Simple Harmonic Motion (Practice Questions)
Simple Harmonic Motion (Practice Questions)
Simple Harmonic Motion (Practice Questions)
2. (a) A copper wire of length 8 m and a steel wire of length 4 m, each of cross section 0.5
cm2 are joined end to end and stretched with a tension of 500 N. If yopper = 1 1011
N/m2,
ysteel = 2 1011 N/m2. Find the elastic potential energy stored in the system ?
(b) At what depth a rubber ball shall be kept in a lake so that its volume decrease by 0.1
%? (Given Bulk modulus of rubber = 10 108 N/m2 )
[2+1=3]
FL 500 8
2. (a) ecopper = = = 0.8 mm
Ay 1 10 11 10 .5 10 − 4
FL 500 4
esteel = = = 0.2 mm
Ay 2 10 (0.5 10 − 4 )
11
dx
5. = 5 cos(20t + / 3) = 0
dt
t= .
120
9. A body of mass 1 kg is executing SHM which is given by x = 6 cos (100 t + /4) cm.
What is the
(i) frequency
(ii) initial phase
(iii) acceleration
(iv) maximum kinetic energy for given SHM?
10. If two SHM's are represented by equations y1 = 10 sin [3t +/4] and y2 = 5[sin 3t +3 cos
3t], find the ratio of their amplitudes.
10. y2 = 5 sin 3 t + 53 cos 3t = A cos sin 3 t + A sin cos 3 t [A = 10 , cos = ½ ]
= 10 sin [3t + /3]
y1 = 10 sin [3 t + /4]
A1 = 10 & A2 = 10
A1 10
= =1
A2 10
11. A particle moves along x-axis such that its acceleration is given by a = -b2x. Find the
time period of oscillation.
[2]
2
11. T= , 2 = b2
b
SHM-II-2Marks
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SIMPLE HARMONIC MOTION (BOARD LEVEL)
[1+2=3]
1. fmax = mBg
amax = g
but in SHM amax = A2
A2 = g
g g
A= 2 =
42 f 2
2. One end of a long metallic wire of length L is tied to ceiling. The other end is tied to mass
less spring of spring constant K. A mass m hangs freely from the end of spring. The area
of cross-section and the Young modulus of wire are A and Y respectively. It mass is
slightly pulled down and released, find the time period of oscillation.
2. T = 2 m / k e
k1 = k, k2 = YA/L
ke = (k1k2 / k1 + k2)
( YA + kL )
T = 2 m .
YAk
F/A
3. Y=
/
Fdx (L − x )Wdx
x
= = L
AY LAY dx
L (L − x )Wdx
The total elongation =
0 LAY
L
W x2 WL
= Lx − = .
LAY
2 0 2AY
k2
m
4. T = 2 (k1 + k 2 )
k1k 2
2k
5. =
m
6. In case of simple harmonic motion (a) at what displacement the kinetic and potential
energies are equal (b) what fraction of total energy is kinetic and what fraction is
potential when displacement is one half of the amplitude.
1 1 1
6. k= m2(A2–y2), u = m2y2 and E = m2A2
2 2 2
(a) According to given condition
1 1
k=u i.e m2(A2–y2) = m2y2
2 2
y = 0.7A
k y2 1 3
(b) fk = = 1 − 2 = 1 − = (y = A/2 given)
E A 4 4
u y2 1
fP = = 2 =
E A 4
7. At the ends of three successive seconds the distance of a point moving with SHM from
its mean position, measured in the same direction are 1m, 5m and 5m. Find the time
period of complete oscillation.
7. From eq x = A sin t
1 = A sin t …(i)
5 = A sin (t + 1) …(ii)
5 = A sin (t + 2) …(iii)
solving above equations
6 sin = 10 sin cos
cos = 6/10
= cos-1(3/5)
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SIMPLE HARMONIC MOTION (BOARD LEVEL)
2
T= −1
cos (3 / 5)
8. A disc is suspended at a point R/2 above its centre. Find its period of small oscillation.
I
8. T = 2
Mg
3R
T = 2 .
2g
9. A horizontal platform vibrates up and down with a simple harmonic motion of frequency
2/ Hz. Find the maximum amplitude so that an object kept on the platform remains in
contact with the platform.
9. mg – F = mA2 F
10 5
Amax = 2
= m
4 8
10. A particle executes SHM in straight line. Starting from rest, it travels a distance ‘A’ in the
1st second, it travels a distance ‘B’ in the next second, in the same direction. Prove that
the amplitude of oscillation is 2A2/(3A – B).
SHM-III-2Marks
1. One end of a long metallic wire of length L is tied to the ceiling. The other end is tied to
a massless spring of spring constant k. A mass m hangs freely from the free end of the
spring. The area of cross-section and the Young’s modules of the wire are A and Y
respectively. Find the time period with which mass m will oscillate if it is slightly pulled
down and released.
F 20
1. v= = 10 m/s.
0 .2
v 10
= = = 1m.
n 10
2 2
k= = = 6.28 m-1
1
t x
(b) y = A sin 2 −
T
t x
Equation of wave y = 0.02 sin 2 −
0 .1 1
at x = 0.2 m and t = 0.2 sec.
y = 0.012 m.
dy
(c) slope at any point is = −kA cos( t − kx )
dx
2
= − (0.02 ) cos( −72 0 )
1
= - 0.039.
2. If a particle of mass m moves in a potential energy field v = u0 – ax + bx2 where u0, a and
b are positive constant. Calculate the force constant, frequency of the particle and point
of equilibrium.
dU
2. F=- = a − 2bx … (i)
dx
Force constant = 2b
1 k 1 2b
f= =
2 m 2 m
at equilibrium net force should be zero
a – 2bx = 0
a
x= .
2b
3. The equation of motion of a particle started at t = 0 is given by x = 5 sin (20t + /3) where
x is in centimetre and t in second. When does the particle.
(a) first come to rest
(b) first have zero acceleration
(c) first have maximum speed.
d2 x
(b) = −5 20 sin( 20 t + / 3 ). 20
dt 2
20 t + /3 =
t = / 30
M
4. (a) T = 2
k
m+6
0.75 = 2
600
m = 2.55 kg.
6. A disc is suspended at a point R/2 above its centre. Find its period of oscillation.
7. A body of mass 1 kg is executing SHM which is given by x = 6 cos (100 t + /4) cm.
What is the
(i) amplitude of displacement
(ii) frequency
(iii) initial phase
(iv) velocity
(v) acceleration
(vi) maximum kinetic energy ?
[6]
7. x = 6 cos (100 t + /4) cm given.
(i) amplitude = 6 cm.
(ii) frequency = 100/sec.
(iii) initial phase = /4
1 k
8. f=
2 M m
1 k M
f =
2 m + M
from COLM
Mv = (m + M)V,
At equilibrium v = A = 2Af
M. 2 Af = (m + M) 2Af
A M f
or =
A m + M f
M
A = A .
m+M
9. A body of mass 1 kg is executing SHM which is given by x = 6 cos (100 t + /4) cm.
What is the (i) Amplitude of displacement (ii) frequency (iii) initial phase (iv) velocity (v)
acceleration (vi) maximum kinetic energy ?
= - 100 6 2
− x2
d2 x
(v) acceleration; 2
= −10 4 [ 6 cos(100 t + /4)]
dt
= - 104 x
1 W2 L 1 W2 L
10. Energy stored = +
2 Y R2 2 Y (2R)2
5W 2L
=
8R2 Y
11. A particle executes S. H. M. with a time period of 4s. Find the time taken by the particle
to go directly from its mean position to half of its amplitude.
11. x = A sin (t + o)
At t = 0, x = 0 A sin o = 0 or o = 0
Hence, x = A sin (t)
or A/2 = A sin (t)
or ½ = sin (t)
1
t = sin-1 =
2 6
.T
t= =
6 6(2 )
as = 2/T t = T/12 = 1/3 s
SHM-II-4Marks
1. Find the distance form the top end of a uniform bar of length 24 cm is to be mounted on
wall about an axis perpendicular to its length, so that its time period of oscillation will be
minimum.
M 2 2
= + Mx 2 = M ( + x 2 ) … (ii) Mg
12 1
I
from (i) and (ii) T = 2
C
2. If the cylinder rotates by about B than centre of mass (O) moves by R, P moves by
3R
2R and point Q moves by .
2
Thus by conserving the energy of the system
2
1 1 3R 1 1
k(2R)2 + 2k + 3k(R)2 + IB2 = cons tan t
2 2 2 2 2
where IB is the moment of inertia of cylinder about point B.
solving we get,
23kR 22 3
+ MR 22 = cons tan t
4 4
Differentiating w.r.t time we get
23 2 d 3 d
kR 2 + MR 2 2 =0
4 dt 4 dt
d 23k
=−
dt 3m
23 k
=
3m
3m
T = 2
23 k
m m
3. T1 = 2 ; T2 = 2
k ke
4. Let spring with constant k1 is extended by a distance y and spring with constant k2 is
compressed by a distance x then
y x ax
= y=
a b b
energy of the system is constant
1 1 1 1
k 2 x 2 + k1(ax / b)2 + I + mv 2 = cons tan t
2 2 2 2
1 2 2
1 a x 1 M 2 1
2
k 2 x 2 + k1 2 + + mv 2 = cons tan t
2 2 b 2 3 2
Differentiating w.r.t. time t we get
k 2 + k1(a 2 / b 2 )
=
M
+m
3
5. If the wall is not there it would have performed oscillation with an amplitude of ‘2a’ on
either side. But because of wall on one side it makes oscillation with amplitude ‘2a’ but
on other side it makes oscillation with amplitude ‘a’ and collides elastically with wall and
returns back.
Time period of oscillation = 2 (Time taken to go from A to B) + 2 (Time taken to go
from B to C)
t
Form B to C : A/2 = A sin t T= sec.
12
4T
T = TAB + TBC = 2(T/4) + 2(T/12) =
6
6. When both the pendulums are displaced in the same direction by same amount, the
spring will neither compress nor stretch, so the restoring torque on each pendulum about
the point of suspension will be due to its own weight only.
i.e., = - mgl sin = -mgL [as for small , sin = ]
L L L L
L L
k k ky
k ky
y y
mg mg
mg mg
(A) (B) (C)
But as by definition
d2
= I = mL2 [as I = mL2 ]
dt
d2 g
so mL2
= −2 with 2 =
dt L
This is the standard equation of angular SHM with time period T = (2/).
L 1 g
So here T1 = 2 , f1 =
g 2 L
(b) When both the pendulums are displaced in opposite directions by equal amount (say
y), the restoring torque on each pendulum will be due to its own weight and also by
elastic force of the spring which is stretched by 2y (=2L sin ). So the restoring torque
on a pendulum about the point of suspension will be
= -[mgL sin + k(2L sin)L] = - [mgL + 2k L2]
But by definition = I = mL2 (d2/dt2)
d2 g 2k
so 2 = − +
dt L m
d2 g 2k
or 2
= - 2 with 2 = +
dt L m
1 g 2k
so f2 = = + (> f1).
2 2 L m
pulleys and a mass m hangs from it. Calculate the period of small
B
vertical oscillations of the mass m under gravity.
(g = 10 m/s2.)
A m
[6]
11. (a) Restoring torque about “O” due to elastic force of the spring
= –FL = –kyL (F = ky)
= –kL
2 (as y = L)
1 d2
= I = ML2 2
3 dt
1 d
2
2 = −kL2
3 ML dt 2
d2 3k
2 = −
dt M
3k M
= T = 2
M 3k
(b) In angular SHM maximum angular velocity
d 3k
= 0 = 0
dt max M
v = r(d/dt)
d 3k
so Vmax = L = L0
dt max M
13. Two particles execute SHM parallel to x-axis about the origin with the same amplitude
and frequency. At a certain instance they are found at distance A/3 from the origin on
opposite side but their velocities are found toe be in opposite direction. Find the phase
difference between two.
[8]
15. I = mgR
I
T = 2 …(i)
mgR
1
I = mR2 + MR 2 … (ii)
2
from (i) and (ii)
(m + (M / 2)R
T = 2 .
mg
(m+ (M/ 2)) R
= .
m
1
PE = kx2
2 v
0
1 3
kinetic energy KE = Ip 2 = mv 2
2 4
PE + KE = constant P
1 2 3
kx + mv 2 = cons tan t
2 4
differentiating w.r.t time
dx 3 dv
kx + m( 2v ) = 0
dt 4 dt
from kinematic constraints for rolling motion
dx d2 x dv
= 2v 2
=2
dt dt dt
d2 x 8k 3m
2
=− x time period T = 2
dt 3m 8k
dx
= 2v0 = A where = circular frequency 8k / 3m and A = amplitude
dt max
of x
Amplitude of the centre of the disc = A/2 = v0/
3m
= v0 .
8k
17. Two rods of equal cross – section, one of copper and the other of steel, are joined to
form a composite rod. The length of the composite rod at 20°C is 2.0 m. The part of the
composite rod which is made of copper has a length 0.5 m at this temperature. When
the temperature is raised to 120°C the length of the composite rod increases to 2.002 m.
If the composite rod is fixed between two rigid walls and is thus not allowed to expand, it
is found that length of the component rods also do not change with the increase in
temperature. Calculate the Young’s modulus and the coefficient of linear expansion of
steel, Given : Young’s modulus of copper = 1.3 1011 N/m2, coefficient of linear
expansion of copper = 1.6 10–5 /°C
F = – Kx – Kx
m2
=–
(m1 + m2 ) Kx
m1 m1m2
d2 x
= –
(m1 + m2 ) Kx
2
dt m1m2
m1m2
T = 2
K (m1 + m2 )
20. The pulley shown in figure has moment of inertia I about its axis
and mass m. Find the time period of vertical oscillation of its k
cetre of mass. The spring has spring constant k and the string
does not slip over the pulley.
I
[10]
2
dU
=0 [ energy is conserved]
dt
I dv
0 = 2 + m v + 4kxv
r dt
dv 4kx
or =-
dt I
+m
r2
4k
or a = - 2x where 2 =
(I / r ) + m
2
(I / r 2 ) + m
T = 2 .
4k
21. A thin uniform metallic rod of length 0.5 m and radius 0.1cm rotates with an angular
velocity 400 radian/second in horizontal plane about a vertical axis passing through one
of its ends. Calculate elongation of the rod. The density of the material of the rod is 104
kg/m3 and Young's modulus is 2 1011 N/m2.
2L3
= =3.33 10-4 m.
3
1 1 1 1
E = k(2R)2 + k(R)2 + Mv 2 + I2 M
k
2 2 2 2
where v is the velocity of centre of mass & is the angular P
velocity of cylinder.
Since E is constant.
dE
=0
dt
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SIMPLE HARMONIC MOTION (BOARD LEVEL)
d d dv d
4kR2 . + kR 2 + Mv + I =0
dt dt dt dt
5kR2 + MR2 + I = 0
5kR 2 MR 2
=- where I =
(MR 2 + I) 2
5kR2
=
3 2
MR
2
10 k 3 M
Compare it with = - 2 Thus 2 = or T = 2 .
3 M 10 K
SHM-III-4Marks
3. Two particles execute SHM parallel to x-axis about the origin with the same amplitude
and frequency. At a certain instance they are found at distance A/3 from the origin on
opposite side but their velocities are found to be in opposite direction. Find the phase
difference between the two. [6]
6. A train P crosses a station with speed 20 ms−1 and whistles a short pulse of natural
frequency
n0 = 1570 Hz at t = 0. Another train B is approaching towards the same station with
retardation 8 ms−2 along a parallel track. Two tracks are d = 99 m apart. When train A
whistles, train B is 140 m away from the station and listen the pulse at t = 0.5 sec. If the
u = 18 ms−1
1
Hence speed of the train at t = sec.
2
1
vt = u – at = 18 – 8 = 14 ms−1
2
2
1 1 1
x =18 - 8 =8m
2 2 2
Hence frequency of pulse heard by driver of train B
330 + 14 cos
= 1570 = 1706 Hz.
330 − 20 cos
dE =
2L2 0
sin L
dx =
4L