Lesson Notes By Weeks and Term - Senior Secondary School 2
SIMPLE HARMONIC MOTION
SIMPLE HARMONIC MOTION
SUBJECT: PHYSICS
CLASS: SS 2
DATE:
TERM: 1st TERM
REFERENCE TEXT
WEEK EIGHT
TOPIC: SIMPLE HARMONIC MOTION
CONTENT
DEFINITION
This is the periodic motion of a body or particle along a straight line such that the acceleration of the body is directed towards a fixed point .
A particle undergoing simple harmonic motion will move to and fro in a straight line under the influence of a force . This influential force is called a restoring force as it always directs the particle back to its equilibrium position.
Examples of simple harmonic motions are:
iimass on a string
iiithe simple pendulum
for a body performing simple harmonic motion, the general equation is given as
y = A sin [ wt ± kx ]
where k = phase constant, w = angular velocity, t = time , A = amplitude,
As the particle P moves round the circle once, it sweeps through an angle θ = 3600 (or 2π radian) in the time T the period of motion. The rate of change of the angle θ with time (t) is known as the angular velocity ω
Angular velocity (ω) is defined by
ω = angle turned through by the body
Time taken
ω = θ ………………………………………… 1
t (rad /sec)
θ = ω t
This is similar to the relation distance = uniform velocity x time (s= =vt ) for motion in a straight line
A = r = radius of the circle
The linear velocity v at any point ,Q whose distance from C the central point is x is given by
V = ω √ A2 – X2 ………………………………………… 2
The minimum velocity ,Vm corresponds to the point at X = 0 that is the velocity at the central point or centre of motion .
Hence
Vm =ω A = ω r …………………………………………. 3
Thus the maximum velocity of the SHM occurs at the centre of the motion (X=0) while the minimum velocity occurs at the extreme position of motion (x=A ).
EVALUATION
RELATIONSHIP BETWEEN LINEAR ACCELERATION AND ANGULAR VELOCITY
X = A COS θ
Θ = ωt
X = A cos ω t
dx = -ωA sin ω t
dt
dv =-ω2 A cos ω t
dt
= - ω2X = - ω2A = - ω2r ………………………………………….. 4
The negative sign indicates that the acceleration is always inwards towards C while the displacement is measured outwards from C.
ENERGY OF SIMPLE HARMONIC MOTION
Since force and displacement are involved, it follows that work and energy are involved in simple harmonic motion.
At any instant of the motion , the system may contain some energy as kinetic energy (KE ) or potential energy(PE) .The total energy (KE + PE ) for a body performing SHM is always conserved although it may change form between PE and KE .
When a mass is suspended from the end of a spring stretched vertically downwards and released , it oscillates in a simple harmonic motion .During this motion , the force tending to restore the spring to its elastic restoring force is simply the elastic restoring force which is given by
F= - ky …………………………………… 5
K is the force constant of the spring , but F = ma
a = ky
m
y
Mg
The total work done in stretching the spring at distance y is given by
W = average force x displacement
W = ½ ky x y = ½ ky2 ………………………………… 6
Thus the maximum energy total energy stored in the spring is given by
W = ½ KA2 …………………………………. 7
A = amplitude (maximumdisplacement fromequilibrium position).
This maximum energy is conserved throughout the motion of the system.
At any stage of the oscillation, the total energy is
W = ½ KA2
W= ½ mv2 + ½ ky2 ………………………………………….. 8
½ mv2 = ½ KA2 – ½ ky2
v2 = k/m (A2 –y2)
V = √k/m(A2-y2)
The constant K is obtained from
Hooke’s law in which
F= mg = ke
Where e is the extension produced in the spring by a mass m
But V= ω√A2-X2
Therefore ω =√k/m
Hence the period T = 2π/ω
T = 2π√m
k
EXAMPLE:
A body of mass 20g is suspended from the end of a spiral spring whose force constant is 0.4Nm-1
The body is set into a simple harmonic motion with amplitude 0.2m. Calculate :
SOLUTION
a T = 2π √m/k
= 2π √ 0.02/0.4
= 0.447 π sec
= 1.41 sec
= 2π x 0.71
= 4.46 rad. S-1
= ½ (0.4) (0.2)2
= 0.008 J
Vm2= 0.008 x 2
0.02
= 0.8
Vm= 0.89 m/s
Or V= ω A
= 4.462 x 0.2
= 3.98m/s2 .
EVALUATION
A body of mass 0.5kg is attached to the end of a spring and the mass pulled down a distance 0.01m. Calculate (i) the period of oscillation (ii) the maximum kinetic energy of mass (iii) kinetic and potential energy of the spring when the body is 0.04m below its centre of oscillation.(k=50Nm)
FORCED VIBRATION AND RESONANCE
Vibrations resulting from the action of an external periodic force on an oscillating body are called forced vibrations. Every vibrating object possesses a natural frequency ((fo) of vibration. This is the frequency with which the object will oscillate when it is left undisturbed after being set into vibration. The principle of the sounding board of a piano or the diaphragm of a loudspeaker is based on the phenomenon of forced vibrations.
Whenever the frequency of a vibrating body acting on a system coincides with the natural frequency of the system, then the system is set into vibration with a relatively large amplitude. This phenomenon is called resonance.
EVALUATION
GENERAL EVALUATION
the initial kinetic energy of the stone.
WEEKEND ASSIGNMENT
(A) v=wr (B) v=w2r (C) v= wr2 (D) v=w/r.
(A) acceleration is always directed towards a fixed point.
(B) path of motion is a straight line .
(c) acceleration is directed towards a fixed point and proportional to its distance from the point.
(D) acceleration is proportional to the square of the distance from a fixed point.
(a) 1 (b) 2 (c) 3 (d) 4 (e) 5
(d) compound vibration.
(A) 1and 5 (B) 2 and 4 (C) 3 only (D) 4 only (E) 5 and 3
THEORY
READING ASSIGNMENT
NEW SCH PHYSICS FOR SSS –ANYAKOHA.Pages 188-197
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