18-2-2015 1FCI

Prof. Nabila.M.Hassan

Faculty of Computer and Information

Basic Science department

2013/2014

18-2-2015 2FCI

•Aims of Course:

•The graduates have to know the nature of vibration wave motions with emphasis on their mathematical descriptions and superposition.

•The fundamental ideas can be introduced with reference to mechanical systems which are easy to visualize.

•The graduates have to know the nature of vibration and wave motions with emphasis on their mathematical description and superposition Developing the graduate's skills and creative thought needed to meet new trends in science.

•Supplying graduates with basic attacks and strategies for solving problems.

18-2-2015 3FCI

1- A particle oscillates with simple harmonic motion, so that its displacement varies according to the expression x = (5 cm)cos(2t + π/6) where x is in centimeters and t is in seconds. At t = 0 find(a) the displacement of the particle,(b) its velocity, and(c) its acceleration.(d) Find the period and amplitude of the motion.

Solution:

The displacement as a function of time is

x(t) = A cos(ωt + φ). Here ω = 2/s, φ = π/6, and A = 5 cm.

The displacement at t = 0

is x(0) = (5 cm)cos(π/6) = 4.33 cm.

(b) The velocity at t = 0 is v(0) = -ω(5 cm)sin(π/6) = -5 cm/s.

(c) The acceleration at t = 0 is a(0) = -ω2(5 cm)cos(π/6) = -17.3 cm/s2.

(d) The period of the motion is T = π sec, and the amplitude is 5 cm.

 

18-2-2015 4FCI

1- An oscillator consists of a block of mass 0.50 kg connected to a spring. When set into oscillation with amplitude 35 cm, it is observed to repeat its motion every 0.50 s. The maximum speed is : (a) 4.4 m/s ,(b) 44.0 m/s ,( c) 0.44 m/s

2- A particle executes linear harmonic motion about the point x = 0. At t = 0, it has displacement x = 0.37 cm and zero velocity. The frequency of the motion is 0.25 Hz. The max speed of the motion equal:(a) 0.59 cm/s ,(b) 5.9 cm/s ,( c) 0.059 cm/s

3- An oscillating block-spring system has a mechanical energy of 1.0 J, amplitude of 0.10 m, and a maximum speed of 1.2 m/s. The force constant of the spring is, (a) 100 N/m ,(b) 200 N/m ,( c) 20 N/m

4- An oscillating block-spring system has a mechanical energy of 1.0 J, amplitude of 0.10 m, and a maximum speed of 1.2 m/s. The mass of the block is, (a) 1.4 kg ,(b) 14.0 kg ,( c) .140 kg

18-2-2015 5FCI

1- An oscillator consists of a block of mass 0.50 kg connected to a spring. When set into oscillation with amplitude 35 cm, it is observed to repeat its motion every 0.50 s. The maximum speed is : (a) 4.4 m/s ,(b) 44.0 m/s ,( c) 0.44 m/s

2- A particle executes linear harmonic motion about the point x = 0. At t = 0, it has displacement x = 0.37 cm and zero velocity. The frequency of the motion is 0.25 Hz. The max speed of the motion equal:(a) 0.59 cm/s ,(b) 5.9 cm/s ,( c) 0.059 cm/s

3- An oscillating block-spring system has a mechanical energy of 1.0 J, amplitude of 0.10 m, and a maximum speed of 1.2 m/s. The force constant of the spring is, (a) 100 N/m ,(b) 200 N/m ,( c) 20 N/m

4- An oscillating block-spring system has a mechanical energy of 1.0 J, amplitude of 0.10 m, and a maximum speed of 1.2 m/s. The mass of the block is, (a) 1.4 kg ,(b) 14.0 kg ,( c) .140 kg

18-2-2015 6FCI

Content: Part II: Waves

Chapter 1

Oscillation Motion- Motion of a spring:

- Energy of the Simple Harmonic Oscillator:-Comparing SHM with uniform motion

- The simple pendulum:- Damped Oscillations:

- Forced Oscillation

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•Objectives:

Student will be able to: -Calculate the periodic time of S H M.

-- Define the damped motion - Define the resonance.

-Compare between free, damped and derived oscillations

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The simple pendulum:

A simple pendulum is a mechanical system that exhibits

motion consists of a small mass, which is suspended from

a light wire or string of length L. The displacement is

defined by the arc S. The net force on the bob (mass) is

tangent to the arc and equals)sin( mgF

For small angles (sin θ ≈ θ ) , then the force equal

mgmgF sin

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A simple oscillation of the pendulum, the resulting force is mgmgF sin

when θ is small

The displacement is directly related to θ

S = L θ so that θ = S/ L , then

gL

mF

Apply Newton's second law for motion in the

tangential direction

L

gm

dt

dmF

2

2

The equation of motion for simple pendulum

becomes L

g

dt

d

2

2

)cos(max t

max is the max angular position and the angular frequency

L

g

gLT 2

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Example (1)

a)How much energy is stored in the spring of a

toy gun that has a force constant of 50 N /m and

is compressed 0.150 m?

Sol: The energy can be found directly from

equation 22 )150.0)(/50(2

1

2

1mmNKXPEel

JmN 563.0.563.0

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b)In absence of friction and neglecting the mass of

the spring, at what speed will a 2.00 g plastic bullet

be ejected from the gun?

Sol: Since there is no friction, the potential energy is

converted into kinetic energy.

 JKXmV 563.

2

1

2

1 22

21

3

21

102

)563.0(22

kg

J

m

PEv l

elF PEKE

sm /7.23

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Problem:

- A 20 g particle moves in simple harmonic motion with a

frequency of 3 oscillations per second and an amplitude of

5cm.

(a) Through what total distance does the particle move

during one cycle of its motion?

(b) What is its maximum speed? Where does that occur?

(c) Find the maximum acceleration of the particle. Where in

the motion does the maximum acceleration occur? 

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Damped Oscillations:

Where the force is proportional to the speed of the moving object and

acts in the direction opposite the motion.

The retarding force can be expressed as:

R = - bv ( where b is a constant called damping coefficient)

and the restoring force of the system is – kx,

then we can write Newton's second law as

xxx mabvkxF 2

2

dt

xdm

dt

dxbkx

When the retarding force is small compared with the max restoring force that is, b is small the solution is,

)cos()( 2

tAetxt

m

b 2)2

(m

b

m

k

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represent the position vs time for a damped oscillation with decreasing

amplitude with time

The fig. shows the position as a function in time of the object oscillation

in the presence of a retarding force, the amplitude decreases in time,

this system is know as a damped oscillator. The dashed line which

defined the envelope of the oscillator curve, represent the

exponential factor

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The fig. represent position versus time:

•A - under damped oscillator

• b- critical damped oscillator

• c- Overdamped oscillator.

as the value of "b" increase the amplitude of the oscillations

decreases more and more rapidly.

When b reaches a critical value bc ( ), the system does

not oscillate and is said to be critically damped.

And when the system is overdamped.

oc mb 2/

oc mb 2/

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Forced Oscillation:

For the forced oscillator is a damped oscillator driven by an external force that varies periodically Where

tFtF o sin)(

where ω is the angular frequency of the driving force and Fo is a constant

From the Newton's second law

2

2

sindt

xdmkx

dt

dxbtFmaF o )cos( tAx

2222 )(

/

m

b

mFA

o

o

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mk

o is the natural frequency of the un-damped oscillator (b=0).

The last two equations show the driving force and the

amplitude of the oscillator which is constant for a given driving

force.

For small damping the amplitude is large when the

frequency of the driving force is near the natural frequency of

oscillation, or when ω ͌ ≈ ωo the is called the resonance and the

natural frequency is called the resonance frequency.

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Amplitude versus the frequency, when the frequency of the

driving force equals the natural force of the oscillator, resonance

occurs. Note the depends of the curve as the value of the

damping coefficient b.

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Note that: The applied force F is in phase with the velocity. The rate at

which work is done on the oscillator by F equals the dot product F. v ;

this rate is the power delivered to the oscillator.

Because the product F. v is a maximum when F and v are in phase,

we conclude that at resonance the applied force is in

phase with the velocity and the power transferred to the

oscillator is a maximum.

Summary of the chapter 1:

1- The acceleration of the oscillator object is proportional to its

position and is in the direction opposite the displacement from

equilibrium, the object moves with SHM. The position x varies with

time according to,)cos()( tAtx

2- The time for full cycle oscillation is defined as the period, /2T.

For block spring moves as SHM on the frictionless surface with a period

m

kT

22

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3- The frequency is defined as the number of oscillation per second, is the

inverse of the period m

kTf

2

11

4- The velocity and the acceleration of SHM as a function of time are

)sin( tAdt

dxv )cos(2

2

2

tAdt

xda

and

We not that the max speed is Aω , and the max acceleration is Aω2 .

The speed is zero when the oscillator is at position of x=± A , and is a

max when the oscillator is at the equilibrium position at the equilibrium

position x=0.

18-2-2015 22FCI

5- The kinetic energy and potential energy for simple harmonic oscillator are

given by,)(sin2

1

2

1 2222 tAmmvK )(cos2

1

2

1 222 tkAkxU

The total energy of the SHM is constant of the motion and is given by

2

2

1kAE

6 -A simple pendulum of length L moves in SHM for small angular

displacement from the vertical, its period is gLT 2

18-2-2015 23FCI

7- For the damping force R = - bv, its position for small damping is described by

2)2

(m

b

m

k)cos()( 2

tAetx

tm

b

8 - If an oscillator is driving with a force tFtF o sin)(

it exhibits resonance, in which the amplitude is largest when driving

frequency matches the natural frequency of the oscillator.

18-2-2015 24FCI

What is the effect on the period of a pendulum of doubling its length?

gLTL /2 LL 2

gLT L /222 LLL TTT 414.122

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Useful website

http://cnx.org/content/m15880/latest/

http://www.acs.psu.edu/drussell/Demos/SHO/mass-force.html

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