# Lesson Notes By Weeks and Term - Senior Secondary School 2

SIMPLE HARMONIC MOTION

SUBJECT: PHYSICS

CLASS:  SS 2

DATE:

TERM: 1st TERM

REFERENCE TEXT

• New School Physics by M.W Anyakoha
• SSCE WAEC Past Questions
• UTME Past Questions

WEEK EIGHT

TOPIC: SIMPLE HARMONIC MOTION

CONTENT

• Definition
• Velocity, acceleration and energy
• Forced vibration

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:

1. loaded test tube  in a liquid

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

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

1. A body of mass 0.2kg is executing simple harmonic motion with an amplitude of 20mm. The maximum force which acts upon it is 0.064N.Calculate (a) its maximum velocity  (b) its period of oscillation.
2. A steel strip clamped at one end , vibrates with a frequency of 20HZ and an amplitude of 5mm at the free end , where a small mass of 2g  is positioned. Find the velocity of the end when passing through the zero position.

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 :

1. The period of the motion
2. The frequency of the  motion
3. The angular speed
4. The total energy
5. The maximum velocity of the motion
6. The maximum acceleration

SOLUTION

a          T = 2π √m/k

= 2π √ 0.02/0.4

= 0.447 π sec

=  1.41 sec

1.   f=1/T  = 1/1.41 = 0.71Hz
2.   ω =2πf

= 2π x 0.71

1. Total energy = ½ KA2

=  ½ (0.4) (0.2)2

=  0.008 J

1. ½ mv2 = /12 KA2

Vm20.008 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

1. Explain the terms forced vibrations, resonance. Give two examples of forced vibrations and     two examples of resonance.
2. Describe an experiment to demonstrate forced vibration and resonance..

GENERAL EVALUATION

1. State the principle of floatation
2. A stone of mass 2.0Kg is thrown vertically upward with a velocity of 20.0m/s. Calculate

the initial kinetic energy of the stone.

WEEKEND ASSIGNMENT

1. Which of the following correctly gives the relationship between linear speed v and angular speed w of a body moving uniformly in a circle of radius r?

(A)  v=wr   (B)  v=w2r  (C)  v= wr2  (D)  v=w/r.

1. The motion of a body  is simple harmonic if the:

(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.

1. The maximum kinetic energy of a simple pendulum occurs when the bob is at position.

(a)  1  (b)  2  (c)  3  (d)  4   (e)  5

1. The vibration resulting from the action of an external periodic force on the motion of a body is called:(a) Forced  vibration.  (b) damped vibration.  (c) natural vibration.

(d) compound vibration.

1. The maximum potential energy of the swinging pendulum occurs  positions

(A) 1and 5  (B) 2 and 4  (C) 3 only  (D) 4 only  (E) 5 and 3

THEORY

1. Define simple harmonic motion(SHM). A body moving with SHM has an amplitude of 10cm and a frequency of 100Hz. Find (a) the period of oscillation (b) the acceleration at the maximum displacement (c) the velocity at the centre of motion.
2. Define the following terms: frequency, period, amplitude of simple harmonic motion. What is the relation between period and frequency.

NEW SCH PHYSICS FOR SSS –ANYAKOHA.Pages 188-197