john g. cramer professor of physics b451 pab [email protected]

Physics 121C - MechanicsLecture 23 (Knight: 14.1 -

14.6)Simple Harmonic Motion

December 7, 2005 (29 Slides)

Physics 121C - MechanicsLecture 23 (Knight: 14.1 -

14.6)Simple Harmonic Motion

December 7, 2005 (29 Slides)

John G. CramerProfessor of Physics

B451 [email protected]

December 7, 2005 Physics 121C - Lecture 23 2

AnnouncementsAnnouncementsTycho is Fixed! Homework Assignment #9 is due

on Midnight on Friday, December 9.

Tycho Survey: Please complete the “Survey” in the Homework section of Tycho as soon as possible. Its completion is required for this course.

The Final Exam will be held in this room on Monday, December 12 at 8:30 AM (Note: NOT 10:30 AM!).

Exam #3 will be returned on Friday. The Exam 3 grades and solutions are now on the web. If your Exam 3 scores are missing, see Laura Clement (C136 PAB). Exam #3 regrade requests will be accepted until noon on Monday, December 12.


20 30 40 50 60 70 80 90Score

0

5

10

15

20

rebmu

N

Exam 3

Exam 3 StatisticsExam 3 Statistics

Average = 57.83Std Dev = 13.91

1.7 2.7 3.7

High = 88Median = 58Low = 14


11

5-Dec-0522

3-D Rotation, Precession Supplemental

Angular Momentum

Make-up labs 7-Dec-05

23Simple Harmonic Motion 14.1 to 14.6

9-Dec-05 R4 Review & Extension-

12-Dec-05 FEFINAL EXAMINATION: Monday, December 12, 8:30 to 10:20 AM

Lecture Schedule (Part 4)

Lecture Schedule (Part 4)

We are here!


OscillationsOscillations

Oscillations (whether sinusoidal or otherwise) have some common characteristics:1. They take place around an equilibrium

position;2. The motion is periodic and repeats with each

cycle.


Frequency and PeriodFrequency and Period

1/ and 1/f T T f

is the frequency (units: Hz oscillations per second)f

is the period (units: s)T


Example: Radio Station Frequency and Period

Example: Radio Station Frequency and Period

What is the oscillation period of an FM radio station that broadcasts at 100 MHz?

8100 MHz 1.0 10 Hzf

88

11/ 1.0 10 s 10 ns

1.0 10 HzT f

Note that 1/Hz = s


Simple Harmonic MotionSimple Harmonic Motion

The special case of an oscillation that has smooth sinusiodal motion is called “Simple Harmonic Motion”, abbreviated SHM. The motion can be described by a period (the time required for one complete oscillation) and an amplitude (the distance from the equilibrium position to a turning point).

An example is a glider attached to a spring and moving on an air track.


Kinematics ofSimple Harmonic Motion

Kinematics ofSimple Harmonic Motion

2( ) cos cos 2

tx t A A f t

T

2(in rad/s) 2 (in Hz)f

T

( ) cosx t A t

max

( )( ) sin sinx

dx tv t A t v t

dt

max

22

Av A fA

T


Example: A System in SHM

Example: A System in SHM

An air-track glider is attached to a spring,pulled 20 cm to the right, and releasedat t-=0. It makes 15 completeoscillations in 10 s.

a. What is the period of oscillation?b. What is the object’s maximum speed?c. What is its position and velocity at t=0.80 s?

15 oscillations

10 s1.5 oscillations/s 1.5 Hz

f

1/ 0.667 sT f

max

2 2 (0.20 m)1.88 m/s

(0.667 s)

Av

T

2 2 (0.80 s)cos (0.20 m)cos 0.062 m 6.2 cm

(0.667 s)

tx A

T

max

2 2 (0.80 s)sin (1.88 m/s)sin 1.79 m/s 179 cm/s

(0.667 s)

tv v

T


Example: Finding the Time

Example: Finding the Time

A mass, oscillating in simple harmonic motion, starts at x=A and has period T. At what time, as a fraction of T, does the mass first pass through x=½A?

11 162

cos2 2 3

T Tt T

1

2

2cos

tx A A

T


SHM and Circular MotionSHM and Circular Motion

Uniform circular motion projected into one dimension is simple harmonic motion.


SHM and Circular MotionSHM and Circular Motion

cosx A

d

dt

0If 0 at 0, t t

( ) cos cosx t A A t


The Phase Constant 0The Phase Constant 0

0t

0

( ) cos ( )

cos

x t A t

A t

0

max 0

( ) sin

sin

v t A t

v t

0 0

0 0

cos

sinx

x A

v A


The Phase Constant 0The Phase Constant 0


Example:Using Initial Conditions

Example:Using Initial Conditions

A mass on a spring oscillates with a period of 0.80 s and an amplitude of 10 cm. At t=0 s it is 5.0 cm to the left of equilibrium and moving to the left. What are its position and direction of motion at t=2.0 s?

21 100 3

(0.05 m)cos cos 120

(0.10 m)

x

A

0The mass is moving to the left, so 120 .

2 27.854 rad/s

(0.80 s)T

0

2

3

( ) cos

(0.10 m)cos (7.854 rad/s)(2.0 s)

5.00 cm

x t A t

0

2

3

( ) sin

(0.10 m)(7.854 rad/s)sin (7.854 rad/s)(2.0 s)

68.1 cm/s ( >0 so moving to the right)

x

x

v t A t

v


Clicker Question 1Clicker Question 1

Shown are four harmonic oscillators at t=0. All are oscillating with amplitude A as shown. Their velocity directions are shown by the green arrows.

Which one has the phase constant ?


Energy and SHMEnergy and SHM1 12 22 2

constantxE K U mv kx

1 22

( )E x A U kA 1 2max2

( 0)E x K mv

1 12 2max2 2

mv kA max

kv A A

m

12

2

k k mf T

m m k

1 1 1 12 2 2 2max2 2 2 2

constantxE mv kx kA mv


Energy and SHMEnergy and SHM1 1 1 12 2 2 2

max2 2 2 2constantxE mv kx kA mv

2 2 2 2kv A x A x

m

222 20 00 0

mv vA x x

k


Example:Using Conservation of

Energy

Example:Using Conservation of

Energy A 500 g block on a spring is pulled a distance of 20 cm and released. The subsequent oscillations are measured to have a period of 0.80 s. At what position (or positions) is the speed of the block 1.0 m/s?

2 2 2 2kv A x A x

m

222 2 (1.0 m/s)

(0.20 m) 0.154 m 15.4 cm(7.85 rad/s)

vx A

2 20.80 s so 7.85 rad/s

(0.80 s)T

T


Clicker Question 2Clicker Question 2

Four springs have been compressed from their equilibrium positions at x=0 cm.

Which system has the largest maximum speed in its oscillation?


The Dynamics of SHMThe Dynamics of SHM

2( sin ) cosxx

dv da A t A t

dt dt

sinxv A t

( ) cosx t A t


The Dynamics of SHMThe Dynamics of SHM

2 cosxa A t

( ) cosx t A t

2xa x


The Equation of MotionThe Equation of Motion

sp( )xF k x

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

2x

x 2

dv d xa

dt dt

2

2

d x kx

dt m


Solving the SHMEquation of MotionSolving the SHM

Equation of Motion2

2

d x kx

dt m

0( ) cosx t A t

0sindx

A tdt

2

202

cosd x

A tdt

20 0cos cos

kA t A t

m

2 The ( ) equation is a solution if k

x tm

What is x?

Let’s make a wild guess:

0Solution: ( ) cos with /x t A t k m


Example:Analyzing an Oscillator

Example:Analyzing an Oscillator

At t=0, a 500 g block oscillating on a spring is observed to be moving to the right at x=15 cm. It reaches a maximum displacement of 25 cm at t=0.300s.

a. Draw a position vs. time graph for one cycle of motion.

b. At what times during the cycle will the mass pass through 20 cm?

1 100

(15 cm)cos cos 0.927 rad

(25 cm)

x

A

0It moves to the right at t=0, so 0.927 rad.

max max 0 max 0cos so 0x A A t t

0 max/ ( 0.927 rad) /(0.300 s) 3.09 rad/st

22.03 sT

10

1

1cos

1 (20 cm)0.927 rad cos

3.09 rad/s (25 cm)

0.092 s and 0.508 s

xt

A


Vertical OscillationsVertical Oscillationssp( ) (because )yF k y k L y L

net sp

eq

For equilibrium ( ) ( )

0

y y yF F w

k L mg

eq eq. Now displace the mass from .mg

L Lk

net sp

eq

eq

( ) ( )

( )

( )

y y yF F w

k L y mg

k L mg ky

net( ) yF ky


Vertical OscillationsVertical Oscillations

0( ) cos( )y t A t

net( ) yF ky


Example:A Vertical Oscillation

Example:A Vertical Oscillation

A 200 g block hangs from a spring with spring constant 10 N/m. The block is pulled down to a point where the spring is 30 cm longer than it’s unstretched length, then released. Where is the block and what is its velocity 3.0 s later?

/ 0.196 m =19.6 cmL mg k

0( 0.30 m) ( 0.196 m) 0.104 m; =0A

/ (10 N/m) /(0.2 kg) 7.07 rad/sk m

0( ) cos( )

( 0.104 m)cos[(7.07 rad/s)(3.0 s)]

0.074 m

= 7.7 cm (above equilibrium position).

y t A t

0sin( ) 0.52 m/s 52 cm/sxv A t


End of Lecture 23End of Lecture 23Tycho is Fixed! Homework Assignment #9 is due

on Midnight on Friday, December 9.

Tycho Survey: Please complete the “Survey” in the Homework section of Tycho as soon as possible.

The Final Exam will be held in this room on Monday, December 12 at 8:30 AM (Note: NOT 10:30 AM!).

Exam #3 should be graded and returned by Friday. The Exam 3 solutions are on the web. Exam #3 regrade requests will be accepted until noon on Monday, December 12.

john g. cramer professor of physics b451 pab [email protected]

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