simple harmonic motion

Definition:
 Such a motion in which acceleration is directly proportional to
the displacement and is directed towards the mean position is
called simple harmonic motion(SHM).

Condition FOR SHM:
The system should haves restoring force.
The system should have inertia.
The system should be frictionless.

Hooke’sLaw
force that is applied to spring is directly proportional to the
displacement.
 If the spring is un stretched, there is no net force on the mass
or the system is in equilibrium.
if the mass is displaced from equilibrium, the spring will exert
a restoring force, which is a force that tends to restore it to the
equilibrium position.
F = kx
where,
F  Elastic force
k  Spring constant
x  Displacement
𝑭 ∝ 𝒙

Expression for acceleration of the body executing
SHM:
Consider a mass ‘m’ attached to one end of
elastic spring which can move freely on a
frictionless horizontal surface.
When the mass is released, it begins to vibrate
about its mean or equilibrium position.
But due to elasticity, spring opposes the applied
force which produces the displacement. This
opposing force is called restoring force.

Expression for acceleration of the body executing
SHM:
The restoring force is written by;
If ‘a’ is the acceleration produced by force ‘F’ in mass-spring system at
any instant, then according to Newton’s law of motion.

Comparing (i) and (ii)
From above equation we can write it as
Solution of the form
∴ 𝜔 =
𝑘
𝑚

Velocity can found by differentiating
displacement
Acceleration can found by differentiating velocity
Simplifying acceleration in terms of displacement:
Acceleration can also be expressed as:

Characteristic of Mass-spring system executing SHM:
When a body is vibrating, its displacement from
the mean position changes with time. The value
of its distance from the mean position at any
time is known as its instantaneous displacement.

 The maximum value of displacement is known
as its amplitude.
 A vibration means one complete round trip of
the body in motion.
The time required to complete one vibration is
called time period

The number of cycles per second. A
cycle is a complete round trip is called
Frequency
If T is time period of a body executing
SHM, its angular frequency can be
written as;

Phase angle:
The angle 𝜃 = 𝜔𝑡 which specifies
the displacement as well as the
direction of the point executing SHM
is known as phase angle.

General Equation
where,
x  Displacement
A  Amplitude of the oscillation
f  Frequency
t  time
ф  Phase of oscillation
If there is no displacement at time t = 0, the phase is ф = π/2

Loudspeaker.
The cone of a loudspeaker oscillates in SHM at a frequency
of 262 Hz. The amplitude at the center of the cone is A = 1.5
x 10-4 m, and at t = 0, x = A. (a) What equation describes the
motion of the center of the cone? (b) What are the velocity
and acceleration as a function of time? (c) What is the
position of the cone at t = 1.00 min (= 1.00 x 10-3 s)?

a. w = 2p f = 2p ×262Hz =1650rad/s,
x(t) = Asin(wt +f),
x(0) = Asinf = A Þ f = p 2,
x(t) = Asin(wt +p 2)
= 1.5´10-4
m( )cos 1650t( ).
Solution:

b. v(t) =
dx
dt
= -0.25m/s( )sin 1650t( );
a(t) =
dv
dt
= -410m/s2
( )cos 1650t( )
c. at t =1.00ms,
x = 1.5´10-4
m( )cos 1650rad/s×1.00´10-3
s( )
= -1.2´10-5
m.
CONT…

Energy conservation in Simple harmonic motion:
if the friction effect are neglected, total mechanical
energy of vibrating mass spring system remains constant
 The velocity and position of the vibrating body are
continually changing
 The kinetic and potential energies also change, but their
sum must have the same values at any instant.
 By hook’s law
 F=-kx
𝑊 = 𝑓 𝑑𝑥
 U=-W
 U=− 𝑓 𝑑𝑥
 𝑈 = − −𝑘𝑥 𝑑𝑥
 𝑈 = 𝑘 𝑥 𝑑𝑥
 𝑈 = 𝑘
𝑥2
2

 Putting value of displacement
 Kinetic energy is given by
Kinetic energy is maximum if x=0 when the mass is at
equilibrium position.
The energy is partly P.E and partly K.E.

Simple pendulum:
 A simple pendulum is idealized model consist of a
point mass suspended by an inextensible string of
length l fixed
 When pulled to one side of its equilibrium position
A to the position B through a small angle θ and
released, it starts oscillating to and fro over the
small.

Simple pendulum performs SHM:
 Condition for SHM is that restoring force F
should be directly proportional to the
displacement oppositely directed. The path
of the bob is not straight line, but the arc of
the circle of radius l

 Let T is the tension in the spring. When the
particle is at point B two forces are acting
on it:
1. mg, the weight of the point bob acting
vertically downward.
2. T, the tension along the string.
 The weight mg can be resolved into two
rectangular components.
1. Component of weight mg along the
spring=mgcosθ
2. Component of weight mg perpendicular to
the string=mgsinθ

 Since there is no motion of bob along the string, so the
component mg cosθ must be equal to tension in the
string T
T=mg cosθ
 Component mg sinθ is responsible for the motion of the
bob towards the mean position. Thus, the restoring force
F is;
F=-mg sinθ ………(iv)
So sinθ=
𝑥
𝑙
F=-mg
𝑥
𝑙
 By 2nd law of motion
F=ma …………….(v)
 Comparing (iv) and (v)
ma=- mg
𝑥
𝑙
a= -
𝑔𝑥
𝑙
a=-(constant)x
a∝-x

Expression for time period:
We know that
a=-x𝜔2
⇨ a= -
𝑔𝑥
𝑙
-x𝜔2
=-
𝑔𝑥
𝑙
⇨𝜔 = 𝑔/𝑙
𝑇 =
2𝜋
𝜔
⇨ T=2𝜋
𝑙
𝑔
Frequency:
F=1
𝑇
F=1
2𝜋 × 𝑔/𝑙

An object in simple harmonic motion has the
same motion as of an object in uniform circular
motion:
Relation between uniform circular motion and SHM

 Consider the particle in uniform circular motion with
radius A and angle φ
 x= A cos φ
 Particle’s angular velocity, in rad/s, is

𝑑φ
𝑑𝑡
=ω
 This is the rate at which the angle φ is
increasing.
If the particle starts from ∅0= 0 at t = 0,
its angle at a later time t is simply
 φ =ωt

As φ increases, the particle’s x-component is
x(t) = A cos ωt
The particle in started at ∅0 = 0. fig shows a
more general situation in which the initial angle
∅0 can have any value. The angle at a later time t
is then
φ =ωt+ ∅0
 v(t) = -ωA sin(ωt+ ∅0) =
v(t) = -𝑣 𝑚𝑎𝑥 sin(ωt+ ∅0)

Definition:
 In reality there is always resistance in
any system. We neglected this when we
are dealing with SHM that is idealistic
case
 The oscillation in which the amplitude
decreases gradually with time are called
damped oscillation.

 The equation of motion for a mass m vibrating on
the end of a spring of force constant k, in the
absence of any damping, is
m 𝑥= − kx
 Damping force is given by
 Applying Newton 2nd law the total force Ftot on
the body is on the body is
 Since Ftot = Fs + Fd
Therefore

 This differential equation may be rearranged into
 𝜔 =
𝑘
𝑚
𝛽=
𝑐
𝑚
 𝑥 + 𝛽 𝑥 + 𝜔2 𝑥 = 0
 The auxiliary equation of differential equation
𝑚2
+ 𝛽𝑚 + 𝜔2
= 0
𝑚 =
−𝛽 ± 𝛽2 − 4𝜔2
2
m=
−𝛽
2
±
𝛽2
𝟒
− 𝜔2

There are three cases
Under damped case(light damping)
(
𝛽2
𝟒
− 𝜔2
) < 0
The system oscillates about the equilibrium position
with decreasing amplitude over a period of time.

Air resistance on a pendulum, air
resistance lowers the amplitude of a
pendulum. This is an example of Light
Damping
Example

Case 1:

𝛽2
𝟒
− 𝜔2 < 0

𝛽2
𝟒
− 𝜔2 = −𝑛2
 m=
−𝛽
2
± 𝑖𝑛
 x=𝑒−
𝛽𝑡
2 asin 𝑛𝑡 + 𝑏𝑐𝑜𝑠 𝑛𝑡
 By putting
 a=dcosφ b=dsinφ
 x=𝑒−
𝛽𝑡
2 𝑑 sin 𝑛𝑡 cosφ + 𝑐𝑜𝑠 𝑛𝑡 sinφ
 x=𝑒−
𝛽𝑡
2 𝑑(sin(nt + φ)
 if 𝛽 is small quantity, then amplitude 𝑒−
𝛽𝑡
2 𝑑 is
gradually decreases with time and tend to zero as t
approaches to infinity

critically damped case
𝛽2
𝟒
− 𝜔2
= 0
The system does not oscillate & damping
is just adequate such that the system returns to its
equilibrium position in the shortest possible time.

 Automatic door closer
 The recoil mechanisms in most guns are
also critically damped so that they return
to their original position, after the recoil
due to firing, in the least possible time..
Examples:


𝛽2
𝟒
− 𝜔2 = 0

𝛽2
𝟒
= 𝜔2
 m=
−𝛽
2
,
−𝛽
2
 m= 𝜔 , 𝜔
 x=(A+Bt)𝑒−𝜔𝑡
 In this case the motion is non-periodic and non-
oscillatory
Also x 0 as t ∞
lim
𝑛→∞
(A+Bt)
𝑒 𝜔𝑡
= lim
𝑛→∞
(Bt)
𝜔𝑒 𝜔𝑡
= 0

 over Damped case
(Heavy damping)
𝛽2
𝟒
− 𝜔2 > 0
The damping is so great that the displaced
object never oscillates but returns to its
equilibrium position very slowly.

 Coiled spring mattress, where the springs slowly return to a
equilibrium position when a person lies on the mattress.
Example:


𝛽2
4
− 𝜔2 > 0

𝛽2
4
− 𝜔2 = 𝜆 2
 m=
−𝛽
2
± 𝜆
 x=C𝑒
−𝛽
2
+𝜆 𝑡
+ 𝐷𝑒
−𝛽
2
−𝜆 𝑡
 In this case the motion is non-periodic and non-
oscillatory. It is called dead beat. Both the
exponentials decay with time. Therefore, it tends
to zero as t ∞.

An application of damped oscillation is the
shock absorber of the car which provides a
damping force to prevent excessive
oscillations. If the shock absorbers are badly
worn, then car becomes too bouncy and this is
uncomfortable.

Another application of damped oscillation is
the shock absorber of the human body
skier’s body moves over the bumpy snow
smoothly while his /her tights and calves act
like a damping spring.

 Swing door need to be damped to prevent them
swinging back on people.
 Imagine walking through a swing door without
dampers; if you walk too slowly through it, it
might bounce back and hit the person.
 A damper is fitted to door slows its motion and
stops it from swinging back. The damper needs
to provide heavier than critical damping so the
door close slowly.
Swing DOOR

Damped force oscillation
In order to overcome the damping effect of
the medium an applied force, called driving
force, is generally applied. In such a case the
system is called damped force oscillator

𝑥 +
𝑐
𝑚
𝑥 +
𝑘
𝑚
𝑚 = 𝐹𝑐𝑜𝑠𝑝𝑡
 𝑥 + 𝛽 𝑥 + 𝜔2 𝑥 = 𝐹𝑐𝑜𝑠𝑝𝑡
(𝐷2+𝛽𝐷 + 𝜔2)𝑥 = 𝐹𝑐𝑜𝑠𝑝𝑡
 (𝐷2+𝛽𝐷 + 𝜔2)𝑥 =0
 Complementary function decays with time.
For this reason this part of solution is called
the transient solution.

simple harmonic motion

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