Shm Lecture 4 Phs1005

TOPIC: Simple Harmonic Motion (SHM)

ENGINEERING PHYSICS 1 (PHS1005)

Developed and presented by: J. Bennett & H. Simpson

Physics Division

School Of Natural & Applied Sciences Faculty of Science & Sport

University of Technology, Jamaica

1 Success =

Faith+Commitment+Effort

Oscillations

When a vibration or an oscillation repeats itself over the same path the motion is periodic.

Examples of periodic motion

Oscillating pendulum.

Mass oscillating at the end of a vertical spring.

Mass oscillating at the end of a horizontal

spring on a frictionless surface.

Success =

Faith+Commitment+Effort 2

Periodic motion

When an object vibrates or oscillates back and forth over the same path, covering each cycle in the same time, the motion is periodic.

Success = Faith+Commitment+Effort 3

Oscillating Pendulum


Mass oscillating at end of horizontal spring on a frictionless surface


Mass oscillating at end of a vertical spring


Oscillations

We need to define a few terms:

The distance x of the body from the equilibrium position is called the displacement.

Amplitude: maximum displacement.

Period: Time for one complete cycle


Oscillations

In the lab you will examine the oscillating pendulum and the oscillation of a mass on the end of a vertical spring.


SHM Definition

Vibratory motion in a system in which the: acceleration is always directed towards a fixed point.

acceleration is directly proportional to its distance from the fixed point.

net restoring force is directly proportional to the negative of the displacement.

motion is periodic, amplitude does not change with time,

and energy is conserved.


SHM Hooks law and Newton 2nd Law

Consider a mass placed on a vertical spring according to Hookes Law:

From Newtons second law:

Success = Faith+Commitment+Effort

kxF

F ma10


Consider a mass placed on a spring according to Hookes Law:

From Newtons second law:


kxF

F ma11



kxF F ma

adt

xd2

2

makxRecall

makx

12

Differential equation for SHO

Combining equations from Hookes law and Newton's law yields a differential equation.

This equation is the differential equation for the simple harmonic oscillator

02

2

xm

k

dt

xd


G.S of DE for SHM

Assume general form of solution of equation is:

Where a and b are arbitrary constants


inx aCos t bS t

14

First Differentiation of G.S.

.

dx d da Cos t b Sin t

dt dt dt

dxa Sin t b Cos t

dt


Second Differentiation of G.S.

22 2

2

d xa Coswt b Sin t

dt

2( cos sin )a t b t


Substitution

Substituting

into

02

2

xm

k

dt

xd

xanddt

xd2

2


Substituting and factorizing

yields

2( cos sin ) ( cos sin ) 0k

a t b t a t b tm

2( cos sin )( ) 0k

a t b tm


02

2

xm

k

dt

xd

18

Angular frequency equation

Thus the equation of motion is satisfied by our trial solution if

2 0k

m

2 k

mk

mSuccess =


GS of DE

x = a Cos wt + b Sin wt

is in fact the general solution to the differential equation.

In real Physical situations the constants a and b are determined by the initial conditions.


SHM with Initial condition set

Suppose the mass is started at is maximum displacement x = A and released without a push v = 0, t = 0 applying initial conditions


SHM with Initial condition set contd

cos sinx a t b t

dt

dxsin0 cos0a b b

At t = 0 and a = A and bw = 0 so b = 0 and the motion is a pure sine curve

cosx a tSuccess =


SHM with Initial condition set contd


cosx a t

23

SHM

There are many other situations in which neither a nor b is zero such as when the spring is stretched a certain distance at t = 0

and given a push so at t = 0 x is less than A.

In all cases specifying two properties such as displacement and velocity at a given instant will uniquely determine a and b


Displacement Equation

x = a Cos wt + b Sin wt can be written in the following more

convenient form

x = A Cos (wt + )


Displacement Equation (contd)

A is defined as the amplitude which occurs when the cosine has a

maximum value of 1 and is called the phase angle. It tells us how long after or before t = 0 the peak x = A is reached for = 0

x = A Cos wt


Angular (w)and Natural frequency (f) of SHM

Recall

so

k

m

M

Kf2

m

Kf

2

1


2 f

27

Period of SHM

2m

Tk

N.B: The period and frequency do not depend on the amplitude.


SHM where mass begin oscillation at the equilibrium position.


SHM with phase angle 0

Displacement time graph for a SHM with phase angle not equal to zero.


Displacement, velocity and acceleration graphs of SHM


31

Maximum velocity and acceleration in SHM


Instantaneous and maximum Velocity of SHM

( )dx d

v ACos tdt dt

Max V occurs when sine function is a maximum of 1 so:


( )v ASin t

maxv A

34

Maximum value of acceleration of SHM

maximum value of acceleration is determine by the equation:


2

maxa A

2 k

m35

QUESTION 1

A spring stretches 0.150 m when a 300 g mass is hung from it. The spring is then stretched an additional 0.100 m from this equilibrium position and released Determine the:

a) values of the spring constant k and the angular frequency.

b) amplitude of the oscillation

c) maximum velocity

d) maximum acceleration of the mass.

e) period and frequency.

f) displacement , velocity and acceleration as a function of time.

g) velocity and acceleration at t = 0.150 s Succeiss =


QUESTION 1


Solution 1

(a) F = kx

K = F x

K= 0.300 kg x 9.80

0.150 m

= 19.6 N/m


Solution 1


K

M

19.6 /

0.300

N m

kg= 8. 08 rad/s

=

Solution 1 (contd)

(b) A = 0.100 m

(c) = (8.08rad/s)(0.100 m) =0.808 m/s


maxv A

40

Solution 1 (contd)

(d) Max acceleration occurs when x = A = +0.100 m

= (8.08)2( 0.100) = 6.53 m/s2

or

a max = kA = (19.6 N/m )(0.100 m) m 0.300 = 6.53 m/s2


2

max maxa v A A

41

Solution 1 (contd)

(e) T = 2

T = 2

f = 1 = 1.30Hz

T

m

k

0.3000.770

19.6s


Solution 1 (contd)

(f) The motion begins at a point of maximum displacement down wards. Taking x as positive upwards

X = -A at t =0

So we need a sinusoidal curve that has a maximum

negative value at t = 0 which is a negative cosine curve

x = - A cos 2f t so at t =0, x = -A

x = - 0.100 mCos 8.08 t


Solution 1 (contd)

Velocity as a function of time

V = dx = d (-0.100 m Cos8.08 t)

dt dt

=0 .808 m/s Sin 8.08 t

at t = 0.150 s

V = 0.808 m/s Sin [(8.08)(0.150)]


Solution 1 (contd)

V= 0.76 m/s

Is the velocity 0.76m/s???

If not what is it?


V = 0.808 m/s Sin (8.08)(0.150)

Solution 1 (contd)

Acceleration as a function of time

a = dV = d (0 .808 m/s Sin 8.08 t)

dt dt

= 6.52m/s2 cos 8.08 t

a =6.52 cos 8.08 (0.150)


Solution 1 (contd)

a =2.29 m/s2


Is the acceleration 2.29 m/s2 ???

Phase shift

shift sine to the left to create cosine

shift cosine to the right to create sine


QUESTION 2

An object oscillates with simple harmonic motion along the x axis. Its position varies with time according to the equation and is in meters.

t is in seconds and the angles in the parentheses are in radians.

a) Determine the amplitude, frequency, and period of the motion.

b) Write equations for the velocity and acceleration of the object at any time t.


( ) 4.00 os4

x t c t

QUESTION 2 (Contd)

(c) Using the results of part (b), determine the position,

velocity, and acceleration of the object at t = 1.00 s.

(d) Determine the maximum speed and maximum acceleration of the object.

(e) Find the displacement, velocity and acceleration of the object between t = 0 and t = 1.00 s respectively.


Solution 2

(a) By comparing this equation with Equation of

simple harmonic oscillator

x = A cos (wt + ), we see that A = 4.00 m and

therefore

T = 1/f = 2.00 s.

Success =


= 2.00s

Problem 3 (part 1)

A 200-g block connected to a light spring for which the force constant is 5.00 N/m is free to oscillate on a horizontal, frictionless surface. The block is displaced 5.00 cm from equilibrium and released from rest.

a) Find the period of its motion.

b) Determine the maximum speed of the block.

c) What is the maximum acceleration of the block?

d) Express the position, speed, and acceleration as functions of time.

Success =


Solution 3 (part 1)

a) 1.26 sec

b) 0.25m/s

c) 1.25m/s2

d) x = A cos wt = (0.0500 m) cos 5.00t

V=- w A sin wt = - (0.250 m/s) sin 5.00t

a= -w2A cos wt = -(1.25m/s2)cos 5.00t


Problem 3 (part 2)

Given that the block is released from the same initial position, x(0) = 5.00 cm, but with an initial velocity of v(0) = 0.100 m/s?

Determine the new equations for:

a) displacement.

b) velocity

c) acceleration.


Solution 3 (part 2)


References

Make reference to the handout on SHM. See two (2) additional questions and solution.

http://dev.physicslab.org/Document.aspx?doctype=3&filename=Momentum_Impulse.xml

http://images.tutorvista.com/content/kinematics/displacement-vectors.gif

A-level physics by Roger Muncaster

Further Mechanics by Jefferson & Beadworths

Physics for Scientists and Engineers (4th Edition) by Giancoli

Success =


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Shm Lecture 4 Phs1005