Oscillations Theory

8/17/2019 Oscillations Theory

1/10

APHY 111

Vibrations

1

General DepartmentFrederick University

1

Periodic Motion• Periodic motion is motion of an object that

regularly returns to a given position after a fixed timeinterval

• A special kind of periodic motion occurs inmechanical systems when there is a restoring force.

• For examples vibration occurs when a body is displacedform its equilibrium position while a restoring force ispresent

Restoring Force


2/10

Motion of a Spring-MassSystem

• A block of mass m isattached to a spring, the

block is free to move on africtionless horizontalsurface

• When the spring is neitherstretched nor compressed,the block is at theequilibrium position x = 0

Hooke’s Law• Hooke’s Law states F s = - kx – F s is the restoring force

• It is always directed toward the equilibrium position• Therefore, it is always opposite the displacement

from equilibrium

– s e orce spr ng cons an – x is the displacement

x>0 Fs


3/10

Acceleration• The force described by Hooke’s Law is the

net force in Newton’s Second Law

kxma

F F

x

Hooke

−=

=Σ

MotionHarmonicSimple

• The acceleration is not constant – Therefore, the kinematic equations cannot be

applied – When the block passes through the

equilibrium position, a = 0

xm

k a

x −=

Simple Harmonic Motion• Acceleration

• We let• Then a = -ω 2 x • In the absence of friction, the motion will continue forever

2

2

d x k a x

dt m= = −

2 k

mω =

• A solution is x (t ) = A cos ( ω t + φ )

A is the amplitude of the motionThis is the maximum position of theparticle in either the positive ornegative direction

ω is called the angular frequencyUnits are rad/s

φ is the phase constant or the initial phaseangle

period


4/10

Frequency and Period

• The period , T , is the time interval required

• The frequency, f represents the numberof oscillations per second.

f=2πω• Units are cycles per second = hertz (Hz)

7

for the particle to go through one full cycleof its motion.

Low frequencyHigh period

High frequencyLow period

π 2=T

Amplitude and phaseThe amplitude of the motion is the maximum position of the particle ineither the positive or negative direction.The phase determines the position at time t=0.

8

φ=0


5/10

Example 1: An object oscillates with simple harmonic motion along the x axis. Its position varies with time according to the equation

where t is in seconds and the angles in the parentheses are in radians.(a) Determine the amplitude, frequency, and period of the motion.(b) Calculate the velocity and acceleration of the object at any time t.(c) Make a sketch of position, velocity and acceleration.(d) Using the results of part (B), determine the position, velocity, and acceleration ofthe object at t =1.00 s.

)cos()00.4()( t mt x π =

9

.

Example 2: A 200g block connected to a light spring for which the force constantis 5.00 N/m is free to oscillate on a horizontal, frictionless surface. The block isdisplaced 5.00 cm from equilibrium and released from rest, as in the figure.

(a) Find the period of its motion.(b) Express the position as a function of time

10

c a s e max mum acce era on o e oc


6/10

Example 3: The cart has a mass of m and is attached to two springs, of stiffness k 1 andk 2. If all springs are unstretched when the cart is in the equilibrium position shown:

(a) Derive the equation of motion for the cart when it is displaced from equilibrium by anamount x. Neglect friction.

(b) Determine the natural frequency of oscillation and make a plot of the cart ’sdisplacement as a function of time.

(c) Determine the period and frequency of oscillation.

The Pendulum

s =LΘ

The period and frequency of a simple pendulum depend only on the length of thestring and the acceleration due to gravity (see Physics Lab 10)


7/10

Energy of the SHM Oscillator

• Assume a spring-mass system is moving on africtionless surface

• This tells us the total energy is constant• The kinetic energy can be found by

• K = ½ mv 2 = ½ mω 2 A2 sin 2 (ω t + φ )

• The elastic potential energy can be found by• U = ½ kx 2 = ½ kA2 cos 2 (ω t + φ )

• The total energy is E = K + U = ½ kA 2

Energy of the SHM Oscillator,cont• The total mechanical

energy is constant• The total mechanical

energy is proportional to thesquare of the amplitude

• Energy is continuouslybeing transferred betweenpotential energy stored inthe spring and the kineticenergy of the block


8/10

• As the motion

continues, theexchange of energyalso continues

Energy of the SHM Oscillator,cont

• Energy can be used tofind the velocity

( )2 2

2 2 2

k v A x

m

A x ω

= ± −

= ± −

Energy in SHM, summary


9/10

Example 4: A 0.500kg cart connected to a light spring for which the force constant is20.0 N/m oscillates on a horizontal, frictionless air track.(a) Calculate the total energy of the system and the maximum speed of the cart if theamplitude of the motion is3.00 cm.(b) What is the velocity of the cart when the position is 2.00 cm?(c) Compute the kinetic and potential energies of the system when the position is 2.00cm.

17

Example 5: A block of unknown mass is attached to a spring with a spring constant of6.50 N/m and undergoes simple harmonic motion with an amplitude of 10.0 cm. Whenthe block is halfway between its equilibrium position and the end point, its speed ismeasured to be 30.0 cm/s. Calculate(a) the mass of the block, (b) the period of the motion, and (c) the maximum accelerationof the block.

18


10/10

Example 6: A 200g block is attached to a horizontal spring and executes simpleharmonic motion with a period of 0.250 s. If the total energy of the system is 2.00 J, find(a) the force constant of the spring and (b) the amplitude of the motion.

19

Oscillations Theory

Documents

Transcript of Oscillations Theory