Physics 102 Waves

Physics 102 Waves

Moza M. Al-Rabban

Professor of Physics

mmr@qu.edu.qa

Lecture 2More Oscillations

Lecture 2More Oscillations

2

The Dynamics of SHMThe Dynamics of SHM

0( ) cosx t A t

2

02

20

2

( ) cos

cos

( )

x

da t A t

dt

A t

x t

The acceleration is proportional to the negative of the displacement at any time.

3

The Equation of Motion in SHMThe Equation of Motion in SHM

This is the equation of motion for the system. It is a homogeneous linear 2nd order differential equation.

sp( )xF k x

net sp( ) ( )x x xF F kx ma

x

ka x

m

2

20

d x kx

dt m

4

Solving the Equation of MotionSolving the Equation of Motion2

20

d x kx

dt m

Q: How do you solve this differential equation?

A: Our method - guess the solution, and see if it works.

0cosx A t

0 0cos sind dx A t A t

dt dt

2

20 02

sin cosd dx A t A t

dt dt

20 0cos cos 0

kA t A t

m

2 or k k

m m

It works!

5

Example:Analyzing an Oscillator

Example:Analyzing an Oscillator

cosAx

rad927.0

At t=0, a 500 g block oscillating on a spring is observed to be moving to the right at x=15 cm. At t=0.30 s, it reaches its maximum displacement of 25 cm.(a)Draw a graph of the motion for one

cycle.(b)At what time in the first cycle is x=20

cm?

0 15 cm, 25 cmx A

1 10 0cos / cos 0.60 0.927 radx A

tAtx cos)(The motion is SHM. The position equ. Of the block is

The object is initially moving to the right, which tells us that the phase constant must be between - and 0 rad. Thus

cosAx

6

Example:Analyzing an Oscillator

Example:Analyzing an Oscillator

)cos(max wtAAx

(a)Draw a graph of the motion for one cycle.

(b)At what time in the first cycle is x=20 cm?

0At 0.30 s, cos 1 t t

0Therefore, / ( 0.927 rad)/0.30 s 3.09 rad/st

2 / 2.03 sT

10cos ( / ) /

0.0917 s and 0.508 s

t x A

The motion is SHM. The position equ. Of the block is

tAtx cos)(

The block reaches its maximum displacement at time t = 0.30 s. At that instant of time

)927.009.3(cos)25()( tcmtx

7

Vertical OscillationsVertical Oscillations

net spAt equilibrium, ( ) ( ) 0y y yF F w k L mg

/ equilibrium positionL mg k

Does the additional force of gravity change the motion?

Now consider a mass m hanging from a spring with constant k. When the mass is attached, the spring stretches by L. Finding L is a static –equilibrium problem in which the upward spring force balances the downward weight force of the block. So,

is a distance and it is a +ve number. is the displacement. In this configuration, the equilibrium position is given by:

LkykFysp

Ly

8

net sp( ) ( ) ( ) ( )y y yF F w k L y mg k L mg ky

net( ) yF ky 0( ) cos( )y t A t /k m

m

Let the block oscillate around this equilibrium position.

The origin of the y-axis at the block’s equilibrium position

If the block moves upward, the spring gets shorter compared to its equilibrium length, but the spring is still stretched compared to its unstretched length.

When the block is in position y, the spring is stretched by an amount

And hence exerts an upward spring force

The net force on the block at this point is

yL

yLkFsp

Because, the motions are the same, everything we have learned about horizontal oscillation is equally valid for vertical oscillations.

9

Example 7: A Vertical OscillatorExample 7: A Vertical Oscillator

2/ (0.20 kg)(9.8 m/s )/(10 N/m) 0.196 mL mg k

A 200 g block hangs from a spring with constant k = 10 N/m. The block is pulled down to a point where the spring is 30 cm longer than its unstretched length, then released.

Where is the block and what is its velocity 3.0 s later?

Although the spring begins by being stretched 30 cm, this is not the amplitude of the oscillation. Oscillations occur around the equilibrium position, so we have to begin by finding the equilibrium point where the block hangs motionless on the spring.

Stretching the spring 30 cm pulls the block 10.4 cm below the equilibrium point,

so A=10.4 cm.

(0.300 m) (0.196 m) 0.104 msA y L

10

Example 7: A Vertical OscillatorExample 7: A Vertical Oscillator

wtcmty cos4.10)(

cmradssradcmy 4.7))0.3)(/07.7cos(()4.10(

/ (10 N/m) /(0.20 kg) 7.07 rad/sk m

At t= 3.0 s the block’s position is:

cosAAy

The initial condition

The block is 7.4 cm above the equilibrium position, or 12.2 cm below the original end of the spring.

Its velocity at this instant is

scmwtwAvx /52)sin(

11

The PendulumThe PendulumThe PendulumThe Pendulum

The pendulum’s position can be described by the arc of length s, which is zero when the pendulum hangs straight down.

Because angles are measured ccw, s and are

+ve when pendulum is to the right of center,

-ve when it is to the left.

12

The PendulumThe Pendulum

net( )

sint t t

t

F F w

mg ma

2

2sin

d sg

dt

2

2sin 0

d s sg

dt L

The sine function makes this equation more complicated than the oscillating spring’s ones. This is not a homogeneous linear 2nd order differential equation describing SHM.

Two forces are acting on the mass:

The string tension and the weight.

Using for acceleration “ around” the circle,

2

2

tdsd

at

Where the angle is related to the arc length by =s/L, so

The equation of motion for an

oscillating pendulum.

13

The Small Angle ApproximationThe Small Angle Approximationrs

2

20

d s gs

dt L

This is a homogeneous linear 2nd order differential equation describing SHM, but only for small swings (s<<L).

2

2sin 0

d s sg

dt L

The pendulum equation is:

The height of the triangle is h=r sin

Suppose that the angle is “small”.

)(sin radiansin

14

The Small Angle ApproximationThe Small Angle Approximation2

20

d s gs

dt L

This is exactly the same equation as the equation for mass oscillating on a spring.

2

20

d x kx

dt m

Lg

fw

wttorwtAts

2

)cos()()cos()( max

The angular frequency

Is determined by the length of the string.

15

Example 8: A Pendulum Clock

Example 8: A Pendulum Clock

What length L must a pendulum have to give an oscillation period T of exactly 1 s?

2 (independent of )L

T mg

2 22 1 s

(9.80 m/s ) 0.248 m2 2

TL g

16

Example 9: The Maximum Angle of a PendulumExample 9: The Maximum Angle of a Pendulum

A 300 g mass on a 30 cm long string oscillates as a pendulum. It has a speed of 0.25 m/s as it passes through the lowest point. What maximum angle does the pendulum reach?

2(9.80 m/s )5.72 rad/s

(0.30 m)

g

L

max / (0.25 m/s)/(5.72 rad/s) 0.0437 mA v

max

(0.0437 m)0.145 rad 8.3

(0.30 m)

A

L

maxv AThe speed of the lowest point is:

17

Conditions for SHMConditions for SHM we have solved all simple harmonic motion problems once we have solved the problem of horizontal spring.

The restoring force of a spring, Fsp = kx, is directly proportional to the displacement x from equilibrium.

The pendulum’s restoring force, in the small angle approximation, is directly proportional to the displacement s.

A restoring force that is directly proportional to the displacement from equilibrium is called a linear restoring force.

For any linear restoring force, the equation of motion is identical to spring equation ( other than perhaps using different symbols).

Any system with a linear restoring force will undergo simple harmonic motion around the equilibrium position.

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Moreover, even when the restoring force is nonlinear (e.g., the pendulum), for sufficiently small displacements it can be treated as a linear system (small angle approximation).

Therefore, we often treat atoms, molecules, and nuclei as small harmonic oscillators, even when the restoring forces involved may be very different from Hooke’s Law forces.

Everything that we learn about an oscillating spring can be applied to the oscillations of any other linear restoring force, ranging from the vibration of airplane wings to the motion of electrons in electric circuits.

19

Identifying and Analyzing SHMIdentifying and Analyzing SHM1. If the net force acting on a particle is a linear restoring force, the

motion will be simple harmonic motion (SHM) around the equilibrium position.

2. The position x as a function of time is x(t) = A cos (t +0). The velocity v as a function of time is v(t) = A sin (t + ). The maximum speed is vmax = A . The equations are given here in terms of x, but they can be written in

terms of any other appropriate variables (y, , etc.)

3. The amplitude A and the phase constant are determined by the initial conditions, through x0 = A cos 0 and v0x = A sin 0.

4. The angular frequency and the period T depend on the physics of the particular situation but are independent of A and .

5. Mechanical energy is conserved. Thus, ½mvx2 + ½kx2 = ½kA2 = ½mvmax

2. Energy conservation provides a relationship between position and velocity that is independent of time.

20

Clicker Question 1Clicker Question 1 One person swings on a swing and finds that the period is 3.0 s. A second person of equal mass joins him on the same swing.

With two people swinging, the period is:a. 6.0 s;

b. 3.0 s;

c. 1.5 s;

d. Between 3.0 s and 6.0 s;

e. Can’t tell without knowing the length.

21

Damped OscillationsDamped Oscillations

Drag force: D bv

net sp( ) ( )x x x x xF F D kx bv ma

2

20

d x b dx kx

dt m dt m 2

0( ) cos( )bt

mx t A e t

2202 2

1

4 4

k b

m m

0 and k m

m b

Note that damping reduces the oscillation frequency.

Ampl. Dampingfactor

Sinusoidalvariation

(damped oscillator)

b: the damping constant

22

Energy in Damped SystemsEnergy in Damped Systemsm

b

1 12 / 2 2

max2 2

1 2 / /02

( ) ( )t

t t

E t kx k Ae

kA e E e

The oscillator’s mechanical energy decays exponentially with time constant .

1

2

3

2.71828...

0.3678... 36.8%

0.1353... 13.5%

0.04978... 4.98%

e

e

e

e

/ 20( ) cos( );tx t A e t

23

Example: A Damped PendulumExample: A Damped Pendulum

A 500 g mass on a 50 cm long string oscillates as a pendulum. The amplitude of the pendulum is observed to decay to ½ of its initial value after 35 s.

(a) What is the time constant of the damped oscillator?

(b) At what time t1/2 will the energy of the system have decayed to ½ of its initial value?

max maxAt 0, , and at 35 s, / 2.t x A t x A / 2 / 2 / 2 ln 2tAe A t

/ 2 ln 2 (35 s) / 2 ln 2 25.2 st

1/ 2 /0 0 1/ 2/ 2 / ln 2tE e E t

1/ 2 ln 2 (25.2 s) ln 2 17.5 st

24

Clicker Question 2Clicker Question 2

Which of these decays has the shortest time constant?

25

Driven Oscillationsand Resonance

Driven Oscillationsand Resonance

2

2cosd

d

Fd x b dx kx t

dt m dt m m

Now, suppose we drive a damped mechanical oscillator with an external force F(t) = Fd cos(dt). Then the equation of motion is:

The system will show the property of resonance. The oscillation amplitude will depend on the driving frequency d, and will have its maximum value when:i.e., when the system is driven at its resonant frequency 0.

0 /d k m

The width of the resonance curves depends on the amount of damping. Wider curves with smaller resonance curves correspond to more damping and larger values of b.

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Chapter 14 Summary (1)Chapter 14 Summary (1)

27

Chapter 14 Summary (2)Chapter 14 Summary (2)

28

Chapter 14 Summary (3)Chapter 14 Summary (3)

End of Lecture 2End of Lecture 2

Before the next lecture, read Knight, Chapters 20.1 through 20.3.

30

Homework #1Homework #11. In taking your pulse, you count 75 heartbeats in 1 min. What are the period

and frequency of your heart’s oscillations?

5. What are the (a) amplitude, (b) frequency, and (c) phase constant of the oscillation shown in the Figure.

14. The position of a 50 g oscillating mass is given by , where t is in s. Determine: a. The amplitude. b. The period. c. The spring constant. d. The phase constant. e. The initial conditions. f. The maximum speed. g. The total energy. h. The velocity at t = 0.4 s.

)4/10cos()0.2()( tcmtx

31

18. A spring is hanging from the ceiling. Attaching a 500 g physics book to the spring causes it to stretch 20 cm in order to come to equilibrium.

a. What is the spring constant?

b. From equilibrium, the book is pulled down 10 cm and released. What is the period of oscillation?

c. What is the book’s maximum speed? At what position or positions does it have this speed?

21. Make a table with 4 columns and 8 rows. In row 1, label the columns (°), (rad), sin , and cos . In the left column, starting in row 2, write 0, 2, 4, 6, 8, 10, and 12.

a. Convert each of these angles, in degrees, to radians. Put the results in column 2. Show 4 decimal places.

b. Calculate the sines and cosines. Put the results, showing 4 decimal places, in columns 3 and 4.

c. What is the first angle for which and sin differ by more than 0.0010?

d. What is the first angle for which cos is less than 0.9900?

e. Over what range of angles does the small-angle approximation appear to be valid?

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29. A 205 g air-track glider is attached to a spring with spring constant 4.0 N/m. The damping constant due to air resistance is 0.015 kg/s. The glider is pulled out 20cm from equilibrium and released. How many oscillations will it make during the time in which the amplitude decays to

of its initial value? 1e