Simple harmonic motion
Periodic motion: A motion which
repeats along the same path after a definite interval of time is called
periodic motion. Earth’s motion around the Sun, motion of hand or wall clock,
and motions of planets are example of periodic motion.
Oscillation/ vibration: When a
body moves back and forth repeatedly about a mean position, its motion is
called oscillation or vibration. Motion of a simple pendulum, vibration of a
tuning fork, motion of a spring etc are examples of oscillation or vibration.
Simple harmonic oscillation: The
type of vibratory motion of a body such that the restoring force or the
acceleration acting on the body is directly proportional to the displacement
from the mean position and always directed towards the mean position is called
the simple harmonic oscillation or motion. Motions of a simple pendulum with
small amplitude, vibration of arms of tuning fork, motion of a spring etc are
examples of simple harmonic oscillation or motion (SHM).
Characteristics of simple harmonic oscillation: A simple
harmonic motion has the following characteristics:
1. Its motion is periodic, 2. It
particular time interval the motion becomes opposite, 3. Its motion is along a
straight line, 4. Its acceleration is proportional to the displacement, 5.
Acceleration is opposite to displacement and 6. Acceleration points towards the
mean position of the object.
Expression of displacement, velocity and acceleration of a particle
executing Simple harmonic motion:
a) Displacement:
Let a particle move around a point O in a circular path ABCD
of radius A at an angular velocity ω in the direction shown by the arrow in
fig-3. Let at time t it comes to point P from point A. From P a normal is drawn
on the diameter DB. Here the displacement of the end point of the normal is, x
= ON.
Now, from fig-3, , here θ is the angular displacement.
We, get, ON/OP = sin θ
or, ON = OP X sin θ or, x = Asin θ
Or, x = Asinωt……………(1). If T is the time period of
oscillation, then ω = 2π/T = 2 πf.
So, x = Asin 2 πft………….(2). Eq. (1) and (2) are the eqs. Of
displacement of a particle executing SHM.
b) Velocity: We know that the rate of change of displacement
is called velocity. It is denoted by v.
Relation between velocity and
displacement: We know, x = Asinωt or, sinωt = x/A and cosωt = √(1-sin2ωt).So, v = Aω cosωt = Aω √(1-sin2ωt)
=
(i) When x = A, then v = 0 and (ii)
when x = 0, then v = Aω
so, at the mid-point of the motion
of N, the velocity will be maximum and with the increase of displacement
velocity will start decreasing. At the extreme ends, i.e., at B or D its
velocity will be zero.
c) Acceleration: We know the rate of change of velocity is
called acceleration. It is denoted by a.
This is (above) the relation
between acceleration and displacement. Negative sign indicates that
acceleration and displacement are opposite to each other.
(i) when x = 0, then a = 0 and (ii)
when x = A, then a= -ω2A. That is at maximum displacement of the
motion of N, acceleration is maximum and at mid point it is zero.
Differential equation of simple harmonic
oscillation
Let a particle of mass m oscillate
in simple harmonic motion. Now, at time t if its displacement is x, then
Velocity,
and acceleration,
. Now, magnitude of force acting on the particle, 



Since force or acceleration is
proportional to displacement and is in opposite direction, so

Again, if the angular velocity of
the particle is ω, then we get,
………………..(2) [By eq.
(1)]

Comparing eq. (1) and (2), we have
……… (3). This is the differential equation of a particle
executing SHM.

Solution of differential equation
of SHO: The differential
equation of a particle executing SHM
is given by




When, x = A, then dx/dt = 0.
Inserting this condition in eq. (3), we get, c = ω2A2.
So, 





Prob-1: The amplitude and
frequency of an object executing SHM
are 0.01 m and 12 Hz, respectively. What is the velocity of the object at
displacement 0.005 m? What is the maximum velocity of the object?
Solution: a) We know,
=0.653m/s.

b) We know, when x = 0,
then v = vmax. so, 

Prob-2: The time period of an
object executing SHM is 0.001s and
its amplitude is 0.005m. Calculate its acceleration at 0.002m from the
mid-point of the motion.
Solution: a) We know, a =
ω2x =

Progressive waves
Definition: If the
wave generated from a source progresses with time from one point to another
through a medium, it is called progressive wave. Progressive wave can be
longitudinal or transverse.
Characteristics of Progressive
waves:
- These waves are generated by continuous disturbance of a portion of a medium.
- These waves travel with a fixed velocity through a uniform medium.
- The velocity of propagation of the waves depends on the density and elasticity of the medium.
- Vibrations of the particles of medium may be transverse or longitudinal.
- As the wave progresses, every point of the medium undergoes same change of pressure and density.
Equation
of progressive waves:
or, 


Stationary Waves
Definition:
The resultant wave produced by the superposition of two progressive waves,
having same wavelength and amplitude, traveling in opposite directions is
called stationary wave. The stationary wave has no forward motion but remains
fixed in peace.
In stationary
waves there are certain points where the amplitude is zero. These points are nodes and there are some
points where the amplitude is maximum. These points are called antinodes. In
fig-1 points A are antinodes and N is nodes.
