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NCERT-XI / Unit- 14 – OSCILLATION
Q.1-What is Simple Harmonic Motion?
A periodic motion confined between two fixed points about a mean position such
that acceleration produced is directly proportional to the displacement , but taking
place in opposite direction in known as simple harmonic motion.
Eg – Motion of a simple pendulum.
Q.2- Prove that projection of uniform circular motion along its diameter is
SHM or Deduce the equation of motion of SHM.
Let us consider a particle under going uniform circular motion with uniform
angular velocity in anticlockwise direction along the circumference of a circle of
radius r . When the particle goes from X to Y, Y to X/, X/ to Y/, Y/ to X/, the
projection of the particle
moves from O to Y, Y to O, O to Y/, Y/ to O,
respectively along the diameter YO Y/ .
So the projection of a uniform circular motion , along a diameter of the circular
path is (i) taking place symmetrically about the mean position O
(ii) is confined between two extreme position Y and Y/
(iii) is periodic , as  = constant as = 2π/T
Let at any instant t, it be at P such that OM = y and POX = 
From ∆OPM,
Sin  = OM/OP = y/R => y = r sin .
=> y = r sin t -------------- (i)
 = /t so = t
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NCERT-XI / Unit- 14 – OSCILLATION
Differentiating equation (i) with respect to time we have velocity as
=> v =
dy d

r sin t = - r  cos t -------- (ii)
dt dx
Differentiating equation (ii) with respect to time we have acceration as
=> a =
dv d

r  cos t. = - 2r sin t
dt dt
=> a = - 2y …………………(iii)
As w is constant => a  – y ---- (iv)
Hence, motion of the projection of uniform circular motion along any of its
diameter is SHM in nature.
(iii) => a + 2y = 0 -------- (iii)
This is the equation of motion of SHM.
d2y
2
In differential form the equation of motion of SHM =>
2 +  y = 0 …(v)
dt
Q.3- Give the condition of maximum and minimum velocity and acceleration
of a body during SHM .
We know , displacement of SHM , y = rsint ……………..(i)
And velocity of SHM , v = r cost ……………..(ii)
And acceleration of SHM , a = - r2sint ……………..(i)
y 2
r2  y2
 r2  y2
(ii) => v  r  1  sin wt  r  1  ( )  r 
2
r
r
2
(iii)=>
a  2y
At extreme position, y = r , so v = 0 and a = 2 r
At mean position , y =0 , so v =  r and a = 0
-----------------------------------------------------------
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NCERT-XI / Unit- 14 – OSCILLATION
Q. 1. Prove that motion of a simple pendulum is simple Harmonic in nature.
Calculate its frequency & time period.
A simple pendulum is a metallic bob suspended by means of a mass less
inextensible string from a rigid support.
As shown in the diagram. Let us consider a simple pendulum of mass (m) be
suspended by means of are string OA of length (l) from a rigid support, and
in equilibrium position.
Let it be displaced from A to B such that its displacement, AB= x. Let  is the
angle made by the string with the vertical.
Resolving its weight mg at B, we have –
1. mg sin , towards the mean position
2.mg cos , in a direction opposite to the tension T in the string.,
As, mg cos T, they will cancelled each other .
Hence driving force in the bob, F = mg sin  -------------- (i)
As  is small so from  AOB, Sin  =
AB x
 ---------- (ii)
OA L
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NCERT-XI / Unit- 14 – OSCILLATION
(i)=> F = mg(- x/L) (negative signal indicates that force and displacement are
always opposite to each other)
=> ma = - mg x/L
=> a = - g x/L
=> a = - g/Lx ------------- (iii)
as g/L is constant
a  – x.
So the motion of the simple pendulum is simple harmonic in nature.
Comparing equation (iii) with standard eqn of simple harmonic motion, a   2 x ,
we have

2

g
L
g
  
 v 

l
1
2π
2π

T
g
l
 T  2 π
L
g
g
L
Q. 3. State the laws of vibration of a pendulum
Ans:- From the explain for time period of a simple pendulum, T  2π L g , we
have
1. Law of length – Time period of a simple pendulum is directly proportional to the
square root of effective length of the pendulum, provided g remains constant.
2. Law of Gravity of acceleration due to Gravity – Time period of a simple
pendulum is inversely proportional to the sqr root of acceleration dire to gravity,
providing its length remaining the same.
3. Law of mass – Time period of simple pendulum is independent of the mass of
the bob.
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NCERT-XI / Unit- 14 – OSCILLATION
Q.4.What is second pendulum ? calculate its length.
Ans:- The simple pendulum of which time period equal to 2 seconds, is known as
second’s pendulum.
We know that
T  2π L
2
2
2
2
g => T = 4  L/g =>L = GT /4 .
Here T = 2 seconds. So L = 9.8 x 4 /4 2.
= 9.8/9.87 = 0.9927 m = 99.27cm.
Q. 2. Prove that motion of a vertically loaded spring is SHM in nature and
also calculate its frequency and time period.
Ans:-
.
Let us consider a spring of spring constant, k. be suspended from a visit support
in a vertically downward direction. Let it be elongated by a amount, l by
suspending a weight at it free end so that restoring force acting on the spring,
F = - k l -------------- (i)
Let the spring be elongated further by a small amount, x by applying a force, F/
so that restoring force is,
F/ = - k (l + x) ------- (ii)
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NCERT-XI / Unit- 14 – OSCILLATION
The acceleration of the body = Net force acting on the spring / mass of the body
a = (F/ - F)/m = [- k (l + x) + kl]/m
=> a = - k x/m ---------- (iii)
So, k/m is constant , a  - x.
So, the motion of a vertically loaded spring is simple harmonic in nature.
Comparing eq(iii) with the standard equation oh SHM , a   2 x , we have
k
2π
   k 
 k  T  2π k
m
m
m
m
T
1 k
 v 
m
2π
 2 
Q.5.Calculate the energy of a body in SHM
Or –Prove that a body in SHM obeys the principal of conservation of energy.
Ans :-
Let us consider a body of mass ‘m’ , connected to a horizontal spring of spring
constant ,k on a smooth horizontal surface , undergoing S.H.M. in between X & X l
about the mean position O, with r as its amptitude.
To calculate its Potential energy (U) –
Let P be a point at a distant of x from the mean position at which we have to
calculate its potential energy. The magnitude of restoring force at P is given as –
F = kx ---------------- (i)
Let Q be a point very much closer to P so that PQ = dx. The amount of work done
in moving it from P to Q.
dW = F. dx ----------- (ii)
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NCERT-XI / Unit- 14 – OSCILLATION
Total amount of work done in moving it from O to P.
W
W 
 dW
0
x
x
  Fdx 
0
 Kxdx
0
x2 x
1
 k[
] 0  kx 2 _(iii)
2
2
This amount of work done will be converted into potential energy at P,
=> UP = ½ k x2 --------------- (iv)
But k = m 2. so
=> UP = ½ m2 x2 -------- (v)
To calculate its Kinetic energy (K) :We know , velocity of the particle at P, v =  r 2  x 2
So KP = ½ mv2 = ½ m 2 (r 2 – x2) ------ (vi)
So total energy at P , EP = UP + KP = ½ m2x2 + ½ m 2 ( r 2 – x2)
EP = ½ m2 r 2 ---------- (vii)
This is the total energy of a particle executing SHM at any point.
At O , x= 0 so, EO = UO + KO = 0 + ½ m 2 ( r 2 – 0) = ½ m2 r 2
At X , x= r
so, EX = UX + KX = ½ m2r 2 + ½ m 2 ( r 2 – r2) = ½ m2 r 2
So , EP = EO = EX = ½ m2 r 2 ,
which is found to be constant at any point on its path . Thus principle of
conservation of energy is obeyed by the body executing SHM
Drawing the graph between KE & PE with distance , it is found that graph between
total energy & distance is a straight line.
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