JJ311 MECHANICAL OF MACHINE Ch 2 Simple Harmonic Motion

7/21/2019 JJ311 MECHANICAL OF MACHINE Ch 2 Simple Harmonic Motion

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SIMPLE HARMONIC MOTION

PREPARED BY:NAZIHAH BINTI MOHD NOOR



2.1 Concept of simple !"monic motion#$HM%

2.1.1 De&ne te te"minolo'ies in $HM

2.1.2 $HM () *sin' ! s*it!(le +i!'"!m

2.1., C!lc*l!te time pe"io+- !mplit*+e-f"e*enc)- /elocit)- m!0im*m /elocit)-!ccele"!tion !n+ m!0im*m

!ccele"!tion of te $HM

2.2 ine!" motion of !n el!stic motion

2.2.1 El!stic s)stem#sp"in' !n+ m!ss%

CONTEN



TERMINOLOGIES OF SHM

Motion in ic ! (o+) mo/es (!c3

!n+ fo"t o/e" ! &0e+ p!t- "et*"nin'to e!c position !n+ /elocit) !fte" !+e&nite inte"/!l of time.

E*il("i*mPosition



4ORM5A

SHM

)(

cos

max

2

max

max

22

2

massmma Inertia

r A

Aa

AV

x Av

A x

xa

→=

=

=

=

−=

=

=

ω

ω

ω

θ

ω

ω

π

π

ω

π ω

2

1

2

2

=

=

=

=

T

f

T

f

f



45NDAMENTADE4INITION$

Displacement that measured from theequilibrium point

x

Time T

Velocity or speedv

Acceleration a

Mass m

Force F

anular !elocity ω

Maximum Velocity"#peed V max

Maximum acceleration amax



EXERCISE 1

Points moving with simple harmonic motion

have acceleration 9m/s2

and velocity 0.92m/swhen it was in 65mm from the centre position.Find

i. Amplitudeii. Periodic time the movementSOLUTIONS

i. Amplitude

ii. Pe"io+ic time te mo/eme



EXERCISE 2 (i)

A particle moving with simple harmonic motion has a

periodic time of 0.s and it was !ac" and forth!etween two points is #.22m. $etermine

i. Fre%uency and amplitude of the oscillation

ii. &elocity and acceleration of the particle when it is00mm from the center of oscillationiii.'he ma(imum velocity and acceleration of the

movementSOLUTIONS

i. Frequency

Amplitude



ii. &elocity and acceleration of the particle when

it is 00mm from the center of oscillation

EXERCISE 2 (ii)

EXERCISE 2 (iii)



iii. 'he ma(imum velocity and acceleration of

the movement

EXERCISE 2 (iii)

EXERCISE 3 (i)



A !ody of mass #.5"g moving with simple harmonic

motion is towards one end of the swing. At the time itwas in A) *60mm from the center of oscillation)velocity and acceleration is 9m/s and ##0m/s2 respectively.

i. Fre%uency and amplitude of the oscillationii. +a(imum acceleration and the inertia of the !ody

when it to the edge of swing

EXERCISE 3 (i)

SOLUTIONS

i. 4"e*enc)

=

Amplit*+e

EXERCISE 3 (ii)



ii. +a(imum acceleration and the inertia of the

!ody when it to the edge of swing

EXERCISE 3 (ii)



'he piston of a steam engine moves with simple

harmonic motion. 'he cran" rotates at #20r.p.m with astro"e of 2 metres. Find the velocity and accelerationof the piston when it is at a distance of 0.*5metre fromthe centre.

SOLUTIONS

i. Velocity of the piston

ii. Acceleration of the piston

SHM !A"#A



SHM !A"#A

M!0. Displ!cement 6 A

M!0. 7elocit) 6 Vmax = ωA

M!0. Accele"!tion 6 amax =

EXERCISE 4



$ Figures ,a- and ,!- are the displacementtime graphand accelerationtime graph respectively of a !ody in

simple harmonic motion. hat is the fre%uency of themotion

EXERCISE 4

SOLUTIONS

Frequency

EXERCISE 5



'he following graphs show the variation of

displacement) x and velocity) v with time) t for a !odyin simple harmonic motion. hat is the value of '

EXERCISE 5

SOLUTIONS

EXERCISE 6



A particle moves in simple harmonic motion along a

straight line a!out point x=0.40cm and the velocity is1ero. 'he fre%uency of the motion is 2.51. 3alculatethe4i. Periodii. Angular velocityiii. Amplitudeiv. $isplacement at time t v. +a(imum velocityvi. +a(imum acceleration

EXERCISE 6

SOLUTIONS



i. Pe"io+

i. An'*l!" /elocit)

iii. Amplit*+e

i/. Displ!cement !t time t

/. M!0im*m /elocit)

/i. M!0im*m !ccele"!tion



A sprin resists bein stretched or

compressed.

INEAR MOTION O4 AN

EA$TIC MOTION

comp"esse+

st"etce



Hoo3e8s !

When a spring is stretched, there is arestorin$ force that is proportional to thedisplacement.

F % &'(

F

x

m



HOOKE'S LAW

The restorin force of an ideal sprin is i!en by%

&here k is the sprin constant and x is the

displacement of the sprin from its

unstrained lenth' The minus sin indicates

that the restorin force al&ays points in a

direction opposite to the displacement of

the sprin'



Te fo"ce +esc"i(e+ () Hoo3e8s! is te net fo"ce in Neton8s$econ+ !

k stiffness of the sprin ("m) sprin constant ("m)

xm

k a

makx

F F Newton Hooke

−==−

=



xa 2ω =

xmk a

−=

m

k

m

k

=

=

ω

ω 2



SHM Mass & SpringSyst!

Si!p" Pn#$"$!

ω

π

π

ω

π ω

2

1

2

2

=

=

==

T

f T

f

f

k

mT

f T

m

k

f

f

m

k

f m

k

π

π

π

π ω ω

2

1

2

2

2*

=

=

=

=

==

g

l T

f T

l

g

f

f

l

g

f l g

π

π

π

π ω ω

2

1

2

2

2*

=

=

=

=

==



e

x

estatic deflectiongra!ity

kemg

keT mg T

=

==11 *

xm

k a

makx

makxkeke

maT mg

ma F

xek T

−=

=−=−−

=−

=

+=

+

1

+

1 )(

g

eT

k

mT

g

e

k

m

kemg

π

π

2

2

=

=

=

=

%ISPLACEMENT IN SHM



%ISPLACEMENT IN SHM

m

x = 0 x = +A x = -A

x

$ $isplacement is positive when theposition is to the right of the e%uili!riumposition ,( 0- and negative whenlocated to the left.

$ 'he m!0im*m displacement is calledthe amplitude A.



ELOCIT IN SHM

m

x = 0 x = +A x = -A

v (+)

$ 7elocit) is positi/e en mo/in' tote "i't !n+ ne'!ti/e en mo/in'

to te left.$ It is 9e"o !t te en+ points !n+ !m!0im*m !t te mi+point in eite"+i"ection # o" ;%.

v (-)



Acceleration in SHM

m

x = 0 x = +A x = -A

$ Accele"!tion is in te +i"ection of te"esto"in' fo"ce. #a is positi/e en 0 is ne'!ti/e- !n+ ne'!ti/e en 0 ispositi/e.%

$ Accele"!tion is ! m!0im*m !t te en+points !n+ it is 9e"o !t te cente" of

oscill!tion.

+x -a

-x +a

F ma kx= = −



EXERCISE 7



A !ody of mass #"g !eing hung with springs straightfrom one end attached to a rigid support. 'he !odyproduced 25mm static deection. 7t was pulled down28mm and then released. Find

i. 'he acceleration !egan to the !ody

ii. Periodic timeiii. 'he spring ma(imum forceiv. &elocity and acceleration the !ody when it is #2mm

from the e%uili!rium positionSOLUTIONS



Te !ccele"!tion (e'!n to te (o+)

ii. Pe"io+ic time



Te sp"in' m!0im*m fo"ce

i/. 7elocit) !n+ !ccele"!tion te (o+) en it is 12mm f"om tee*ili("i*m position

THE SIMPLE PEN%ULUM



,n order to be in #-M% the restorinforce must be proportional to the

neati!e of the displacement' -ere

&e ha!e.

&hich is proportional to sin θ and

not to θ itself'

-o&e!er% if the anle is small%sin θ / θ '

THE SIMPLE PEN%ULUM



Therefore% for small anles% &e ha!e.

&here

The period and frequency are.

EXERCISE 7



A mass is suspended from a string 60mm long. 7t isnudged so that it ma"es a small swinging oscillation.

$etermine the fre%uency and periodic time.

SOLUTIONS



T(an) y*$ +

JJ311 MECHANICAL OF MACHINE Ch 2 Simple Harmonic Motion

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