01 Simple Harmonic Motion.pptx

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I can . . .

• recognize when an object is in

Simple Harmonic Motion

• distinguish:

• relate SHM to circular motion

• predict the position x, velocity v, and acceleration a of an object in SHM for any given time.

Period, T Amplitude, A

Position, x Velocity, v

Centripetal Accel, a_c

from from from from from Frequency, f Position, x Angle, θ

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S

imple

H

armonic

M

otion

is that motion in which a body

moves back and forth over a

fixed path, returning to each

position

,

velocity

, and

acceleration

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(5)

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I can . . .

• recognize when an object is in Simple Harmonic Motion

• distinguish:

• relate SHM to circular motion

• predict the position x, velocity v, and acceleration a of an object in SHM for any given time.

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F

o

rc

e

(N

)

Distance (m)

Hooke’s Law

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When is the Force_spring, Velocity, and Acceleration

the greatest versus least? WHY?

When is the Velocity, and the Acceleration in the same direction versus opposite direction? WHY?

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Knowing:

the

spring constant

,

k

the

mass

,

m

and the Amplitude, A

Find an equation that would predict how

fast

,

v

, the

mass is traveling when it is at

position

,

x

.

Amplitude, A

Position, x

Conservation

of

(14)

½

m

v

2 + ½

kx

2 = ½

k

A

2 Knowing:

the

spring constant

,

k

the

mass

,

m

and the Amplitude, A

How

fast

,

v

, is it traveling when it is at

position

,

x

?

Amplitude, A

Position, x

Conservation

of

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I can . . .

• recognize when an object is in Simple Harmonic Motion • distinguish:

• relate SHM to circular motion

• predict the position x, velocity v, and acceleration a of an object in SHM for any given time.

Period, T

Amplitude, A

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Clock (seconds)

Estimate:

Period, T

Frequency, f

Amplitude, A

5 s

0.2 Hz

1.9 cm

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Period, T

Time to make one complete

cycle

.

Frequency, f

N

umber of

cycles

made each second.

Amplitude, A

Maximum displacement, x from equilibrium

position.

(

seconds

/

cycle

)

(

cycles

/

second

) (Hertz)

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Example 1: The suspended mass makes 30 complete

oscillations in 15 s. What is the period and frequency of the motion?

x F

Period: T = 0.500 s

Frequency: f = 2.00 Hz

15 s

0.50 s

30 cylces

T



1

1 0.500 s

f

T

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6.28 seconds

Angle (Radians) = Arc Length (Meters)_{Radius (Meters)}

How many radians are in ONE rotation?

2·π·R

R = 2·π

An angle of ONE Radian is when

the

arc length

equals the

radius

.

ONE Radian (57.3°) 2 3 4 5 6

How many arc lengths (radii) fit into the

circumference?

If your Angular Velocity is

1 radian per second, how long will it take to rotate

once?

2 radians per second,

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ω =

2π

f

ω =

2π radians

_cycle

_second

cycles

θ

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http://mathprecalculus.wikispaces.com/Chapter+5.1+Unit+Circle+Approach

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π 3π/2 2π π/2

0

x = cos(θ)

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The reference circle compares the circular motion of an object with its horizontal projection.

The Reference Circle

x = Horizontal displacement. A = Amplitude (x_max).

q = Reference angle.

ω = 2π

f

x = Acosωt

t = Time.

f = frequency

cos

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LEARNING GOAL

for

S

imple

H

armonic

M

otion

1. Recognize

when an object is in

SHM

.

2. Know how one is

to the other:

3. See how

SHM

is related to Circular Motion

4. Be able to

predict

the

position, x, velocity,

v,

and

acceleration, a

of an object in

SHM

for any time.

Period, T Amplitude, A Position, x Velocity, v Centripetal Accel, a_c versus versus versus versus versus Frequency, f Position, x Angle, θ

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Does this equation model reality?

Δx

Δt

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y

=

¼a

· x

+

b

vertical axis

horizontal axis

constant Slope

constant Intercept

= · +

10

20

Slope = 2

y

=

2

· x

+

6

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y

=

2a

· x

2 y

x

y

x

2

Use a graph to find

the value of “

a

”.

y

=

2a

·(

x

2 ₎

= ·

vertical axis horizontal axis constant Slope

Slope = 0.5

0.5 = 2a

0.25 = a

LINEARIZE THE DATA

100

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𝒚

𝟐 =

𝟑 (

𝒂 𝒙

+

𝟐 𝒃

)

𝒚

=

𝟐 𝒂

𝒙

+

𝒃

𝒚 +𝟐= 𝒂 𝒙

𝟑 +

√

𝒃

What variables should be graphed in order to linearize the data?

Find the values of a and b in terms of slope and

intercept.

LINEARIZING DATA

𝒚

=

−

𝟓 𝒂 𝒙

𝟐

+

𝒃

𝒚

=

−

𝟓 𝒂 𝒙

𝟐

+

𝒃

𝑦

2 =

3 𝑎 𝑥

+

6 𝑏

𝑦 = 𝑎

3 𝑥 +

√

𝑏 − 2

𝑦

=

2 𝑎

1 𝑥

+

𝑏

a = slope/(-5) b = intercept

a = 3·slope

b = (intercept+2)2

a = 1/3·slope b = 1/6·intercept

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x y y/x 1 1.6 1.60 2 6 3.00 3 6 2.00 4 10 2.50 5 12 2.40 6 13 2.17 7 16 2.29 8 21 2.63 9 18.1 2.01 10 19.2 1.92

Average = 2.25

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Work and Kinetic Energy

A resultant force changes the velocity of an object and does work on that object.

m v_o m v_f x F F

(

) ;

Work



Fx



ma x

2 2 0

2

f

v

a

x





2 2 1 1 0

2 f 2

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Work Done in Stretching a Spring

F

x

m

Work done

ON

the spring is

positive

;

work

BY

spring is

negative.

From Hooke’s law the force F is:

F (x) = kx

x

₁

x

₂

F

To stretch spring from

_x

1

to x

2

, work is:

(Review module on work)

2 2

2 1

½

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Example 2: A 4-kg mass suspended from a spring produces a displacement of 20 cm. What is the spring constant?

F

20 cm

m

The stretching force is the weight (W = mg) of the 4-kg mass:

F = (4 kg)(9.8 m/s2) = 39.2 N

Now, from Hooke’s law, the force constant k of the spring is:

k = =

D

F

D

x

39.2 N

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Example 2(cont.: The mass m is now stretched a distance of 8 cm and held. What is the potential energy? (k = 196 N/m)

F

8 cm

m

U =

0.627 J

U =

0.627 J

The potential energy is equal to the work done in stretching the spring:

0

2 2 2 1

½

Work



kx



kx

2 2

½

½(196 N/m)(0.08 m)

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Displacement in SHM

m

x = 0 x = +A

x = -A

x

• _{Displacement is}_positive_{when the position is}

to the right of the equilibrium position (x = 0) and negative when located to the left.

• _The_maximum_{displacement is called the}

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Velocity in SHM

m

x = 0 x = +A

x = -A

v (+)

• _{Velocity is positive when moving to the right}

and negative when moving to the left.

• _{It is zero at the end points and a maximum}

at the midpoint in either direction (+ or -).

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Acceleration in SHM

m

x = 0 x = +A

x = -A

• _{Acceleration is in the direction of the}

restoring force. (a is positive when x is

negative, and negative when x is positive.)

• _{Acceleration is a maximum at the end points}

and it is zero at the center of oscillation.

+x -a -x

+a

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Maximum Force

Maximum Acceleration

Maximum Force

Maximum Acceleration Net Force = Zero

Maximum Velocity

m·g

m·g m·g

F_spring

F_Net

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Acceleration vs. Displacement

m

x = 0 x = +A

x = -A

x a v

Given the spring constant, the displacement, and the mass, the acceleration can be found from:

or

Note: Acceleration is always opposite to displacement.

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Example 3: A 2-kg mass hangs at the end of a spring whose constant is k = 400 N/m. The mass is displaced a distance of

12 cm and released. What is the acceleration at the instant the displacement is x = +7 cm?

m

+x a = -14.0 m/s2

a = -14.0 m/s2

a

Note: When the displacement is +7 cm

(downward), the acceleration is -14.0 m/s2

(upward) independent of motion direction. (400 N/m)(+0.07 m)

2 kg

a  

kx

a

m

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Example 4: What is the maximum acceleration for the 2-kg

mass in the previous problem? (A = 12 cm, k = 400 N/m)

m

+x The maximum acceleration occurs

when the restoring force is a

maximum; i.e., when the stretch or compression of the spring is largest.

F = ma = -kx x_max =  A

a

_max

= ± 24.0 m/s

2

a

_max

= ± 24.0 m/s

2

Maximum Acceleration:

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Conservation of Energy

The total mechanical energy (U + K) of a vibrating system is constant; i.e., it is the same at any point in the oscillating path.

m

x = 0 x = +A

x = -A

x a v

For any two points A and B, we may write:

½mv_A2 + ½kx

A 2 = ½mvB2 + ½kxB 2

½mv_A2 + ½kx

A 2 = ½mvB2 + ½kxB 2

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Energy of a Vibrating System:

m

x = 0 x = +A

x = -A

x a v

• _{At any other point:}_{U + K = ½mv}2 + ½kx2

U + K = ½kA2 x =  A and v = 0.

• _{At points}_A_and_B_{, the velocity is zero and the}

acceleration is a maximum. The total energy is:

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Velocity as Function of Position.

m

x = 0 x = +A

x = -A

x a v

v_max when x = 0:

k v A m  2 2 k

v A x

m

 

2 2 2

1 1 1

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Example 5: A 2-kg mass hangs at the end of a spring whose constant is k = 800 N/m. The mass is displaced a distance of

10 cm and released. What is the velocity at the instant the displacement is x = +6 cm?

m

+x

½mv2 + ½kx 2 = ½kA2

v = ±1.60 m/s

v = ±1.60 m/s 2 2

k

v A x

m

 

2 2

800 N/m

(0.1 m) (0.06 m) 2 kg

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Example 5 (Cont.): What is the maximum velocity for the previous problem? (A = 10 cm, k = 800 N/m, m = 2 kg.)

m

+x

½mv2 + ½kx 2 = ½kA2

v = ± 2.00 m/s

v = ± 2.00 m/s

0

The velocity is maximum when x = 0:

800 N/m

(0.1 m) 2 kg

k

v A

m

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Velocity in SHM

The velocity (v) of an oscillating body at any instant is the horizontal

component of its tangential velocity (v_T).

v_T = A;

v = -v_T sin q

v = -2f A sin 2f t

-v_T = 2f A

q = t = 2f t

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a_c  vt2

R

The acceleration (a) of an

oscillating body at any instant is the horizontal component of its

centripetal acceleration (a_c).

Acceleration Reference Circle

a = -a_c cos q

R = A

a = -2A _cos₍_t₎

q = t = 2ft

a = - 42f 2A cos(2ft)

 2_A

(R)2

R

  2_R

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The Period and Frequency as a

Function of

a

and

x

.

T = 1/f

a = - 42f 2 x

Recall that

F = ma = -kx

-a

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Example 6: The frictionless system shown below has a 2-kg

mass attached to a spring (k = 400 N/m). The mass is displaced a distance of 20 cm to the right and released. What is the frequency of the motion?

m

x = 0 _{x = +0.2}_m

x a v

x = -0.2 m

f = 2.25 Hz

f = 2.25 Hz

1 1 400 N/m 2 2 2 kg

k f

m

 

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Example 6 (Cont.): Suppose the 2-kg mass of the previous problem is displaced 20 cm and released (k = 400 N/m). What is the maximum acceleration? (f = 2.25 Hz)

m

x = 0 _{x = +0.2}_m

x a v

x = -0.2 m

Acceleration is a maximum when x =  A

a

=  40 m/s2

a

=  40 m/s2

2 2 2 2

4 4 (2.25 Hz) ( 0.2 m)

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Example 6: The 2-kg mass of the previous example is

displaced initially at x = 20 cm and released. What is the velocity 2.69 s after release? (Recall that f = 2.25 Hz.)

m

x = 0 _{x = +0.2}_m

x a v

x = -0.2 m

v = -0.916 m/s

v = -0.916 m/s

v = -2f A sin 2f t

v = -2f A sin 2f t

(Note: q in rads)

The minus sign means it is moving to the left.





2 (2.25 Hz)(0.2 m)sin 2 (2.25 Hz)(2.69 s)

v    

2 (2.25 Hz)(0.2 m)(0.324)

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Example 7: At what time will the 2-kg mass be located 12 cm to the left of x = 0?

(A = 20 cm, f = 2.25 Hz)

m

x = 0 _{x = +0.2}_m

x a v

x = -0.2 m

t = 0.157 s

t = 0.157 s

-0.12 m

cos(2 )

x  A  ft

1

0.12 m

cos(2 ) ; (2 ) cos ( 0.60)

0.20 m x ft ft A  _ _   _  _ 2.214 rad 2 2.214 rad;

2 (2.25 Hz)

ft t



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The Simple Pendulum

The period of a simple

pendulum is given by:

mg L

For small angles q.

1

2 g

f

L



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What must be the length of a simple pendulum for a clock which has a period of two seconds (tick-tock)?

L

L = 0.993 m

2 L

T

g





2 2 2 2

4 ; L =

4 L

T g

T

g





2 2 2

(2 s) (9.8 m/s )

4 L



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http://pirt.asu.edu/news%20Pendulum%20Wave.asp?index=0

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𝐿

=

𝑇

2

_∙

_𝑔

4 𝜋

2

p 2

L T

g



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I can . . .

• recognize when an object is in

Simple Harmonic Motion

• distinguish:

• relate SHM to circular motion

• predict the position x, velocity v, and acceleration a of an object in SHM for any given time.

(65)

Summary

Simple harmonic motion (SHM) is that motion in which a body moves back and forth over a fixed path, returning to each position and velocity after a definite interval of time.

F

x

m

The frequency (rev/s) is the reciprocal of the period (time for one revolution).