Motion then its velocity is maximum.
A motion in which a body moves back and forth =0A
max (maximum velocis
repeatedly about a fixed point (called mean position) is " When the particle is at the extreme position.
called oscillatory or vibratory motion.
i.e., x=tA, then its velocity is zero.
Simple Harmonic Motion v=04 - 4' =0
Simple Harmonic Motion (SHM) is that type of
oscillatory motion in which the particle moves to and Therefore, the velocity of a particle executmy
SHM is zero at the either of its extreme posi"
fro about a fixed point under a restoring force whose
magnitude is directly proportional to its displacement Acceleration in SHM
and directed towards mean position, i.e.
F -yor F= -ky The acceleration of SHM at an instant is detmtu
where, k is a positive constant called force constant or the time rate of change of velocity at that instt
spring factor. a = -0
" Mathematically a SHM can be expressed as " When the particle is at the mean positionie.
2r then acceleration is zero.
J=A sin o = Ain -

(iminimum acceler
2Tt ean position =0
y= Acos 0r= 4cos positioni.:
" When the particle is at the extreme
where, y = displacement from mean position at time t. X= A,hen acceleration is maximum.
A= amplitude or maximun displacement, (extrene position) = -0*A(maximumacceled
0= angular frequency and T=time period.
 ULATONS

325

Displacement. Velocity and Acceleration of a Body

D'gplacementv(t)=A
cos ay
Velocitv r(t)=-(04 sin ont
Executing SHM
Velocity Acceleration a ) (nA cos (of
+oA

-oA

variesfrom -
nature,
difYerence
Ato Sinusoidal in nature, v () varies from- (04 to o4.
dalin phase It has a phase difference of Sinusoidal in nature, a (t) varies from
t has zero
2 w.r.t. v() o4 to )A.
It has aphase difference of IW.r.t. vt).

SHM
law in " The frequency of oscillation of potential energy and
e simpleeharmonic motion the
statesthat
:ereCuted motion kinetic energy is twice as that of displacement or velocity
iby aparticle subjected to a force, which is or acceleration.
roportionalto the displacement of the particle and is
alwaysdirected towards the mean position.
System Executing SHM
Restoringforce,F=-mo»y=-ky Aspring block system and simple pendulum come under
SHM.
here,
Simple Pendulum
2T Asimple pendulum, in practice, consists of a heavy but
)=
T m small sized metallic bob suspended by a light, inextensible
and flexible string.
inertia factor
.. Time period, T= 2r, The motion of a simple pendulum is simple harmonic whose
spring factor time period and frequency are given by
For linear SHM.
T=2T and v
displacement
T= 2r acceleration " Ifapendulum of length l at temperature ° Chas atime
Ergy in SHM period T, then on increasing the temperature by A9°Cits
time period changes to T x AT.
la particle of mass mexecutes SHM, then at a AT 1
displacement yfrom mean position, the particle possesses where,
potential and kinetic energy.
At any " If the length of a simple pendulum is increased to such an
displacement y, 1 extent that /’ o, i.e., l>> R, then its time period is
()Potential energy, U= mo
2 2 T=2r R/g = 84.6 min
(ii Kinetic energy,
where, R= radius of the earth.
K!
2 ma° (4 - y' )=-KA'-y'
2
) Spring Mass System
(ii) Total energy, If the mass is once pulled so as to stretch the spring
E= U+ K= and released, the spring pendulum oscillates simple
2
mo4'=2myA harmonically having time period and frequency.
At mean position,
potential energy is kinetic energy 1S maximum and
T= 2 and v =

"At zero.
Kinetexticreenergy
me position,
is zero.
potential energy iS maximum and Angular frequency, @=
The time period of potential and kinetic energies is T/2.
326 CHAPTERWISE &TOPICWISE - MEDICAL
The mechanical encrgy E of the
oscillator is given by Important Points Regarding
SOLVED
Spring Block System
ht/m
(i) Ifthe length of the
spring
is
Time period of the oscillations of spring-tmass system are shown in
following table
the effective force constant
and the time period ma de n i m
becometismiens
becomesnt
(i) If aspring of spring constant kis
e
Various Cases equal parts, then the
Figures spring
() Spring is light becomes nk and time period constant
T = 2n, V00000m
(iii) The force constanttof astiffer
becones
m
X=0(mean)
than that of a soft spring. sping is hi
position
(ii) Spring isnot light oo0000so0000000
LLmw M
Simple HarmonicMotion
but has mass m, ML in Special Cases
T=2r| The time period of SHM due to the
li motiinon Utgi
incompressible and non-viscousliquid
given as
m

T= 2r,
[m
where, h = height of undisturbed liquid in ext
(iiil) Spring connected m mT and L= 2h=total length of liquidcolumn
with two masses
(m, and m,) T=2; k
00000000OO000 . The time period of a ball performing SHM:
hemispherical bowl is expressed as
(iv) Springs with
00O000000000 R-r
T= 2T
spring constants
(k, and k,)
connected in
m(k +k,): where, R and rare radii of bowl and ball.
series k ky
kik, OO000000 O000000 m " The time period of a ball executing SHMin:
tunnel through the earth is expressed as
T= 2r,
o200000000
00000000 where, R =radius of earth.
(v) Spring connected
in parallel T= 2 " The time period of an elastic wire executng
SHM, due to the restoring force F= -A|
=2
k + k, expressed as
|m ml

kk,
and
T=21AY
k=

where, = = reduced mass of the system, k,

m+ m where, Y= Young's modulus,
and k, = k +kz. l=length of wire. A/=elongatiol
and A=area of cross-section.
aSCLLATTONS 329

20 Two particlesexecute SHM of the same arnplitude and

4 To
harnonic motions
simple
mo are represcntcd by Irequency along the came straight line If they pass one
=5/sin 2ti + v3cos 2r1 another when going in opposite directions. each time their
displacement is half ther arnpl1tude. the phase d1fference
'; =Ssin|2r/+ between then is MHT CET,
and
T

The
ratio) of theiramplitudes is (KCET) (a)
(b) 2:1 (c) 1:3 (d) 3:1
(a)1:1 (c)
simpleharmonic motions Aand Bare given
T#o
bythe following equations
15respectively [CG PMT) 21 A particle executes SHM. its time period is 16s If t
J =asin ot + passes through the centre of oscillation. then its velocity is
6 AMU)
2ms at time 2 s. The amplitude will be
(b) 4 cm
V; =asin of + 6 (a) 7.2 m
(c) 6 cm (d) 0.72 cm
difference between the waves is
Thephase 2005
(b) (d) zero 22 Which of the following functions represent asimple
6
harmonic oscillation? TAIIMS

(a)sin ot - cos t (b) sin

2009
blockissresting on a piston which is moving vertically (c) sin ot + sin 20t (d)sin t - Sin 2 or
16 A
aith SHM of period 1S. At what amplitude of motion. 23 The displacement of a particle pertoming simple harmonic
willthe block and piston separate? [UP CPMTJ
motion is given byx= 8sin 0- 6cos 0, where distance
(a)0.2m (b) 0.25 m is in cm and time is in second. The amplitude of motion is
(c)0.3 m (d) 0.35 m (a) 10 cm (b) 2 cm [MHT CET
17 The phase difference between the instantaneous velocity (c) 14 cm (d) 3.5 cm
and acceleration of a particle executing simple harmonic 24 The minimum phase difterence between two simple
motion is [AlIMS] harmonic oscillations. (MHT CET
(a) 0.5r (b) T (c) 0.707 I (d) zero V =(1/2)sin or + (v3 2)cos o:.
18 The displacement equation of a simple harmonic
oscillator is given by [MHT CET]
(a) (b) (d)
y= A
sin 0t - Bcos ot 12 12
The amplitude of the oscillator will be 25 Which of the following is not characteristic of simple
(a) A- B (b) 4+B harmonic oscillation? [Kerala CEE)

(c)4' +B? (d) (4' +B?) (a) The motion is periodic

(b) The motion is along straight line about the mean
2006 position
19 The moion of aparticle varies with time according to the (c) The acceleration of the paricle is directed towards
relation [BHU) the extreme positions
y= a(sin Qt + cos Ot) (d) The oscillations are responsible for the energy
(a) the motion is oscillatory but not SHM transportation
(b) the motion is SHM with amplitude a (e) The period is given by T=2, -where
" the svmbols
(c) the motion is SHM with amplitude av2
(d) the motion is SHM with amplitude 2a have usual meaning

Answers
1(
2 (a) 3 () 4 (e) 5 (b) 6 (b) 7 (a) 9 ) 10
11 (d)
12 (a) 13 (b) 14 (b) 15 (c) 16 (o) 17 (a) 18 c) 19 c 20
21 (o)
22 (a) 23 (a) 24 (b) 25 ()
 Topic 1
Displacement and Phase in SHM
2019 8 What is the phase differencee
1 The distance covered by a particle
between
undergoing SHM m harmonic motions represented by x
one time period is =Asin
(a) zero
(amplitude =A) [NEET (Odisha)]
and x, = Acos ot ?
(b)4 (c) 24 (d) 44
2 The displacement of aparticle executing Tt
simple
motion is given by y= A, + 4sin aot + B cos ot harmonic[NEET] 2
Then, the amplitude of its oscillation isgiven by 2011
(a) v4+B? (b) 4 +(A +B) 9 Out of the following functions
(c) 4+ B () 4, +4' +B2 particle which represents SHM representing r
1. y=sin Qt - cos Mt
2017 II. y=sin 0t
3 The displacement of a
particle along the X-axis is given (3r
- 3or
byx=a sin ot. The motion of the particle corresponds III. y=5cos
4
to

(a) simple harmonic motion of

frequencyo/T IPMER. V. y=1+ ot + o12
(b) simple harmonic motion of frequency 3o/2r (a) Only IV does not represent SHM
(c) non-simple harmonic motion
(d) simple harmonic motion of frequency o/2t (b) Iand III
(c) Iand II
2014
(4) Only I
4 A body oscillates with SHM according to the 10 The motion which is not simple harmonic is ea
equation
.(in SI unit), x =Scos |2rt +* Its instarntaneous (a) vertical oscillations of a spring
(b) motion of simple pendulum
displacement at t = ls is [Kerala CEE) (c) motion of a planet around the sun
a m (b)
1 (d) oscillation of liquid column in a U-tube
5 V3
m m
(a)m (e) vertical oscillation of a wooden plank loai;
liquid
(e m
V2 11 The amplitude and the time period in a SHM Is U3
and 0.4 s respectively. If the initial phase is 12
5 Estimate the tine taken by the oscillating pendulum to the equation of SHM will be
shift from x = 0to A/2, where Ais the amplitude. [WB JEE)
(a) y=0.5sin 5nt (b) y=0.5sin 4t:
(a (c (d) (c) y=0.5sin 2.5tt (d) y=0.5cos 5t!
4 12 16
12 x(t)= Acos (ot + )is the cquation of'simpleharn
6 Two pendulums have time period T and 5T/4. They start
SHM at the same time from the mean position. What will motion. In this equation,ois called
be the phase difference between them after the bigger (a) phase constant (b) frequency
pendulum completed one oscillation? [UP CPMT] (c) amplitude (d) displacement
ha
(a)45° (b) 90° (c)60° (d) 30° 13 Aparticle executing asimple harmonic motionm:
to
7 The amplitude of aparticle executing SHM is 4cm. Atthe period of6s. The time taken by the particle th
mean position, the speed of theparticle is l6 cm/s. The the mean position to half the ampliude, starting
distance of the particle from the mean position atwhich mean position is
the speed of the particle becomes 8/3 cm/s will be 3
(a) s
(a) 2/3 cm (b) W3cm [AFMC]
(c) l cm (d) 2 cm () s
asCMLATIONS
329

20 Two particles exccute SHM of the same amplitude and

1010 sinple,harmonic motions are represcnted by Irequency along the same straight line. If they pass one
2rul another when going in opposite directions, cach time their
4
Two
.= 5|sin 2nr+ v3cos
displacement is half their amplitude, the phase difference
and
=Ssin|2rr+ 4 between them is
T
(MHT CET]

amplitudes is (a) (b)

ratioofftheir [KCET] 3 4
The (b) 2:1 (c) 1:3 (d) 3:1 2Tt
(a)1:1 (c) (d)
simple harmonic motions Aand Bare given 6
Two
15 rspectively bythe following cquations [CG PMT) 21 Aparticle executes SHM, its time period is 16 s. If it
)} =a sin or + passes through the centre of oscillation, then its velocity is
(AMU]
2 ms- at time 2 s. The amplitude willbe
(a) 7.2 m (b) 4 cm
=asin or + (c) 6 cm (d) 0.72 cm
difterence between the waves is
Thephase 2005
(b) (c) (d) zero 22 Which of the following functions represent a simple
3
harmonic oscillation? [AIIMS]
(a) sin 0t- cos Ot (b) sin ot
2009
Ablock is resting on a piston which is moving vertically (c)sin ot + sin 2ot (d) sin 0t - sin 20t
with SHM of period l s. Atwhat amplitude of motion. 23 The displacement of a particle performing simple harmonic
will the block and piston separate? [UP CPMTJ
motion is given by x = 8sin ot + 6cos t , where distance
(a) 0.2 mn (b) 0.25 m is in cm and time is in second. The amplitude of motion is
(c) 0.3 m (d) 0.35 m (a) 10 cm (b) 2cm [MHT CET]
17 The phase difference between the instantaneous velocity (c) 14 cm (d)3.5 cm
and acceleration of a particle executing simple harmonic 24 The minimum phase difference between two sinmple
motion IS [AlIMS] harmonic oscillations, [MHT CET]
(a) 0.57 (b) T (c) 0.707 (d) zero y, =(1/2) sin ot + (V3/ 2) cos ot,
V = Sin ot + cos 0I is
18 The displacement equation of a simple harmonic
oscillator is given by [MHT CET]
(a)
12
(b)
12
(c)
6
()6
'=Asin cot Bcos ot
The amplitude of the oscillator will be 25 Which of the following is not characteristic of simple
(a) 4 - B (b)4 + B harmonic oscillation? [Kerala CEE]
(c) y4 +B? (d) (4? + B²) (a) The motion is periodic
(b) The motion is along straight line about the mean
2006 position
19 The motion of aparticle varies with time according to the (c) The acceleration of the particle is directed towards
relation [BHUJ the extremepositions
y= a(sin (0t + cos O) (d) The oscillations are responsible for the energy
(a) the motion is oscillatory but not SHM transportation
m
(b) the motion is SHM with amplitude a (e) The period is given by T = 2, ,where the symbols
(c)the motion is SHM with amplitude av2
(a) the motion is SHM with amplitude 2a have usual meaning

1
Answers
2 (G) 3 (c) 4 (e) 5 (b) 6 (b) 7 (d) 8 (b) 9 (b) 10 (c)
12 (o) 13 (b) 14 (b) 15 (c) 16 (b) 17 (a) 18 (c) 19 (c) 20 (d)
22 (a) 23 (a) 24 (b) 25 (c)
 Explanations
1(d) In a simple
harmonicmotion
(SHM) the particle oscilates atut
mean position on a straight Ine. its = 2aoi-2sin cx)
and
I:=cos tor
The particle Oves fom its mean
position (O) to an extreme sitOn (P) I-=4 sin
and then retunm to its mean
Phase difference.
covering same distance of SIIOn
4.
Then by the conserv atie forve. it is -..(11)
moved in opposite direction to a pont From Eqs. () and ( . the motion of
Obv distance 4and then back to
mean particle isDoT-Simple harOnic motion.
position covering a distance of4. This
comprises of one time erii as shOwI 4 re Given. r =5 cos 2 | 6
beiow
(where, r is a displacement) 9 b For a simple harmon
(at =1s)
[:: cos (360 0) = cos ] Hence. equation = 0
and 37.
exteme
JOsitor
mean
DOStOn
in one Dme 3eroc
y=5cos-3
are satisfing this condiion bg
Hence. in one time peioi tcoves 2 5 b Initial position at =0 is r=0
distance of The euation of SHM.I=4 sin ex periodic and ;= in'isenr
DOt SHM.
I= OP- PO- 00 - 90 When.
10 c) The motion of plzns
2 a) The Sun is periodic but Dot sie
is
displacement of give paricia moion Simpl harmoc TTT:
Special simpliieicase zsrE
V=4, -4 sincx Bcoscx sina-= some effecz in2t TestOrs
Thegeneral equaiot of SHM ca be 6
miidle position
gIven as =
6 11 d, =in (o-0
I=aSin bcosor
So. ifom Egs. (i and (iü). ecar sa =a in
that 4, be the value of mea posiion. 6x
at which y=0.
..Amplirude. = =05sn
6 b) When bigge pendulm of ime
R=,4- B- 2AB cost period (5I 4) completes one vibratio J=05 in
As two function sine andcosie beve he smalla peatulum al coplee
phase shift to 90. S 4)brations. It mezs the smaler
penduln lbe leaiing he bigzer
T se=12 rad =9
3c For a SHM.

According question
I=Sn G= _16
==4
Differentiating wrIL 13 b) Ecuation for sinpi
= osin cxcosr
= c02snOCOsr
192 =16 l6
Again differennaing *r: .
12=16
Topic 2
Velocity, Acceleration and Energy of SHM
2019 6 When a particle executing SHM
1 Average velocity of a particle
executing SHM in one
frequency , then the kinetic energy of the Oscil ates itha
(a) changes periodically with: a
complete vibration is
(a).40 (b) Ao
[NEETI
(b) changes periodically with a
(c) changes periodically with a
quencypatofice2y
ffrreequency
2018
2
(c) zero (d) 2 (d) remains constant
7Ifa body is executing simple
frequency of v
2 Aparticle doing SHM having harmonic motlon at
current displacement is V3/2 times:the
and angular frequency Srad's isamplitude Scm, mass 0.5 kg
at lcm frommean
position. Find potential energy and kinetic energy. JIPMERJ
mean position, then the ratioi
and kinetic energy is between potameplntitiuadel
(a)KE = 625x 10 J, PE= 150×10" J (b) 2:3
(b)KE = 150x 10 J, PE= 625x 10 J (a) 3:2 (c) V3:1
8 A10 kg metal block is attachedIto a
(d)3:1
(c) KE = 625x 10 J, PE = 625 x10 J spring of s
(d) KE = 150x 10J, PE= 150x10 J constant 1000
Nm-.Ablockis displaced from
2014 equilibrium position by 10cm and released. The
maximum acceleration of the block is
3 The oscillation of a body on a
smooth
IS represented by the equation, x= Ahorizontal surface (a) 10 ms (b) 100ms 2 (c) 200 ms 2 (d) 0.1x
cos (Ot
where, x=displacement at time t 9 A body of mass 1kg is executing simple harmoni:
0=frequency of oscillation motion. Its displacement y (cm) at t seconds is giver
Which one of the following graphs shows corectly the T
y=6sin 100r + Its maxinmum kinetic energyis
variation of a with t? [CBSE AIPMT)
(a) 6J (b) 18 J (c) 24 J (d) 36J
1
a 2013
(a) ot (b) t’ 10 The KE and PE of a particle executing SHM of amp
awill be equal, when displacement is
2
(b) ay2
(c) 2a
(c) ot (d) ol t’
17 A long spring, when stretched by a distance xias
Here, a =acceleration at time t potential energy U. On increasing the stretching to t
T=time period potential energy of the spring will be
(a)
4 When the displacement of a particle executing simple (b) nU
harmonic moion is half its amplitude, the ratio of its
(c) n'U
kineticenergy to potentialenergy is [Kerala CEE)
(a) 1:3 (b)2:1 (c) 3:1
(d) 1:2 (e)2:3 2012
12 Abody is motion wih
5 The displacement of a particle having SHM is executing simplesharmonic theh
velocity of m
angular frequency of 2 rad The otthe
x=10sin | 10xt + 4 mand its angular frequency. at 20 mm
displacement, when the amplitude
is 60 mm is
(a)6x (b)8x
(d) 10x (a) 131 mms (b) 118 mms
(c) 2x (c) 113 mms -1
(d) 90 mms
SCNLATIONS 333

displacerment of a particle executing SHM is given by

r=0.25
a
The sin 200rcm. The maximum specd of the particle 22 The maximum vclocity of aparticle in SHM is v. If the
3 amplitude is doublcd and the time period of oscillation
[MHT CET] leereascd to of its original valuc, the maximum velocity
(b) 100 cms-l
(a)200cms
-1 (d) 5.25 cms becomes (WB JEEJ
(c) 50
cms
(a) 18 (b) 121 (c) 6v (d) 31
ofthe second pendulum is decrcased by 0.3
Thelengthit shiftedto
is Chennai from London. If the 23 The total cnergy of the body executing simple harmonic
cm when due to gravity at London is 981 cm/s.the
acceleration motion (SHM) is E. Then, the kinetic energy when the
displacement is half of the amplitude is [CG PMTI
aceleration due to gravity at Chennai is (assume n = 10)
3E v3E
(b) 978 cm/s² (a) (b) (c) (d)
(a)981cms? [Manipal) 2 4 4
(c)984cm/s (d) 975cm/s?
24 A particle is vibrating in a simple harrnonic motion with
pertorming SHM, at adistance A/2, the an amplitude of 4 cm. At what displacement from the
15 Foraa
body equilibrium position, is its energy half potential and half
between KE and PE will be
correctrelation [OJEE]
kinetic? [VMMC]
to PE (b) KE is 2 times of PE
(a) KE is equal (a) 1 cm (b) 2 cm
of PE (d) KE is half of PE
(c) KE is 3 times
(c) 3 cm (d) 2/2 cm
spring is stretched by 10 cm, the potential energy
16 When a
storedis E. When the spring is stretchedi by 10 cm more, 2009
the potential energy stored in the spring becomes [WB JEE] 25 Which one of the following equations of motion
(a)2E (b)4 E (c) 6E (d) 10 E represents simple harmonic motion? [CBSE AIPMTJ
17 Uis the PE of an oscillating particle and F is the force (a) Acceleration =- kox + k, x
acting on it at a given instant. Which of the following is (b) Acceleration=-k(x +a)
true? [MHT CET]
(c) Acceleration = k(x+ a)
(a)+x=0 (b) +x=0 (d) Acceleration = kx
F F
F where, k, ko, kË and a are all positive.
+x=0 (d) +x=0
2U 26 A particle is executing linear simple harmonic motion.
The fraction of the total energy that is potential, when its
18 For a particle in SHM, if the amplitude of the displacement is 1/ 4 of the amplitude is [J&K CET]
displacement is a and the amplitude of velocity is v, the 1
amplitude of acceleration is [MHT CET] (a) (b) (d)
16 4
(c)
2
(a) va (b) (d)
a .2a a
27 A particle performing SHM has time period 2/ 3 and
path length 4 cm. The displacement from mean position at
I9 The average acceleration of a particle performing SHM which acceleration is equal to velocity is [MHT CET]
Over one complete oscillation is [MHT CET]
(a) zero (b)0.5 cm (c) l cm (d) 1.5 cm
(a) (b) 0
2
28 A body executing linear simple harmonic motion has a
V2 velocity of3 ms when its displacement is + cm and a
(c) zero (d) A 2 velocity of 4ms when its displacement is 3 cm. What is
20 In SHM, the the amplitude of oscillation?
acceleration is ahead of velocity by a phase (BCECE]
angle [OJEE] (a) 5 cm (b) 7.5 cm
(a) 0° (b) T/ 2 (c) Tt (d) 2Tt (c) 10 cm (d) 12.5 cm
21 A
particle is executing linear simple harmonic motion of 29 A simple pertorms SHM about n = 0with an annplitude a
amplitude A. At what displacement, is the energy of the and time period T. What is the speed of the pendulum at
Particle half potential and half kinetic'? [WB JEEJ =A/2? [Manipal]
(a)4 V34T V3An
4 (b) (a) (b)
T
(c)4 A
(d) 2An (d) VSAn
(c)
V3 T
 -
334 CH4TERWSE &TOCwiSE MEDICAL

5TE r

2006
=

=]

Puat PWET

2007

45 4b :
SCLLATONS 335

energy ofa body executing SHM is /3

thekkinetic The
49 The maximum velocity of a simple harmonic motion
potentialenergy.
When displacement of the body is x
the oftheamplitude,
of
where xis [RPMT) represented by y= 3sin| 1001+m
r
per
cent (b) 87 (BCECEJ
(a)33 (d) 50 is given by
3T -I
( c6
)7 ms (c) 100 ms (d) ms
(a) 300 ms (b) 6
6
2005
simple
hamonic motion, maximum velocity is at 50 Asimple pendulum performs simple harmonic motion
47(a)
Inextremepositions [Punjab PMET) about x= 0with an amplitude a and time period 7T. The
[CBSE AIPMT]
halffofextreme positions speed of thependulum at x = al 2 willbe
ic)equilibriumpositions
(b) Tav3 37a Trav3
and l equilibrium (a) (b) (c) (d)
between extreme positions 2T T T
(d)
statements.
Considerthe following 51 A coin is placed on a horizontal platform which undergoes
Thetottal
energy of a particle executing simple harmonic vertical simple harmonic motion of angular frequency (0).
on its
motion depends [BCECEJ The arnplitude of oscillation is gradually increased. The
()amplitude coin will leave contact with the platformn for the first time
() period (a) at the mean position of the platform [AIMS]
displacement, offthese statements (b) for an amplitude of
(1II)
correct
(a) Iand IIare
correct
(b)II and 1Iare (c) for an amplitude of
(c)land IIIare correct
d) 1. IJ and III are correct (d) at the highest position of the platform

Answers
2 (b) 3 (a) 4 (a 5 (G) 6 (b) 7 (d) 8 (a) 9 (b) 10 (d
12 (b) | 13 (b) 14 (b 15 (a) 16 (o) 17 (b) 18 (b) 19 (c) 20 (b)
11 C)
21 ic) 22 (c) 23 (c) 24 (d) 25 (b 26 (a) 27 (c) 28 (a) 29 (b) 30 (c)
31 (b) 32 (d) 33 (a) 34 (æ 35 (c) 36 (d) 37 (a) 38 (d) 39 (c) 40 (a)
41 (b) 42 (d) 43 (a) 44 (d) 45 (0) 46 (b) 47 48 (a) 49 (a) 50 (d)
51 (b)

Explanations
1c) The average velocity ofa particle 2 (b) Given, m =0.5 kg, 0 = 5rad/s, If ? then =0
executing simple harmonic motion X=10m 4
(SHM) is If t= then =-4
A = 5x 10m
1
Total displacement . Potential energy, PE =-mo'x We can see that only graph (a) will
Time interval satisfy the above results.
Where, x and x, are the initial and final PE -x0.5 x(5)' x(10)° L+A
position of the particle executing SHM. 23
x 10J = 6.25 x 10J
As, in vibrational motion, the 4
xecutes SHM about its mean particle
position.
1
: Kinetic energy, KE =mo(4-x) -A}
So, after one complete vibration of the
particle, it will reaches its initial or K -x0.5 x (5)°(25 1)x 10 4 (a)For simple harmonic motion, the
position, i.e. 2
= 150xl0J
displacement,
Displacement, x, X, = 0
3 (a) We can find the correct graph by
I = A cos 0t

Vav
Hence, the average velocity is zero.
putting different values of t in the
given expression x = A cos ot
(*)
COS Ot = ’ 0 = 30°
ift= 0, then =+4
 341

bn under the action is t times that

27 The magnitude of maximumacccleration oscillator. The
force is changed to harmonic
otmaximum velocity of a siple (Punjab PMET)
time period of the oscillator in sceond is
-n with time period (b) 2
(a) 4
Aneously in the same (c) 1 (d)0.5
riod will be [AFMC 2007
with an
28 A particle executes simple harmonic oscillation minimum
amplitude a.The periodof oscillation is T. The
the amplitude
time taken by the particle to travel half of[CBSE AIPMT, JIPMER)
from theequilibrium position is T
T
simple hamonic (a) (b) (d)
4 2
go from its mean
ent (amplitude) is 29 The maximum displacement of the particle executing
SHM is lcm and the maximum acceleration is
S [MHT CET
(1.57)' cms.Its time period is [AMU]
(a) 0.25 s (b) 4.0 s
(c) 1.57 s (d) 3.14 s
SHM is given by
ere x is in metre and tin 2006
on (in second) is
0.02
30 A rectangular block of mass m and area of cross-section
(MGIMS] Afloats in a liquid of density p. If it is given a small
0.2 vertical displacement from equilibrium it undergoes AIPMT]
[CBSE
oscillation with a time period T. Then,
(a) T o o (b) T o«
[AIMS]
harmonic motion with a (c) T l ()T« m
harmonic motion with a a
31 Aparticle executes simple harmonic motion with
frequencyf. The frequency with which the potential
on with a period 2T/o [AMUJ
energy oscillates is
on with a period /o
(a)f (b) f/2
cuting oscillations about the (d) None of these
ential energy isU(r) = k[x]', (c) 2,f
time
ant. If the amplitude of 32 A particle executes SHM of amplitude 25 cm and
for the
period 3s. What is the minimum time required
e period Tis [AlIMS]
particle to move between two points 12.5 cm on
either
side of the mean position? [BCECEJ
(b) independent of a
(a) 0.5 s (b) 1.0 s
(c) 1.5 s (d) 2.0 s
(d) proportional to a2

Answers
6 (a) 7 (c) 8 (d) 9 (a) 10 (c)
4 (b) 5 (a)
16 (d) 17 (b) 18 (b) 19 (d) 20 (b)
d) 14 (b) 15 (b)
27 (b) 28 (c) 29 (b) 30 (b)
(b) 24 (b) 25 (b) 26 (a)
aLATONs

339

of particle in SHM.
energy 49 (a) The
Rrtential,
given cquation is
8a U=-mo2 written as 51 (h)As the amplitude is increased, the
y= 3sin| 1001 4 ..i) maximum acceleration of the platform
U=zm2r), The general cquation of (along with coin as long as they

U=2r'mf'r? ...()
harmonicmotion is writtcnsimple
as
doesn't get separatcd) increases.
Coir
V=Asin (ot +) ...(ii)
particle in SHM.
Ainetic energyof Equating Eqs. (i) and (i), we get Platforrn

A=3. (0) =100 Equilibriun pos1tion

K=2rimf4-') ...(ii)
MaximumV=velocity.
Ao) Perform1ng SHM
=3x 100= 300 ms If we draw the free body diagram for
Hence,totalenergy
coin at one of the extrerne positions as
E=K+U 50 (d)As we know that shown, below
the
particle executing SHM isspeed of
given by
v=0d' -r' and o=2T
2n'm4 where, x= displacement mg N

Then, from Newton's law

2r
V=
(at x =al2) mg - N = mo'4
obvious that total energy of
Thus, itis
narticle executing simple harmonjc For loosing contact with the platform.
notion depends on amplitude (4) and 2r N3a) V3a N=0
period(7) 7
So, A=

Topic3
Time Period and Frequency
2018 2015
1Ablock of rectangular size of mass mand area of 4 Aparticle is executing SHM along a straight line. Its
crOss-section 4, floats in a liquid of density p. If we give a velocities at distances x, and x, from the mean positions
Smallvertical displacement from equilibrium, it are v and V,, respectively. Its time period is [CBSE)
undergoes SHM with time period T, then [AlIMS]
1 (a) 2Tt (b) 2Tt
(b) T² ocp (c) T'm (d) T² -

2Y=5sin (100t- 2x) what is time period? (c) 2r (d) 2r

(JIPMER] x +x
(a) 004s (b) 1s (c) 0.06 s (d) 002 s
2014

3A particleexecutes 5 Ablock resting on the horizontal surface executes SHM

linear simple harmonic motion with in horizontal plane with amplitude A. The frequency of
a1 amplitude of 3 cm. When the particle is at 2
cm rom oscillation for which the block just starts to slip is
e mean position, the magnitude of its velocity is equal (u = coefficient of friction andg =gravitational
to that of its
acceleration. Then,its time period in seconds acceleration) (MHT CET)
[NEET] A
(a) (b) (c) 2 (d) 4
4T 27 2ru VA 4V A
(b (c) (d) VHg
2T
 ~
CHAPTERWISE &TOPICWISE MEDICAL
340

2013 2010
14 Abodly is exccutingSHM when its
SoLWED
O Prequeney of oscillation of abody is 7Hz when force mean position are 4cm and 5 it cm,
iS applied and 24 Hz when E, is applied. lfboth fones ?; as
and
willbe
, are applicd together, then frequency of oscilation and 8cms,respectively. lts
(a)
2Tt
S
periodic velncity
(b) TtS
(a) 25 Hz 3
(b)31H7 (e) 17Hz (d) 24 11z 3Tt
7 t the maximum velocity and acccleration of a particle (c) (d) 27r s
2
CXecuting SHM are cqual in magnitude, the tinme period
willbe 15 Abody is executing SM of
(a) 1.57 s (b) 3.17s (c) 6.28 s (d) 12.56 s
(UP CPMT)
wile passing through the mean amplposition,
itude Im.is )
8 A simple pendulum of fiequency n falls fieely under frequency.
(a) 1.442 Hz (b) 1.592 Hz
gravity from certain height from the ground levcl. Its (c) 1.322 Hz
frequency of oscillation will (Kerala CEE) (d) 1.132 Hz
(a) remain unchanged (b)be grcater than n 16 Averticle U-tube of iiniform cross section
(c) be less than
(e) beconme infinity
(d) become zero upto a heignt of 20 cm. Calculate
Oscillation of water when it is disturbed
the time Cperorntiaomds
Aparticle executing simple harmonic motion has atime (a) 0.11s (b) 0.25 s
period of4 s. After how much interval of time from t=0 (c) 0,40 s (d) 0.9 s
will its displacement be half of its amplitude? (KCET) 17 The equation of aparticle executing SHM s
1
2 2d'x +
(a)
3 (6) ()
6 dt
32r= 0. Then, the time period of thebody:
2012 (a) zero (b) T/2
10 The displacement equation ofa body (c) T (d) 2r
performing SHM is 18 A 8kg body performs SHM of amplitude 30 cm. T
given as y'=200sin 180r/ +-Thetime period of
4 restoring force is 60N when the displacement is 9
body is Find its time perio.
(MPPMT]
1 (a) 1.432s (b) 1.256s

2011
(a)
65 (b)s (c)
90
(d)s
80
(c) 2.424 s (d) 2.412s
19 The time period of the variation of potential energy
11 A particle of mass m is located in a one-dimensional particle executing SHM with period Tis [Keraa
T 1
potential field, where potential energy is given by (a) (b) T (c) 2T (d)
V(x)= (1- cos px) where Aand pare constants. The
period of small oscillations of the particle is [WB JEE] e)
3
(a) 2r
m
(b) 2r (d)Ap
Ap 2Tr V m 20 The motion of a particle executing SHM in one
12 A harmonicoscillation is represented by dimension is described by x=- 0.3sint7
y= 034cos (30001 +0.74). Find its frequency. is in metre and t in second. The frequency ofoscyila
1000 1500 750 3000
(a) (b) (c) (d) Hz is
T
(a) 3 (b)
13 Asertion (A) Water in a U-tube executes SHM, the time 2T
period for mercury filled upto the same height in te
U-tube be greater than that in case of water.
Reason (R) The amplitude of an oscillating pendulum 2009
(c)% (d)

goes on increasIng. [AlIMS)

(a) Both A and R are correct and R is the corret 21 Two bodies performing SHM are
explanation of A
expresseu
(b) Bot A and R are correct but R is not the correct y = SOsin 100K +and y, =1Osin S0nt
explanation of A difference of their oscillating frequency IS
(c) A is correct but R is incorrect (d)25H
(d) Both A and R are incorrect (a) 5 Hz (b) 50 Hz (c) 20 Hz
 341
sULATONS

simple harmonic motion under the action accclcration is t times that

vexecutes 4 27 Themagnitudc of maximum The
velocity of asimple harmonic oscillator.
body
period s. If the force is changed to of maximum
2A
F witha time 5 [Punjab PMET)
offore
harmonic motion with time period time period of the oscillator in second is
executes sinple (a) 4 (b) 2
it
Ifbothforces F, and,F, act simultaneously inthe same (c) I (d) 0.5
thebodythen, itstime period will be [AFMCJ 2007
directionon oscillation with an
28 Aparticle executes simple harmonic
24
(b). minimum
a)
12 25 amplitude a. The period of oscillation is T. The arnplitude
the
(d) time taken by the particle to travel half of[CBSE AIPMT, JIPMER]
12 from the equilibrium position is
T
24 T T
periodictimeof aparticle doing simple harnonic 4
(b)
8
23 The Thetime taken by it to go from its mean
motionis4s. maximum particle executing
ttothe
position displacement (amplitude) is 29 The maximum displacement of the
acceleration is
(b) I s [MHT CET| SHM is lcm and the maximum
(a)
2s (1.57) cms, Its time period is [AMU]

3 (a) 0.25 s (b) 4.0 s

(c) 1.57 s (d) 3.14 s
motionofa particle executing SHMIis given by
24 The 01 sin 100t (1 + 0.05), where xis in metre and in 2006
second) is
second. The time period of motion (in 30 A rectangular block of mass mand area of
cross-section
(a)0.01 (b) 0.02 [MGIMS]
Afloats in a liquid of density p. Ifit is given a small
(c)0.1 (d) 0.2 vertical displacement from equilibrium it undergoes [CBSE AIPMT]
oscillation with a time period T. Then,
2008 (b) T oc
(a) T« p
25 The function sinoÍ represents [AIMS)
1
(a) aperiodic but not simple harmonic motion with a (c)Ta! (d) T«F
period 2t/
(b) aperiodic but not simple harnmonic motion with a 31 A particle executes simple harmonic motion with a
period /aO frequency f The frequency with which the potential
(c) asimple harmonic motion with a period 2r/o energy oscillates is [AMU]
(d) asimple harmonic motion with a period /o (b) f/2
(a)f
26 Aparticle of mass mis executing oscillations about the (c) 2f (d) None of these
origin on the X-axis. Its potential energy is U(r) =k[x], 32 A particle executes SHM of amplitude 25 cm and time
where kis a positive constant. If the amplitude of period 3 s. What is the minimum time required for the
oscillation is a, then its time period Tis [AlIMS] particle to move between two points 12.5 cm on either
1 [BCECE)
(a) proportional to (b) independent of a side of the mean position?
(a) 0.5 s (b) 1.0 s
(c) proportional to ya (d) proportionalto a
3/2 (c) 1.5 s (d) 2.0 s

Answers
1(a) 5 (a) 6 (a) 7 (c) 8 (d) 9 (a) 10 (C)
2 (a) 3 (c) 4 (b)
11 (d 12 (b) 14 (b) 15 (b) 16 (d) 17 (b) 18 (b) 19 (d) 20 (b)
13 (d)
21 (d 22 (a) 23 (b) 24 (b) 25 (b) 26 (a) 27 (6) 28 (c) 29 (b) 30 (b)
31 (d) 32 (a)
 opic4
systemExecuting SHM
SimplePenduluma and Spring Mass System)
2018
pendulumis hung from the roofof a
1A sufticiently
building and is moving freely to and fro like a
high 6 The time periodof a simple pendulum of length Las
haronic oscillator. The acceleration simple
of the bob of the measured in an elevator descending with acceleration g 3
pendulumis 20 ms at a distance of 5 mfromthe mean [BHUJ
sition. The time period of oscillation is
[NEET] (a) 2r |3L (b)
3L 3/ 2L
(a) 2s (b) T s (c) 2 (d) 2r
(c) 2rs (d) 1s
7 Apendulum has time period T in air when it is made to
2011 Oscillate in water:, it acquired a time period 7T' = v27T.
2 Two simple pendulums first of bob mass M, (bob 4) and The density of the pendulum bob is equal to (Take density
of water = 1)
length L,. second of bob mass M, (bob B) and length [Haryana PMT)
M.=M,and L =2L,. If the vibrational energies of bothL,. (a) 2 (b) 2
are same. Then, which is correct?
[AlIMS]
(c) 2V2 (d) None of these
(a) Amplitude of B is greater than 4 8 A pendulun is made to hang from the ceiling of an
b)Amplitude of B is smaller than 4 elevator. It has period of Ts (for small angles). The
(c) Amplitude will be same elevator is made to accelerate upwards with 10 ms.The
(d) None of the above
period of the pendulum now willbe (Take.g = 10 ms)
3The mass and diameter of a planet are twice those of the (a) TV2 (b) infinite [DUMET
earth. The period of oscillation of pendulum on this planet (c) T/N2 (d) zero
will be (ifit is a second's pendulum on earth) [AlIMSJ 9 If pendulum bob on a 2 m string is displaced 60° from the
(a) 1/42 s (b)2N2 s vertical and then released. What is the speed of the bob
(c) 2s (d) 1/2s
as it passes through the lowest point in its path?
4 Assertion (A) The percentage change in time period [VMMC]
is 1.5%. If the length of simple pendulum increases (a) 2 ms -! (b) /2x 9.8 ms
by 3%. (c) 4.43 ms-!
Reason (R) Time period is directly proportional to (d) V/2 ms
length of pendulum. [AlIMS] 10 Two pendulums of lengths l00 cm and 121 cm start
(a) Both A and R are correct and R is the correct vibrating. At some instant the two are at the mean
explanation of A. position in the same phase. After how many vibrations of
(b) Both A andR are correct but R is not the correct the longer pendulum will the two be in the same phase at
explanation of A. the mean position again? [MGIMS]
(c) A is correct but R is incorrect. (a) 10 (b) 11 (c) 20 (d) 21
(d) Both A and R are
incorrect. 2010
5 A
clock S is based on oscillation of a spring and a clock P 11 In the figure, S, and S, are identical springs. The
1S based on pendulum motion. Both clocks run at the
Same rate on the earth. Ona planet having the same oscillation frequency of the mass n is f. If one spring is
density as earth but twice the radius removed, the frequency willbecome [VMMC]
(AFMC]
(a) S will run faster than P A B
(b) P will run faster
than S O000000m000000
S, S
DOth will run at the same rate as on the eartn
(d) both will run at the same rate which will be different
from that on the earth (a) f (b) 2/ (c) v2/ (d)!
346 CHAPTERWISE &TOPICWISE ~
MEDICAL SOLVED
12 An iron ball of mass Mis 19 Aclock pendulum made of;
hanoed firom the invar has aperiod
spring with aspring constant k. It executes ceiling
aSUM bywitha a 20° IT the clock is used ina
Period P. If the mass of the ball is inercased by four tenperature averages 30"C, climate where the
clock losein cach oscillation?how muh time d
the new period will be times,
[DUMETI
(a) 4P P (for invar 4,
(b) (c) 2P (d) P and g= coDstant)
4
(a) 2.25 10 s
2009 (by 2.5/ \0
(c) 5x 10 s
13 As shown in figure, a simple (d) 1.125 \0
harmonic motion oscillator
having identical four springs has 20 Asimple pendulum is made by
time period 5 mlong copper wire. Its pcriod atistachi
Tng
[BVP) a

osNow,cil atiifom,lkg%
replaced by 10 kg bob, the period of
(a) remains T'
(b) becomes greater than T
(c) becomes less than T
m
(d) Any of above depends on locality
21 The length of a second's pendulum is
2Ak m
(a) T=
(b) T =2r4 2 (a) 99.8 cm (b)99 cm
(c) 100 cm (d) None of these
(c)T =2,T (a) T=2n 2m 22 Apendulum suspended from the
ceiling of atrain k..
a period T when the train is at
14 Two masses m, and m, rest. When the train
are suspended together accelerates with a uniform acceleration a, the reriod
by a massless spring of of oscillation will
force constant k, as
shown in figure. When the masses are in (a) increase
equilibrium, mass m, is removed without (b) decrease
disturbing the system. The angular frequency of | m (c) remain unaffected
Oscillation of masS m, is (BCECE] (d) become infinite
k 2008
(a) (b) (c) |km |km
m, m, 23 The time period of a simple
m pendulum of infinite leng
[take, g = 9.8 m/s and radius of earth (R, )= 6400e
15 Ifa simpie penduium oscillates with an
50 mm and timeperiod of 2 s, then its amplitude of
maximum (a) 86.4 min
velocity is [MHT CETJ (b) 84.6 min
(a) 0.19 ms (b) 0.15 ms (c)0.8 ms (d) 0.26 ms-! (c) infinite (d) zero
24 Ahollow sphere is filled with
16 If the length of second's pendulum is water through the st
increased by 2% hole in it. It is then hung by a long thread and mace
then in a day the pendulum
(a)loses 764 s
[Kerala CEE] Oscillate. As the water slowly flow out of the hoit
(b) loses 924 s bottom, the period of oscillation will MHT
(c)gains 236 s (d) loses 864 s (a) continuously decrease
(e) gains 346 s (b)continuously increase
17 Maximum time period of any simple pendulum on the (c) first decrease, then
(d) first increase, then
increase
earth is decrease
(BVP]
(a) 180.5 min (b) 100 min (c) 90.5 min (d) 1.5 nin 25 Aheavy small-sized sphere isS:suspended by a
honzontal
18 A simple pendulum is executing SHM with a period of 6s
length . The sphere rotates uniformly in a!Vertical.
Tk
with the string making an angle with the
between two extreme positions B and C about a point O. the time period of this conical pendulum S
If the length of the arc BC is 10 cm, how long will the Isin8
(a) t= 2r
pendulum take the move from position Cto a positionD (b) t= 2n,
towards O exactly midway between Cand 0? (EAMCET)
(a) 0.5 s (b) 1 s (c) I.5 s (d) 3 s (c) = 2T, lcos (d) = 2r,
 347
LATIONs

constant kis cut intotwo picces such that 2007

offorce length of the other. The force with a
is ddoublethe S4 A bob of simple pendulum exccutes SHM in water
piece
one ofthelonger pieve will be [Kerala CEE) period T. The period of oscillation of the bob in air is lo
The relation between Tand T, is [take, density of bob
constant
(b)3 k (c) 2k (d)
(a)
1.54 400
(p)= 3
kg/m'] (KCET]
(a) T, =T (b) T, = 2T (c) T= To (d) T = 2T,
pendulumhas atime period 7; onthe surface of
35 The time period of a simple pendulurn in a stationary van
Asimple radius R. When taken to a height of Rabove the
earthof is T. The time periodof a mass attached to a spring is also
's: surface,itstime period is
I,. Then,the ratio T. The van accelerates at the rate 5
ms.If the new time
earth
[Kerala CEE]
,his periods of the pendulum and spring be T, and T,
respectively, then (AFMC]
1 (6)W2 (c) 2 (d) 4
(a) 7, =T, (b) T, >T,
(d)Cannot be predicted
(c) T, <T,
36 One body is suspended from a spring of lengthl, spring
divided
ofa simple pendulumnexecuting simple constant k and has time period i. Now if spring is
Thelengthhmnotion
28harmonic is increased by 21%. The percentage in two equal parts which are joined in parallel
and the
new
increase in the time period of the pendulum of
increased same body is suspended from this arrangement, then
time period of body is [UP CPMT]
[RPMT)
lengthis T
(a)11% (b) 21% (c) 42% (d) 10.5% (c) 2T
(b) T
mass.
0 Amass Mis suspended trom a spring of negligible =60°, The
The spring is pulled a little and then released so that the 37 A pendulum oflength 1m is released from
mass is
rate of change of speed of the bob at = 30° is
mass executes SHM of time period T. If the
increased by m, the tine period becomes 5T/ 3, then the (Take, g= 10 ms) [AFMCJ
ratio of m/Mis [RPMT]
(a) 10 ms (b) 7.5 ms (c) 5 ms -2 (d) 5/3 ms-2
25 16
(b) (d) -2
3 (e) 2.5 ms
30 The angular amplitude of asimple pendulum is ,. The 38 A simple pendulum has a time period Tin vacuum. Its
maximum tension in its string will be [BCECE) time period when it is completely immersed in a liquid of
density one-eighth of the density of material of the bob is
(4) mg (1- ) (b) mg (1 +,) [Kerala CEE)
(c) mg (1-0,) (d) mg (1+ 0%) 8
T
31 Time periodof a simple pendulum of length Iis 7, and
(4),V8 (b). 7
time period of a uniform rodof the same length l
pivoted about one end and oscillating in a vertical plane (e),
IS1,. Amplitude of oscillations in both the cases is small. 39 A body of mass 20 g connected to spring of constant k
Then, T, /T, is [BCECE) executes simple harmonic motion with afrequency of
4
(b) 1 |Hz.The value of spring constant is [Kerala CEE)
3 (d) S Nm-!
(a) 4 Nm (b) 3 Nm-l (c) 2Nm-!
Agrl swings on a cradle in sitting position. If she stands, (e) 2.5 Nm!
the time period of cradle [EAMCET]
(a) decreases (b) increases 40 Ifk, and k, respectively are effective spring constants in
(c) remains constant series and parallel combination of springs as shown in
(0) Tirst increases. then it remains constant [Haryana PMT]
33 Asimple figure, find
pendulum is suspended from the ceiling of alift.
When the lift is at rest its time period is T. With what
k
2k
acceleration should the lift be accelerated upwards in Fooo0000000
uer to reduce its period to T/2? (g is acceleration due to 000000
2k
gavity). [KCET] 9
(a) 2g (b) 3 g (c) 4 g (d) g
(a)
348 CHAPTERWISE & TOPICWISE ~ MEDICAL
41 What effect occurs on the
SOLVED PA
Auniform spring of force eonstant k is cut nto two cees
whose lengths are in the ratio of 1:2. What is the force
49
is taken from the earth's frequencyto deepof a
surface
constant of sccond picce in temms of ? (J&K CET]
(a) Incrcases
(b) Decreases
inpieondulaurrIE
(a) (b)
26
(c) (d)
4 (c) First increases. then decreases 1Puni
(d) No cffect
42 To make the frequency double of a sring oseillator. we 50 Abody of mass 5 kg executes SHM
have to If the force constant is 100
of amplitude rf
(a) reduce the mass to
(Punjab PMET] N/m. calculate its
(b) quadruple the massone-fourth
(c) double the mass
(a) 1.41
(c) 040
(b) 0.46
(d) 0.37
ime.f
(d) half the mass A8 kg body perform SHM
43 What is the velocity of the bob of a
51
restoring force is 60 N,
amplitude 30 cm. The
when the
its mean position, if it is able to rise simple
10 cm?(Take. g=9.8
ms)
pendulum at
to vertical height of
[BCECE)
Find its time period.
(a) 1.236 (b) I.256
displacement is 3y.
(c) 1.212
(a) 2.2 ms-l
(b) 1.8 ms -1 2005
(c) 1.4 ms -1
(d) 0.6 ms 52 If the length ofa pendulum is made9
2006 bob is made 4times,then the value of times and mass
44 3 time period he-
A spring balance has a scale that reads 0 to 50 kg. The (a) 3T (b) (c) 4T
length of the spring is 20 cm. A body suspended from 2T (d) 27
spring when displaced and released. Oscillates with this 53 The masses and length of tWo Simple penduums ar.
period of 0.60 s. What is the weight of the body? J&K CET) given as respectively m4, mg andA: l¡. If fregusr
(a) 22.40 kg A is double that of B, then relation between . ani:
(b) 22.35 kg
(c) 25.30kg (d) 22.36 kg given as
(UP
45 What is the effect on the time period of a (a) l, = (b) , =21,
simple
pendulum, if the mass of the bob is doubled?
2
[Kerala CEE) 4
(a) Halved
(c) Becomes eight times
(b) Doubled (c) l4 = ()7, =4
(d) Becomes zero
(e) No effect 54 The simple pendulum acts as second's pendulum on
46 Choose the correct
statement. J&K CET) earth. Its time on a planet, whose mass and diameter :
(a) Time period of a simple pendulum depends on twice that of the earth is [MHTE
amplitude. (a) 2s (b) 2V2 s (c) 2s
(b) Time shown by aspring watch varies with
acceleration due to gravity. 55 A simple pendulum of length and mass (bob) mis
(c) In a simple pendulum timne period varies linearly suspended vertically. The string makes an angletw
with the length of the pendulum. vertical. The restoring force acting on the penduium
(d) The graph between length of the pendulum and time (a) mg tan [MHTE
period is a parabola. (b)-mg sin 0
(c) mg sin 9 (d) -mg cos
47 Time period ofa spring mass system is T. If this spring is S6 Apoint mass m is suspended at the end of a massie
cut into two partswhose lengths are inthe ratio 1:3 and
of length L and cross-section area A. If Yis the Y'ouny:
the same mass is attached to the longer part, the new time
period will be [J&K CET) modulus for the wire, then the frequency of oseillati
[MHT
T for the SHM along the vertical line is
(b)
V3 (a) |mL (c) (d)
2r V mL. (b) 2T,VY4
V37 (d)V37 the
(c) 57 Asimple CquatortoIKerala
2
Its period
pendulum is taken from the
48 The length of second's pendulum is I m on the earth. It (a) decreases
mass and diameter of a planet is doubled than that of the (b) increases
earth. then its length bccomes [Punjab PMET]
(b) 2m (c) remains the same
(a) l m
(c) 0.5 m (d) 4 m (d) decreases and then increases
(e) becomes intinity
 349
scLANONS

spring
resultant
constant of the system cf springs 65 Aheavy brass sphere is hung from a
weightless inelastic
of
The beloy
shown [Kerala CEE] spring and as a simple pendulum its time period
Oscillation is T When the sphere is immersed in a
nÍn-viscous liquid of density 1/10 that of brass, it will act
as a simple pendulum of period (JCECE)

(a) T (6)9

(b) (4k +2%, Xk; )

k, +k, +k,
66 The period of oscillation ofa simple pendulum of period
(d) surface of the earth is T. Its time
constant length at
[KCET]
(CxXk, +k; +k; ) inside a mine will be
+k, \k, ) (a) cannot be compared
(b) equal to T
(c) less than T
simple pendulums of lengths 1.44 mand l mstart
59Two together. After how many vibrations willthey (d) more than T
Swinging of
again startswinging
together? [J&K CET) 67 The period of oscillation of a simple pendulum moves
of a vehicle, which
pendulum
(a) 5 oscillations of smaller
length /suspended from the roof is
plane of inclinationa,
b)6 0scillations of smaller pendulum without friction down an inclined
WB JEE]
ic)4 oscillations of bigger pendulum given by
()6 0scillations of bigger pendulum (a) 2 (b) 2r. g sina
If the
60 Two pendulums begin to swing simultaneously.
the two is 7:,
Vg cosa
ratio ofthefrequency of oscillations of
be
then the ratio of lengths of the two pendulums will (d) 2r Vg tan O
(a) 7:8 (b) 8:7 (J&K CET]
(c) 49:64 (d) 64:49 68 A simple harmonic oscillator consists of a particle of
mass
k. If spring is mand an ideal spring with spring constant k. The particle
61 Aspring has length / and spring constant
constant of each
divided in two equal parts, then spring oscillates with a time period T. The spring is cut into two
part is (Punjab PMET] equal parts. If one part oscillates with the same particle,
the time period will be [AlIMS]
(a) k (c) 2k (d) 4k
(a) 2T (b) V2T
T
62 The amplitude of an oscillating simple pendulum is 10 cm
and its time period is 4 s. Its speed after 1 s,when it
passes through its equilibrium position is [AMUJ
69 One-fourth length of a spring of force constant k is
(a) zero (b) 2.0 ms (c) 0.3 ms (d) 0.4 ms cut away. The force constant of the remaining spring
03 Asimple second pendulum is mounted in a rocket. Its will be [AFMC]
time period will decrease when the rocket is [AMU] 4
(b)k
3
(a) moving up with uniform velocity
(b) moving up with uniform acceleration (c) k (d) 4k
(c)moving down with uniform acceleration 70 Abody of mass 500 gis attached to ahorizontal spring of
(d) moving around the earth in geostationary orbit spring constant 8r Nm.If the body is pulled to a
lwo simple pendulums whose lengths are 100 cm and distance of 10 cm from its mean position, then its
121 cm are suspended side by side. Their bobs are pulled frequency of oscillation is [Kerala CEE)
lOgether and then released. After how many minimum (a)2 Hz (b)4 Hz
0scillations of the longer pendulum, willbe the two be in (c) 8 Hz (d) 0.5 Hz
phase again? [DUMET]
(e) 4n Hz
(a) 11 (b) 10 (c) 21 (d) 20
Topic 5
Free, Damped, Forced Oscillations
&Resonance andSuperposition of SHM
2015 state of maximum amplitude of
1 When two displacements represented by
measure of the oscilatis
; =a sin (or) and 1', =bcos (o/ ) are superimposed, the (a) free vibration (b) damped vibratior
motion is
[CBSE AIPMT]
(c) forced vibration (d) resonance
(a) not a simple harmonic 2007
(b) simple harmonic with amplitude 5 In case of aforced vibration, the
very sharp when the resonance Wwavei
(c) simple harmonic with amplitude ya' + b (a) applied periodic force is small
(b) quality factor is small
(d) simple harmonic with amplitude (a+b)
(c) damping force is small
2014 (d) restoring force is small
2 The displacement of a particle in a 2006
periodic
given by y=4 cos(t/2) sin (1000t )Thismotion
is
6 Aparticle is executing twodifferent simple harmot
displacement may be considered as the result motions,mutually perpendicular,of different amnl
superposition of n independent harmonic oscillations. and havingphase difference of TU2. The path of the
Here, n is particle will be
[WB JEE]
(a) 1 (b) 2 (a) circular (b) straight line
(c) 3 (d) 4 (c) parabolic (d) elliptical
2009 7 During the phenomenon of resonance
3 Two simple harmonic motions are given by (a) the amplitude of oscillation becomes large
(b) the frequency of oscillation becomes large
x= A
sin (ot +)and y = Asin0 +ð+act on a (c) the time period of oscillation becomes large
(d) All of the above
particle simultaneously, then the motion of particle will be
(a) circular anti-clockwise [UP CPMT 8 The equation of a damped simple harmonic motion s
(b) elliptical anti-clockwise d'x
m + kr= 0. Then, the angular frequency?
(c) elliptical clockwise dt? dt
(d) circular clockwise oscillation is Man

4 If the differentialequation given by y1/2

(a) o'=| (b) a'=
d'y + 2* dy 4m 4m)
dt
+o»y=F, sin pt 2 )/2
describes the oscillatory motion of body in a dissipative (c) o'=| k (d) o'=
4m 4
medium under the influence of a periodic force, then the

Answers
1 () 2( 3 (d) 4 (d) 5 (c) 6 (d) 7 (a) 8 (a)
 Explanations
the
exccuting resonant vibrations. Less
=asin(or
Gven.}
damping. greater will he sharpness.
or
7=hoosor= bsin 6() If the cquations of two mutually
perpendicular SHM of same frequency
resultant
displacement is given by L712)A be
- þ sin(o + o) A
The sinlot + o)
X= u, sin ot and y= a,
motionof superimposed General equation of Lissajous' figure is
Hene.the harmonicwith 2xy cos o= sin
Neis 1a+6simple
auplitude which is the equation of a circle.
a
Now, At (O + 8) = 0,x=0, y= 0
sin (1000) At (or + 8) =n/2, x= A, y= 0 =1
'=4cos| When Ø = then
2c Given. At (ot + S)= I.N= 0, y=- A 2
sin (1000/) ellipse.
- 2x2cos|;
At (ot + S)= 3n/2,x=-4, y= 0 which represents the path of
At (ot + ð)= 2+, x=A, y=0 phenomenon of
From the above data, the motion of a 7(a) During the
2cos=(l+ cos resonance, the amplitude of oscillation
particle is a circle transversed in
becomes large. Because applied
frequency isequal to natural frequency.
clockwise direction.
sin (1000/)
= (1+cos t)
t- sin (1000) 4 (d) The differential equation, 8 (a) Damping force, F, =
-by
=2 sin (1000r) + 2 cos
(10001)- cos dy + 2k + o'y= F sin pt
=2 sin (10001)+ 2 sin where, b= damped constant,
(1000; + ) dt d
=2sin (10001) + sin V= velocity of oscillation
and
+ sin (10001 ) shows oscillatory motion (forced). The k where
B)+ sin (.4 - B)) state of maximum amplitude of the Oo = 2V, =2t. 2T Vm Vm
.'sin d-cos B = sin (4 + oscillation is the measure of resonance.
(999r)
=2sin (10001) +sin (1001z)+ sin body, Oo =angular frequency of
free
n=3 5 (c) Inresonant vibrations of a oscillation.
So.
the frequency of external force applied
(.: above equation shows three on the body is equal to its natural Displacement of damped oscillator is
inside independent oscillations)
frequency. If on increasing and given by
...i) decreasing the frequency of external r=X,e ot msin (ot+ o),
31diGiven. x =A sin(ot +) force from the natural frequency by a
vibrations where, o' = angular frequency of
and y= Asin o+8 + factor, the amplitude of sharp damped oscillator.
reduces very much. In this case,
=Acos (ot + Ó) ...(ii) resonance will take place.
factor, then o=
On squaring and adding Eqs. () But if it reduces by a small Vm 4m'
place. The
and (ii), we get flat resonance will take
sharp and flat resonance willdepend
x*+ = 4? the body
on damping present in
[sin(o + 8)+ cos'(o +8 )]
 Topic 1
Basic of Mechanical Waves
2011
2018 8 Each of the properties of soundl in Column
and in water is depends on one of the quantitiesin
1 If spced of sound in air is 340 1m/s
kHz. then find
1480m's. If frequency of sound is 1000 (JIPMER]
correct answer (matching Column I with
Colu
the value of wavelength in water. code given below the columns.
(a) 2.96 mm (b) 1.48 mm
(d) 1mm
ColumnI Column II
(Haryana Pin
(c)0.74 mm
P Loudness A. Waveform
2014
[Kerala CEE) Q. Pitch B. Frequency
2 ldentify the corrcct statement.
(a) Transverse wave can propagate in gases. R Quality C. Intensity
(b) Transverse wave consists of compressions and
rarefactions. Codes
(c) Longitudinal wave can propagate in solids, liquids P R QR
and gases. (a) A C (b) C A
(d) In alongitudinal wave, particles of the medium (c) B C A (d) B A C
vibrate perpendicular tothe direction of
propagation.
9 The intensity level of a sound wave is defined bv z
(e) In a longitudinal wave, the higher density corresponds arbitrary scale. The zero of the scale is taken at te
to rarefactions. wave intensity
3 The speed of sound in air [Kerala CEE) (a) 1 x1010 Wm? (b) 1x 10" Wm
(a) decreases with temperature (c) 1x1014 Wm? (d) 1x10-ló Wim
(b) increases with pressure 10 A theatre of volume 100x 40x10m' can accommAC
(c) increases with humidity
(d) decreases with pressure 1000 visitors. The reverberation time of the theatre
empty is 8.5 s. If the theatre is now filled with 500
(e) increases with density
Visitors, occupying the front-half seats, the reverbr
2013 time changes to 6.2 s. The average absorption coe:
4 Which of the following is different from others? [WB JEE] of each visitor is nearly
(a) Wavelength (b) Velocity (a) 0.6 (b) 0.5
(c) Frequency (d) Amplitude (c) 0.45 (d)0.7
5 Whena certain volume of water is subjected to increase of 11 A sound absorber attenuates the sound level by 20
100 kPa pressure, the volume of waterdecreases by The intensity decreases by a factor of
0.005%. The speed of sound in water must be [UP CPMT) (a) 1000 (b) 10000
(a) 140m/s (b)300 m/s (c) 10 (d) 100
(c) 1400 m/s (d) 5000 m/s 2009
6 Oxygen is 16times heavier than hydrogen. Equal of'soud
The pressure variations in the propagation Kerae
volumes of hydrogen and oxygen are mixed. The ratio of 12 waves in gaseous medium are
speedof sound in the mixture to that in hydrogen is (b) isothermal
(a) v& (a) adiabatic
(b) 2/17 [Manipal] (c) isobaric (d) isochoric
(o) /v8 (d) 32/17 (e) cyclic
7Asound has an intensity of 2x 10* W/m². Is intensity 13 When awave undergoes refraction,
level (in decibel) is (log 1o 2= 0.3) [WB JEE) (a) its frequency changes
(a) 23 (b) 3 (b) its amplitude changes
(c) 43 (d) 4.3 (c) its velocity changes
(d) Both amplitude and frequency chane
wAVES
365

temperatureat which the speed of sound in air

The doubled of;its value at 0°C is 23 The gas having average speed four times as that of SO,
s
becomes [VMMC]
1092°C (b) 819K (c) 819°C (d) 546C (molecular mass 64) is [MHT CET]

(a) transfer (a) He (molecular mass 4)

waves [KCET) (b) 0, (molecular mass 32)
Sound
s energy not momentum (c) H, (molecular mass 2)
) onnly
(a)energy
(b) (d) CHa (molecular mass 16)
ic)momentum
(c) 24 Which one of the following statements is true? [MP PMTJ
and
id)Both(b) (a) Both light and sound waves in air are transverse.
pontsource of sound emits sound unifornly
A.stationary (b) The sound waves in air are longitudinal while the
16 directionsin a non-absorbing medium. Two pointsP light waves are transverse.
inall. mand 9 mrespectively from the
are at a distance of4
and O (c) Both light and sound waves in air are longitudinal.
Source. Theratio of amplitudes of the waves at P and OtsCET] (d) Both light and sound waves can travel in vacuum.
4 9
(e) 25 The speed of sound waves in a gas (J&K CETJ
(a) (b) 3
(d)
4
(a) does not depend upon density of the gas
(b) depend upon changes in pressure
2008
time offreverberation of a room Ais one second.
(c) does not depend upon temperature
17 The What will be the time (in second) of reverberation of a (d) depends upon density of the gas
-!
room, having all the dimensions double of those of room 26 A boat at anchor is rocked by waves of velocity 25 ms
[Haryana PMT] having crests 100 m apart. They reach the boat once every
(b) 4 (c) 1/2 (d) 1 (a) 4.0 s (b) 8.0 s [AMU]
(a)2
18 Reverberation time does not depend upon (c) 2.0 s (d)0.25 s
[BCECE]
(a) temperature (b) volume of room 27 Compressional wave pulses are sent to the bottom of sea
from a ship and the echo is heard after 2 s. If bulk
(c) size of window (d) carpet and curtain
modulus of elasticity of water is 2 x 10 Nm and mean
10 How many times more intense is a 60dB sound than a temperature is 4°C, the depth of the sea will be [BCECE]
30 dB sound? [KCET]
(a) 1014 m (b) 1414 m
(a) 1000 (b) 2 (c) 100 (d) 4 (c) 2828 m (d) None of these
20 Assertion The change in air pressure affects the speed of 2006
sound.
Reason The speed of sound in gases is proportional to 28 When sound waves travel from air towater, which one of
the square of pressure. [AIIMS] the following remains constant? [Manipal]
(a) Both Assertion and Reason are correct and Reason is (a) Time period (b) Frequency
the correct explanation of Assertion (c) Velocity (d)Wavelength
(b)Both Assertion and Reason are correct but Reason is 29 Velocity of sound waves in air is 330 ms -1 Fora
not the correct explanation of Assertion particular sound in air, a path difference of 40 cm is
(c) Assertion is correct but Reason is incorrect equivalent to a phase difference of 1.6 . The frequency
(d) Both Assertion and Reason are incorrect of the wave is [Manipal]
2007 (a) 165 Hz (b) 150 Hz
(c)660 Hz (d) 330 Hz
<1 The intensity of sound increases at night due to [AFMC)
(a) increase in density of air 30 The velocity of sound isvat 273 K. The temperature at
which it is 2v is [Kerala CEE)
(b) decrease in density of air
(c) low temperature (a) 2 x 273 K (b) 4 × 273 K
(d) None of the above (c) 8x273 K (d) 16x 273 K
22 (e)/2 x 273 K
Sound
because waves
are not transmitted overa long distance,
[MHT CET) 31 Distance between successivecompression and rarefaction
(a) they are absorbed by the is Im and velocity of sound is 360 ms. Find frequency
(b) they have atmosphere of the sound wave. [RPMTJ
(c) tthe constant frequency
height of antenna
(d) velocity of sound required should be very high (a) 180 Hz
(c) 120 Hz
(b) 45 Hz
(d) 90 Hz
waves is very less
366 CHAPTERWISE & TOPICWISE ~M
MEDICAL
32 A bomb explodes on the moon. IHow long 34 Aparticle on the trough of a
SOLWED
come tothe mearn position waVea at any
the sound to reach the will it take for
carth? (Punjab PMET) after time
(a) 10 s (b) 1000 s (a) T/2 (b) T/4 (T
(c) 1day (d) None of these (c) T (d) 2T
2005 35 Sound travels faster after rain than
on a
33 The ratio of velocity of sound in dry dyhe
hydrogen and oxygen at (a) the temperature ofthe
STP is
(a) 16:1 (b) 8:1
[KCET)
(c) the humidity in the air
atrmosphere iNCreasR
(b) the density of aif increases after rai
(c) 4:1 (d) 2:1 (d) None of the above increases
after

Answers
1 (b) 2 (c) 3 (c) 4 (d) 5 (c) 6 (b) 7 (c) 8 (b)
11 d 12 (a) 9 b)
13 (c) 14 (c) 15 (d) 16 (d) 17 (a) 18 (a)
21 (a) 22 (a) 19 ()
23 (a) 24 (b) 25 (d) 26 (a) 27 (b) 28 (b)
31 (a) 32 (d) 33 (c)
29 (
34 (b) |35 )

Explanations
1 b) Given. v
=1480 m/s, plp remain constants. So, velocity of Mass of 0, +Mas
f=1000 kHz Pmixture
sound is independent of pressure. Volume of O,+ Volr:
Speed. v, =f. Also, vo T, so with rise 32 + 2 34 17
1480 temperature, velocity of sound decreases.
1000x 10 1
2
Also, v oc E, S0 with increase in
=1480 mm Also, PH,
2 (c) The correct statement is density, the velocity of sound decreases.
longitudinal wave can propagate in With increase in humidity, density of air Now, velocity
solids, liquids and gases. Transverse
wave does not require medium to decreases. So, with rise in humidity, 1
travel, it consists of crests and troughs. velocity of sound increases.
In a longitudinal wave, particles of the Hence, option (c) is correct.
medium vibrate to the direction of 4 (d) Amplitude is independent of Vmixture -
propagation of wave. In longitudinal wavelength, velocity and frequency of VH, VPmixtue
wave, higher density corresponds to oscillation.
compressions. Because, velocity (v) = frequency O)x
Hence, options (a), (b), (d) and (e) are wavelength (2) hence these are
incorect. related with each other. Whereas 7(c) We know that, intensity lre
amplitude is the maximum decibel,
3 (c) The velocity of sound in a gas or displacement
of the
suffered by the particles
air is given by medium from mean position.
5 (c) Volume elasticity,
V= YP YRT
Vp V M |B|= Ap 100 x 10
AVIV 0.005/100 :.L=10 logio
where symbols have their usual
=2x 10°
meaning.
Thus, from Eq. (i), we can conclude that Speed of sound, v= B2x 10° L=10log,n (2 x10)
L=10[log,n2+log,(0)"1
=1.4x10
As the pressure of gas or air increases, = 1400 m/s L=10(4.3)L=43 decibel
is of's
then density also increases and hence 6 (b) Let one mole of 8 (b) Loudness ofsound
strengih
each gas has
volume as V. When they are same mixed,
term describingSOundintensity.
then density of mixture is It is related to