Physics 1C: Simple Harmonic Motion - Main Page - CASS

23
Physics 1C: Simple Harmonic Motion Friday, 3 April 2015

Transcript of Physics 1C: Simple Harmonic Motion - Main Page - CASS

Physics 1C: Simple Harmonic Motion

!!

Friday, 3 April 2015

Important Info

!

• course website: http://ted.ucsd.edu!

• backup course website: http://cass.ucsd.edu/~rskibba/work/Teaching.html !

!

• http://www.webassign.net !

! self-enroll with our 1C class key: ucsd 2397 3544!

“Chapter 12 Homework Assignment” and “Chapter 12 Suggested Problems” are available on WebAssign. They’re not due for grades but you should work through them. !

Chapter 12 Reading Quiz will be up on WebAssign; it’s now at least on the Ted site (along with syllabus and first two lectures)

Reminders

!

• make sure to register your clicker! next week, your class participation will count.

• next week’s homework assignment will need to be completed on WebAssign by next Friday at 1pm

• the first test will be next Friday at 1pm

• if after reading the book and working on problems you have difficulty with anything, take advantage of office hours and the TA’s problem sessions (Thursdays)

Physics News

• particle physics: Large Hadron Collider particle accelerator (discoverer of Higgs boson) restarts after two-year hiatus

• optics: 2015 is the International Year of Light; Physical Review journals highlight important breakthroughs in optics

• relativity: scientists celebrate Einstein’s publication of General Relativity theory 100 years ago

• astrophysics: Dark Energy Survey scientists discover nine new dwarf galaxies orbiting our Milky Way

Simple Harmonic Motion (SHM)

Welcome to Physics 1C

Friday, April 5

Physics 1C

! if you haven’t done the reading quiz, you have until midnight

! Week 1 Problems due before class on Monday

! Starting next week, no extension. And you need your clicker by then.

! turn off your laptops, phones, etc...

Physics 1C

x = A cos(⇥t + �0)

d2x

dt2= ��2x

Simple Harmonic Motion

positiondx

dt= �A⇥ sin(⇥t + �0)velocity

d2x

dt2= �A⇥2

cos(⇥t + �0)acceleration

“linear restoring force”

Fnet = md2x

dt2= �m�2x

Physics 1C

Fnet = �m�2x

Simple Harmonic Motion

equilibrium

~n

~W

~Fs

Fnet = �kx

� =r

k

mT = 2�

rm

k

Simple Harmonic Motion

Welcome to Physics 1C

Friday, April 5

Physics 1C

! if you haven’t done the reading quiz, you have until midnight

! Week 1 Problems due before class on Monday

! Starting next week, no extension. And you need your clicker by then.

! turn off your laptops, phones, etc...

Physics 1C

x = A cos(⇥t + �0)

d2x

dt2= ��2x

Simple Harmonic Motion

positiondx

dt= �A⇥ sin(⇥t + �0)velocity

d2x

dt2= �A⇥2

cos(⇥t + �0)acceleration

“linear restoring force”

Fnet = md2x

dt2= �m�2x

Physics 1C

Fnet = �m�2x

Simple Harmonic Motion

equilibrium

~n

~W

~Fs

Fnet = �kx

� =r

k

mT = 2�

rm

k

Simple Harmonic Motion

Physics 1C

Fnet = �m�2x

Simple Harmonic Motion

~W

~Fs

� =r

k

mT = 2�

rm

k

unstretched

equilibrium y=0

�L =mg

k

Fnet = �mg + k(�L� y)

Fnet = �ky + (k�L�mg)

Fnet = �ky

Physics 1C

Fnet = �m�2x

Simple Harmonic Motion

Fnet,t = �mg sin �

small angle approximationfor small angles, ! < 10°,

! measured in radianssin ✓ ⇡ ✓

Physics 1C

Fnet = �m�2x

Simple Harmonic Motion

Fnet,t = �mg sin �

Fnet,t ⇡ �mg

Ls

⇥ =r

g

LT = 2�

sL

g

Physics 1C

y

v

a

y = A cos(⇥t + �0)

phase = ⇥t + �0

Clicker-3A

What is the phase when the position of the mass is at its minimum value?

A. 0B. "/2C. "

D. 3"/2E. more than one of these

Simple Harmonic Motion: Pendulum

Physics 1C

Fnet = �m�2x

Simple Harmonic Motion

~W

~Fs

� =r

k

mT = 2�

rm

k

unstretched

equilibrium y=0

�L =mg

k

Fnet = �mg + k(�L� y)

Fnet = �ky + (k�L�mg)

Fnet = �ky

Physics 1C

Fnet = �m�2x

Simple Harmonic Motion

Fnet,t = �mg sin �

small angle approximationfor small angles, ! < 10°,

! measured in radianssin ✓ ⇡ ✓

Physics 1C

Fnet = �m�2x

Simple Harmonic Motion

Fnet,t = �mg sin �

Fnet,t ⇡ �mg

Ls

⇥ =r

g

LT = 2�

sL

g

Physics 1C

y

v

a

y = A cos(⇥t + �0)

phase = ⇥t + �0

Clicker-3A

What is the phase when the position of the mass is at its minimum value?

A. 0B. "/2C. "

D. 3"/2E. more than one of these

Simple Harmonic Motion: Pendulum

Physics 1C

Fnet = �m�2x

Simple Harmonic Motion

~W

~Fs

� =r

k

mT = 2�

rm

k

unstretched

equilibrium y=0

�L =mg

k

Fnet = �mg + k(�L� y)

Fnet = �ky + (k�L�mg)

Fnet = �ky

Physics 1C

Fnet = �m�2x

Simple Harmonic Motion

Fnet,t = �mg sin �

small angle approximationfor small angles, ! < 10°,

! measured in radianssin ✓ ⇡ ✓

Physics 1C

Fnet = �m�2x

Simple Harmonic Motion

Fnet,t = �mg sin �

Fnet,t ⇡ �mg

Ls

⇥ =r

g

LT = 2�

sL

g

Physics 1C

y

v

a

y = A cos(⇥t + �0)

phase = ⇥t + �0

Clicker-3A

What is the phase when the position of the mass is at its minimum value?

A. 0B. "/2C. "

D. 3"/2E. more than one of these

Simple Harmonic Motion: Pendulum

Physics 1C

Fnet = �m�2x

Simple Harmonic Motion

~W

~Fs

� =r

k

mT = 2�

rm

k

unstretched

equilibrium y=0

�L =mg

k

Fnet = �mg + k(�L� y)

Fnet = �ky + (k�L�mg)

Fnet = �ky

Physics 1C

Fnet = �m�2x

Simple Harmonic Motion

Fnet,t = �mg sin �

small angle approximationfor small angles, ! < 10°,

! measured in radianssin ✓ ⇡ ✓

Physics 1C

Fnet = �m�2x

Simple Harmonic Motion

Fnet,t = �mg sin �

Fnet,t ⇡ �mg

Ls

⇥ =r

g

LT = 2�

sL

g

Physics 1C

y

v

a

y = A cos(⇥t + �0)

phase = ⇥t + �0

Clicker-3A

What is the phase when the position of the mass is at its minimum value?

A. 0B. "/2C. "

D. 3"/2E. more than one of these

Simple Harmonic Motion: Pendulum

Physics 1C

y

v

a

Clicker-3B

What is the phase when the position of the mass

is at its equilibrium (y=0) value?

A. 0B. "/2C. "

D. 3"/2E. more than one of these

y = A cos(⇥t + �0)

phase = ⇥t + �0

Physics 1C

y

v

a

What is the phase when the position of the mass

is at its equilibrium (y=0) value; and moving

upward?

A. 0B. "/2C. "

D. 3"/2E. more than one of these

y = A cos(⇥t + �0)

phase = ⇥t + �0

Clicker-3C

Physics 1C

y

v

a

What does the phase constant mean?

The phase constant tells us the initial phase:

where to start the graph at t = 0.

y = A cos(⇥t + �0)

phase = ⇥t + �0

phase constant = �0

�0 =⇡

2for example, when

0 Physics 1C

A simple harmonic oscillator has amplitude 2 cm and phase constant "/3. Draw a graph for the position of the oscillator as a function of time. Physically interpret (i.e.,

what is going on in terms of position, velocity, acceleration) the motion at t = 0 s.

A simple harmonic oscillator has amplitude 2 cm. At t = 0 s, it is located at x = 1 cm and is moving in the +x direction. Draw a graph of the motion and determine the

phase constant.

Simple Harmonic Motion: PendulumPhysics 1C

y

v

a

Clicker-3B

What is the phase when the position of the mass

is at its equilibrium (y=0) value?

A. 0B. "/2C. "

D. 3"/2E. more than one of these

y = A cos(⇥t + �0)

phase = ⇥t + �0

Physics 1C

y

v

a

What is the phase when the position of the mass

is at its equilibrium (y=0) value; and moving

upward?

A. 0B. "/2C. "

D. 3"/2E. more than one of these

y = A cos(⇥t + �0)

phase = ⇥t + �0

Clicker-3C

Physics 1C

y

v

a

What does the phase constant mean?

The phase constant tells us the initial phase:

where to start the graph at t = 0.

y = A cos(⇥t + �0)

phase = ⇥t + �0

phase constant = �0

�0 =⇡

2for example, when

0 Physics 1C

A simple harmonic oscillator has amplitude 2 cm and phase constant "/3. Draw a graph for the position of the oscillator as a function of time. Physically interpret (i.e.,

what is going on in terms of position, velocity, acceleration) the motion at t = 0 s.

A simple harmonic oscillator has amplitude 2 cm. At t = 0 s, it is located at x = 1 cm and is moving in the +x direction. Draw a graph of the motion and determine the

phase constant.

SHM: spring example

Physics 1C

A simple harmonic oscillator has amplitude 2 cm and phase constant "/3. Draw a graph for the position of the oscillator as a function of time. Physically interpret (i.e.,

what is going on in terms of position, velocity, acceleration) the motion at t = 0 s.

Physics 1C

A mass on a spring oscillates with period 0.80 s and amplitude 10 cm. At t = 0 s, it is 5.0 cm to the left of equilibrium and

moving to the left. What are the position and velocity of the mass at time t = 2.0 s?

x = A cos(⇥t + �0) t = 0 x = �5.0 cmA = 10 cm ⇥ =

2�

T= 7.854 rad/s

�5.0 cm = (10 cm) cos �0

x = (10 cm) cos

✓(7.854 rad/s)t +

2�

3

⇥0 =2�

3,4�

3

2�

3,

4�

3

Physics 1C

A mass on a spring oscillates with period 0.80 s and amplitude 10 cm. At t = 0 s, it is 5.0 cm to the left of equilibrium and

moving to the left. What are the position and velocity of the mass at time t = 2.0 s?

x(t) = (10 cm) cos

✓(7.854 rad/s)t +

2�

3

v(t) = �(78.54 cm/s) sin✓

(7.854 rad/s)t +2�

3

x(2 s) = 5.0 cm

v(2 s) = 68 cm/s

Physics 1C

For Next Time

enroll in our course on MasteringPhysics

Syllabus Quiz due Friday (MasteringPhysics)Reading Quiz for Friday’s Class (MasteringPhysics)

Weekly Problems due Friday (MasteringPhysics)

Metacognition Note due Monday (ted)Reading Quiz for Monday’s Class (MasteringPhysics)

obtain your i>clicker and register it

SHM: spring example

Physics 1C

A simple harmonic oscillator has amplitude 2 cm and phase constant "/3. Draw a graph for the position of the oscillator as a function of time. Physically interpret (i.e.,

what is going on in terms of position, velocity, acceleration) the motion at t = 0 s.

Physics 1C

A mass on a spring oscillates with period 0.80 s and amplitude 10 cm. At t = 0 s, it is 5.0 cm to the left of equilibrium and

moving to the left. What are the position and velocity of the mass at time t = 2.0 s?

x = A cos(⇥t + �0) t = 0 x = �5.0 cmA = 10 cm ⇥ =

2�

T= 7.854 rad/s

�5.0 cm = (10 cm) cos �0

x = (10 cm) cos

✓(7.854 rad/s)t +

2�

3

⇥0 =2�

3,4�

3

2�

3,

4�

3

Physics 1C

A mass on a spring oscillates with period 0.80 s and amplitude 10 cm. At t = 0 s, it is 5.0 cm to the left of equilibrium and

moving to the left. What are the position and velocity of the mass at time t = 2.0 s?

x(t) = (10 cm) cos

✓(7.854 rad/s)t +

2�

3

v(t) = �(78.54 cm/s) sin✓

(7.854 rad/s)t +2�

3

x(2 s) = 5.0 cm

v(2 s) = 68 cm/s

Physics 1C

For Next Time

enroll in our course on MasteringPhysics

Syllabus Quiz due Friday (MasteringPhysics)Reading Quiz for Friday’s Class (MasteringPhysics)

Weekly Problems due Friday (MasteringPhysics)

Metacognition Note due Monday (ted)Reading Quiz for Monday’s Class (MasteringPhysics)

obtain your i>clicker and register it

Conservation of Energy

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

• This is an isolated system, so the total energy is constant • The kinetic energy is !!

• The elastic potential energy is

total energy E = K + U = (1/2)kA2 = constant!

Conservation of Energy

Physics 1C

Conservation of Energy

K =12mv

2Us =

12k(�x)2

equilibrium, "x = 0

Physics 1C

equilibrium, "x = 0

block released from restA = Amplitude

K Us

12kA2

Physics 1C

equilibrium, "x = 0

block released from restA = Amplitude

K Us

12kA2

K Us

12mv2

max

Physics 1C

equilibrium, "x = 0

block released from restA = Amplitude

K Us

12kA2

K Us

12mv2

max

K Us

Conservation of Energy

Physics 1C

Conservation of Energy

K =12mv

2Us =

12k(�x)2

equilibrium, "x = 0

Physics 1C

equilibrium, "x = 0

block released from restA = Amplitude

K Us

12kA2

Physics 1C

equilibrium, "x = 0

block released from restA = Amplitude

K Us

12kA2

K Us

12mv2

max

Physics 1C

equilibrium, "x = 0

block released from restA = Amplitude

K Us

12kA2

K Us

12mv2

max

K Us

Conservation of Energy

Physics 1C

Conservation of Energy

K =12mv

2Us =

12k(�x)2

equilibrium, "x = 0

Physics 1C

equilibrium, "x = 0

block released from restA = Amplitude

K Us

12kA2

Physics 1C

equilibrium, "x = 0

block released from restA = Amplitude

K Us

12kA2

K Us

12mv2

max

Physics 1C

equilibrium, "x = 0

block released from restA = Amplitude

K Us

12kA2

K Us

12mv2

max

K Us

Conservation of Energy

Physics 1C

Conservation of Energy

K =12mv

2Us =

12k(�x)2

equilibrium, "x = 0

Etotal

=12mv2

max

=12kA2

kinetic energy = potential energy?Physics 1C

A mass affixed to the end of a horizontal spring oscillates with amplitude A. When the mass is

located at which position below will the potential energy off the mass-spring system be the largest?

xA = -0.6A xB = -0.2A xC = +0.3A xD = +0.8A

A B C D

E. All four configurations have the same potential energy

Clicker-4A

Physics 1C

*credit for correct answer

A mass affixed to the end of a horizontal spring oscillates with amplitude A. At which location will

the mass have the largest speed? (Hint: think conservation of energy)

xA = -0.6A xB = -0.2A xC = +0.3A xD = +0.8A

A B C D

E. The speed is the same at all four points

Clicker-4B

Physics 1C

x

v

a

K " Us # K # Us " K " Us # K # Us "

K=0 K=0U=0 U=0

Conservation of Energy

Conservation of Energy: Example

Physics 1C

Conservation of Energy

K =12mv

2Us =

12k(�x)2

equilibrium, "x = 0

Etotal

=12mv2

max

=12kA2

kinetic energy = potential energy?Physics 1C

A mass affixed to the end of a horizontal spring oscillates with amplitude A. When the mass is

located at which position below will the potential energy off the mass-spring system be the largest?

xA = -0.6A xB = -0.2A xC = +0.3A xD = +0.8A

A B C D

E. All four configurations have the same potential energy

Clicker-4A

Physics 1C

*credit for correct answer

A mass affixed to the end of a horizontal spring oscillates with amplitude A. At which location will

the mass have the largest speed? (Hint: think conservation of energy)

xA = -0.6A xB = -0.2A xC = +0.3A xD = +0.8A

A B C D

E. The speed is the same at all four points

Clicker-4B

Physics 1C

x

v

a

K " Us # K # Us " K " Us # K # Us "

K=0 K=0U=0 U=0

Conservation of Energy: Example

Physics 1C

Conservation of Energy

K =12mv

2Us =

12k(�x)2

equilibrium, "x = 0

Etotal

=12mv2

max

=12kA2

kinetic energy = potential energy?Physics 1C

A mass affixed to the end of a horizontal spring oscillates with amplitude A. When the mass is

located at which position below will the potential energy off the mass-spring system be the largest?

xA = -0.6A xB = -0.2A xC = +0.3A xD = +0.8A

A B C D

E. All four configurations have the same potential energy

Clicker-4A

Physics 1C

*credit for correct answer

A mass affixed to the end of a horizontal spring oscillates with amplitude A. At which location will

the mass have the largest speed? (Hint: think conservation of energy)

xA = -0.6A xB = -0.2A xC = +0.3A xD = +0.8A

A B C D

E. The speed is the same at all four points

Clicker-4B

Physics 1C

x

v

a

K " Us # K # Us " K " Us # K # Us "

K=0 K=0U=0 U=0

For Monday:

1. go through chapter 12 reading quiz on course website:

ted.ucsd.edu (backup:cass.ucsd.edu/~rskibba/ )

2. self-enroll on www.webassign.net using our class key

(ucsd 2397 3544) and FINISH CHAPTER 12 HOMEWORK

if you haven’t already

3. make sure you understand: position, velocity, acceleration;

frequency and period; force and energy!

4. start reading chapter 13 (Mechanical Waves): at least

sections 13.1 and 13.2