Download A motion that repeats itself in equal interval of time is called periodic

A motion that repeats itself in equal interval of time is called periodic motion. The
motion of the earth about its own axis, the motion of the earth round the sun, the
motion of electrons in an atom are few common examples of periodic motion.
When a particle in periodic motion moves to and fro along the same path, the
motion is called as oscillatory or vibratory motion. The oscillation of a simple
pendulum, the oscillation of a mass attached to a spring etc. are the few examples of
oscillatory motion.
It may be said from the above discussion that any oscillatory motion should
always be periodic, but a periodic motion may not always be oscillatory in nature.
The time required to complete one oscillation is called time period
and the
number of oscillations made per second is called the frequency
of the motion. So,
1
the relation between time period
and frequency
is given by,
The position at which
on an oscillatory particle is called its
equilibrium or mean position. The distance of a particle at any instant from the
equilibrium position is known as displacement of the particle. The maximum
displacement on either side of the equilibrium position is called amplitude of
the motion.
If a particle moves to
and fro in such a way that the
acceleration of the particle is
always directed towards a
fixed point in its path of
motion and is proportional
to the displacement of the
particle from the fixed point
at any instant, the motion of
the particle is called simple harmonic motion (S.H.M.).
The graphical representation of simple harmonic motion is given in Fig. 1.
1
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A Textbook of Basic Engineering Physics
If a particle of mass ‘ ’ executes S.H.M. and be the displacement of it from the
equilibrium position, the acceleration of the particle is given by,
1.1
where
is a proportionality constant.
Hence, the force ‘ ’ acting on the particle is given by,
1.2
Here,
is called
of the particle.
representing
The characteristics of S.H.M are,
The motion is linear.
The motion is periodic and oscillatory.
The restoring force acting on the body (
the acceleration of the body) is
proportional and oppositely directed to its displacement measured from a
fixed point (
mean position) in the path of the motion.
If
be the mass of the particle executing
S.H.M. and
be its displacement at any instant
from the mean position ‘ ’ (as shown in Fig. 2),
the force acting on the particle at any point
is
given by,
2
1.3
2
where is the constant of proportionality (force constant) and negative sign is taken,
as the restoring force is opposite to the direction of displacement.
The equation 1.3 may be written as,
2
2
or,
or,
2
2
2
2
2
2
,
2
, where
= natural angular frequency
0
This is
1.4
.
The time period of a S.H.M. is given by,
2
1.5
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Simple Harmonic Motion
Multiplying both sides of equation 1.4 by 2
2
2
2
2
2
0
, we get,
2
or,
2
2
0
Integrating , we get,
2
2 2
Now, the velocity
displacement
, where
is the constant of integration.
1.6
of the simple harmonic particle is zero at its maximum
at
(amplitude of the particle).
2 2
Hence, we get from the equation 1.6 , 0
So, the equation 1.6 takes the form,
2
2
2
2
or,
2
or,
Proceeding with the positive sign, we have,
2
2 2
2
2
On integration,
2
or,
sin
or,
2
1
[where
= integration constant]
sin
sin
or,
1.7
Here
is the phase of the
vibrating particle and is the epoch, or initial phase of the particle (
it is the phase
0 ). Here the natural cyclic frequency of vibration 0
of the particle at
. Here if
2
sin
we proceed with negative sign, the solution will be of the same form
.
To express the solution of equation 1.7
1.7 as,
=
=
=
sin
cos
sin
cos
cos
sin
sin
sin
cos
cos
This is the general solution of equation
sin
in a more general form, let us write relation
are two constants such that
1
1.5
of a S.H.M., where
2
2
and
cos
and
tan 1
Phase is the state of motion of the particle executing S.H.M. and it indicates about the
position and direction of motion at any time.
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A Textbook of Basic Engineering Physics
Now, we get from equation 1.5 by ignoring negative sign,
2
2
2
or,
1.8
acceleration ,
displacement
[
time period of oscillation]
displacement
acceleration
2
We know, the displacement
given by,
1.9
of a particle executing S.H.M. at any time
is
sin
Velocity
=
or,
cos
1.10
When the particle is at its mean position,
cos
1
at time
2
2,
2
2
0,
where
0.
0 so,
sin
, cos
1
=2
2
2
1.11
Now, when the particle is at the mean position of the path
for
0,
max
At the extreme positions of the path
for
We have, from equation 1.10 ,
cos
,
min
0.
2 sin
Acceleration
1.12
When the particle is at the mean position,
0.
2 sin
2
or,
[
at
0,
0 so,
When the particle is at extreme positions (
max
2
2
for
]
), the acceleration
(neglecting the negative sign)
When the particle is at the mean position (
min
sin
0
0
for
0 ), the acceleration,
1.13
Simple Harmonic Motion
5
We know that the displacement of a particle executing S.H.M. at any instant of time
sin
is
1
velocity
cos
Comparing equations 1
sin
and 2
2
2
we get that, the volocity of the particle executing
S.H.M. at any instant leads the displacement by a phase difference
in quadrature)
Now differentiating eqn. 2
2
radian (
two are
we get acceleration
cos
2 sin
2 sin
3
Comparing equations 2 and 3 we find that the acceleration at any instant leads the
velocity by
radians in phase.
2
Similarly comparing equations 1 and
displacement by radian. [
antiphase]
3
we find that acceleration leads the
A particle executing a simple harmonic motion possesses both potential and
kinetic energy at any instant.
Kinetic energy (
with velocity
) The kinetic energy of a particle of mass
executing S.H.M. is given by,
1
2
But,
and moving
2
sin
cos
1
2
1
2
1
2
2
2
2
2
2
2
cos
1
2
2
2
1.14
Potential energy ( ) The potential energy of a vibrating particle of mass at
any instant is calculated from the total amount of work done in overcoming the
2 . The negative sign is taken
restoring force through a distance . Here, force =
as the force is opposite to the direction of motion of the vibrating particle. So, the
potential energy,
2
0
1
2
2 2
1.15
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A Textbook of Basic Engineering Physics
So, we may conclude from equation 1.14 and 1.15 that, for
min
0 , but
1
2
max
2 2
1
2 2 , but
min
2
So, K.E. is maximum when P.E. is minimum and vice versa.
Similarly, for
0,
max
Total energy Now, the total energy,
or,
=
or,
1
2
2
2
2
1
2
2
2
2
1
2
2
0
=
2 2,
2 2
,
where
2
constant
1.16
So, the total energy is constant in S.H.M. and is proportional to the square of the
amplitude.
In case of free vibration, the restoring
force is proportional to its displacement
and it is the only internal force acting
on the body. Hence, we can say from
1.16 that the
equation 1.7 and
maximum amplitude of free vibration
and the total energy of the (free) simple
harmonic motion remain constant with
time [Fig. 4].
1
The displacement of a particle of mass 0.2 kg executing S.H.M. is indicated by
10 sin
m. Calculate
3
12
i
the amplitude
iv
the maximum velocity and maximum acceleration
v
the epoch
ii
the angular velocity
iii the time period
vi the phase difference after 1 second
vii the displacement, velocity and acceleration of the particle after one second
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Simple Harmonic Motion
viii the kinetic energy and potential energy of the particle at displacement
0.04 m .
ix the total energy of the particle.
The general equation of a simple harmonic motion is given by,
sin
1
10 sin
Here,
3
2
12
Comparing equations 1 and 2 , we get
i
10 m
ii
iii
3
2
,
2
6s
3
iv
2
executing S.H.M. is,
0
for
max
10
3.14
3
10
3
The
10.46 m s
2
for
3.14
3
2
1
2
,
max
2
10
10
3
2
10.95 m s
v
3
Now,
is the initial phase for
We get from equation 2 ,
vi
The phase after
when
0.
12
second
1 , phase
1
3
4
1s =
So,
vii
3
12
12
12
4
12
4
3
=
12
=
2
180
= =
= 30
12
6
6
1 second,
10 sin
The
10 sin
12
10
2
4
5 2m
1 second
2
The
1
3
2
3
10 2
5 2
1 second
2
3
2
50
2
3
7.4 m s
5 2
1
7.75 m s
2
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A Textbook of Basic Engineering Physics
1
2
viii K.E.
2
2
2
So, at displacement
1
2
0.2
3.14
3
P.E. =
1
2
0.2
3.14
3
1
2
2 2
0.04 m
K.E.
ix
1
2
and P.E.
2 2
2
2
1
2
10 2
0.04
0.04
2
10.95 joule
0.0001 joule
3.14
3
0.2
2
2
10
2
10.955 joule
2
The differential equation of motion of a freely oscillating body is given by
2
2
2
2
18
0 . Calculate the natural frequency of the body.
According to the problem, 2
2
9
3
2
or,
2
3
or,
2
18
2
2
or, 2
0
2
or,
2
9
2
3
1.5Hz .
3
A weight glass tube is floating in liquid with 20 cm of its length immersed. It is
pushed down a certain distance and then released. Compute the time period of
its vibration.
Let the density of the liquid be
the glass tube be .
g cm
3
and the area of cross-section of
The volume of the liquid displaced by the glass tube
= mass of the displaced liquid = 20
Mass of the tube =
cm 3
20
g
Now, if we consider the tube is pushed by a further length of
upthrust
acting on the glass tube,
2
or,
or,
cm, the additional
2
2
2
2
,
This equation implies that the motion is simple harmonic.
The time period,
2
2
=2
20
=2
20
980
0.89s .