Waves and SoundChapter 14

Review- Chapter 13 – Oscillations about Equilibrium

Mass (M) on an ideal spring with a spring constant o (k) on a Frictionless Table

M

k

x=0

Review- Chapter 13 – Oscillations about Equilibrium

Mass (M) on an ideal spring with a spring constant o (k) on a Frictionless Table

M

k

x=0

Amplitude – A – Max displacement of mass. (m)Period – T – Time for one oscillation. (s)frequency – f – Number of oscillations per second. (Hz)Angular Frequency – w – 2 * PI divided by the period.

Review- Chapter 13 – Oscillations about Equilibrium

Mass (M) on an ideal spring with a spring constant o (k) on a Frictionless Table

M

k

x=0

Amplitude – A – Max displacement of mass. (m)Period – T – Time for one oscillation. (s)frequency – f – Number of oscillations per second. (Hz)Angular Frequency – w – 2 * PI divided by the period.

f= 1T

=2T

Mk

x=0 x=+Ax= -A

M

M

M

M

M

M

M

M

Mx=0 x=+Ax= -A

M

M

M

M

M

M

M

M

t=0

t=T4

t=T2

t=3T4

t=T

Mx=0 x=+Ax= -A

M

M

M

M

M

M

M

M

t=0

t=T4

t=T2

t=3T4

t=T

x=A

v=0 ms

a=−amax

x=0mv=−vmax

a=0m

s2

x=Av=vmax

a=0m

s2

x=−A

v=0 ms

a=amax

x=A

v=0 ms

a=−amax

x=0 x=+Ax= -A

t=0

t=T4

t=T2

t=3T4

t=T

x=A

v=0 ms

a=−amax

x=0mv=−vmax

a=0m

s2

x=Av=vmax

a=0m

s2

x=−A

v=0 ms

a=amax

x=A

v=0 ms

a=−amax

FS

FN

W

v

v

Mx=0 x=+Ax= -A

t=0

x=A

v=0 ms

a=−amax

x=A cos tv=−vmax sin ta=−amax cos t

∣vmax∣=A∣amax∣=A2

Mx=0 x=+Ax= -A

t=0

x=A

v=0 ms

a=−amax

F net=−kxma=−kx

m −A2cos t=−k A cos tm −A2cos t=−k A cos t

m2=k

= km

Mx=0 x=+Ax= -A

t=0

x=A

v=0 ms

a=−amax

= km

T=2=2m

k

f = 1T

Mx=0 x=+Ax= -A

t=0

x=A

v=0 ms

a=−amax

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

-3

-2.5

-2

-1.5

-1

-0.5

0

0.5

1

1.5

2

2.5

3

Position vs. Time

t(s)

x(m

)

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10 10.5 11 11.5 12 12.5 13 13.5 14 14.5 15 15.5 16 16.5 17 17.5 18

-4-3.5

-3-2.5

-2-1.5

-1-0.5

00.5

11.5

22.5

33.5

4

Velocity vs. Time

t(s)

v(m

/s)

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6 6.5 7 7.5 8 8.5 9 9.5 10 10.5 11 11.5 12 12.5 13 13.5 14 14.5 15 15.5 16 16.5 17 17.5 18

-5

-4

-3

-2

-1

0

1

2

3

4

5

Acceleration vs. Time

t(s)

a(m

/s2)

Mx=0 x=+Ax= -A

t=0

x=A

v=0 ms

a=−amax

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

-3

-2.5

-2

-1.5

-1

-0.5

0

0.5

1

1.5

2

2.5

3

Position vs. Time

t(s)

x(m

)

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

-4-3.5

-3-2.5

-2-1.5

-1-0.5

00.5

11.5

22.5

33.5

4

Velocity vs. Time

t(s)

v(m

/s)

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

-5

-4

-3

-2

-1

0

1

2

3

4

5

Acceleration vs. Time

t(s)

a(m

/s2)

Mx=0 x=+Ax= -A

t=0

x=A

v=0 ms

a=−amax

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

-3

-2.5

-2

-1.5

-1

-0.5

0

0.5

1

1.5

2

2.5

3

Position vs. Time

t(s)

x(m

)

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

-4-3.5

-3-2.5

-2-1.5

-1-0.5

00.5

11.5

22.5

33.5

4

Velocity vs. Time

t(s)

v(m

/s)

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

-5

-4

-3

-2

-1

0

1

2

3

4

5

Acceleration vs. Time

t(s)

a(m

/s2)

Mx=0 x=+Ax= -A

t=0

x=A

v=0 ms

a=−amax

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

-3

-2.5

-2

-1.5

-1

-0.5

0

0.5

1

1.5

2

2.5

3

Position vs. Time

t(s)

x(m

)

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

-4-3.5

-3-2.5

-2-1.5

-1-0.5

00.5

11.5

22.5

33.5

4

Velocity vs. Time

t(s)

v(m

/s)

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

-5

-4

-3

-2

-1

0

1

2

3

4

5

Acceleration vs. Time

t(s)

a(m

/s2)

Mx=0 x=+Ax= -A

t=0

x=A

v=0 ms

a=−amax

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

-3

-2.5

-2

-1.5

-1

-0.5

0

0.5

1

1.5

2

2.5

3

Position vs. Time

t(s)

x(m

)

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

-4-3.5

-3-2.5

-2-1.5

-1-0.5

00.5

11.5

22.5

33.5

4

Velocity vs. Time

t(s)

v(m

/s)

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

-5

-4

-3

-2

-1

0

1

2

3

4

5

Acceleration vs. Time

t(s)

a(m

/s2)

Mx=0 x=+Ax= -A

t=0

x=A

v=0 ms

a=−amax

A mass of 20 kg is attached to a spring of k = 100 N/m and pulled out from the equilibrium position to a point of x = + 0.8 m and released from rest. The mass oscillates back and fourth in simple harmonic motion.

a. Find the angular frequency of the motion.b. Find the period of the motion of the motion.c. Find the frequency of the motion.d. Find the maximum velocity and acceleration.e. Write specific equations for x, v, and a as a function of t.f. Draw simple graphs of x, v, and a versus time.

Mx=0 x=+Ax= -A

t=0

x o≠A

vo≠0ms

x=A cos t v=−Asin t

Mx=0 x=+Ax= -A

t=0

x o≠A

vo≠0ms

x=A cos tat t=0.0 s

x o=A cos

v=−Asin tat t=0.0 s

vo=−Asin

Mx=0 x=+Ax= -A

t=0

x o≠A

vo≠0ms

x=A cos tat t=0.0 s

x o=A cos

v=−Asin tat t=0.0 s

vo=−Asin −vo

=A sin

[−vo/]xo

=A sinAcos

=tan

Mx=0 x=+Ax= -A

t=0

x o≠A

vo≠0ms

x=A cos tat t=0.0 s

x o=A cos

v=−Asin tat t=0.0 s

vo=−A sin −vo

=A sin

[−vo/]xo

=A sinA cos

=tan

=arctan −vo

xo

Mx=0 x=+Ax= -A

t=0

x o≠A

vo≠0ms

x=A cos tat t=0.0 s

x o=A cos

v=−Asin tat t=0.0 s

vo=−A sin −vo

=A sin

xo2[−vo

]2

=A2cos2A2 sin2

x o2[−vo

]2

=A2[cos2sin2]

A= x o2[−vo

]2

Mx=0 x=+Ax= -A

t=0

A mass of 20 kg is attached to a spring of k = 100 N/m and pulled out from the equilibrium position. The initial position of x = + 0.8 m and the initial velocity is 1.2 m/s. The mass oscillates back and fourth in simple harmonic motion.

a. Find the angular frequency of the motion.b. Find the period of the motion of the motion.c. Find the frequency of the motion.d. Find the maximum velocity and acceleration.e. Write specific equations for x, v, and a as a function of t.f. Draw simple graphs of x, v, and a versus time.

Mx=0 x=+Ax= -A

t=0

Energy

E total=E potentialEkinetic

E total=UK

ETotal=12

k x212

m v2

12

k A2=12

k x 212

m v2

Mx=0 x=+Ax= -A

t=0

Magnitude of Velocity as a Function of Position.

∣v∣= km[ A2−x2]

Mx=0 x=+Ax= -A

t=0

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 50

25

50

75

100

125

150

175

200

225

Energy vs Time

K

U

Etotal

t(s)

Ene

rgy(

Joul

es)

Mx=0 x=+Ax= -A

t=0

-3 -2 -1 0 1 2 30

25

50

75

100

125

150

175

200

225

Energy vs. Position

U

Etotal

x(m)

En

erg

y(J)

U

K

Pendulum

Θ

T

w=mg

Pendulum

Θ

T

w=mg

Θw y=mg cos

w x=−mg sin

Pendulum

Θ

T

w=mg

Θw y=mg cos

w x=−mg sin

T=2 Lg

L

Physical Pendulum

T=2 lg I

ml 2

ΘCenter of Mass

lL

Physical Pendulum

T=2 lg I

ml2

ΘCenter of Mass

lL

I=13

mL2

Physical Pendulum

T=2 lg I

ml2

ΘCenter of Mass

lL

I=13

mL2

l=12

L

Physical Pendulum

T=2 lg I

ml2

ΘCenter of Mass

lL

T=2 lg I

ml2

Physical Pendulum

T=2 lg I

ml2

ΘCenter of Mass

lL

I=13

mL2

l=12

L

T=2 lg I

ml2=2 L

2g 13 mL2

m L22

Physical Pendulum

T=2 lg I

ml2

ΘCenter of Mass

lL

I=13

mL2

l=12

L

T=2 lg I

ml2=2 L

2g 13 mL2

m L22=2 L

g 23

Damped Harmonic Motion

M

F net=−kx−bvma=−kx−bv

Damped Harmonic Motion

M

F net=−kx−bvma=−kx−bv

A=Ao e−bt2m

x=Ao e−bt2m cos t

Damped Harmonic Motion

M

F net=−kx−bvma=−kx−bv

A=Ao e−bt2m

x=Ao e−bt2m cos t

= km−[ b2m]2

=o2−[ b

2m]2

o= km

Damped Harmonic Motion – under damped

M

bvmaxkA

Damped Harmonic Motion – under damped

M

0 25-10

-7.5

-5

-2.5

0

2.5

5

7.5

10

Damped Harmonic Oscillator

x(m)

t(s)

bvmaxkA

Forced Damped Harmonic Motion

M

F applied=F o sin t

F net=F osin t −kx−bv

ma=F osin t −kx−bv

Forced Damped Harmonic Motion

M

F applied=F o sin t

F net=F o sin t −kx−bv

ma=F osin t −kx−bv

x=A cos t

A=F o /m

2−o22

bm2

Forced Damped Harmonic Motion

M

F applied=F o sin t

F net=F o sin t −kx−bv

ma=F o sin t −kx−bv

x=A cos t

A=F o /m

2−o22

bm2

Forced Damped Harmonic Motion

M

F applied=F o sin t

A=F o /m

2−o22

bm2

Resonance – small amplitude driving force produces large amplitude oscillations.

Forced Damped Harmonic Motion

M

F applied=F o sin t

A=F o /m

2−o22

bm2

Resonance – small amplitude driving force produces large amplitude oscillations.Resonates when:

=o

Chapter 14 - sound

Transverse Waves – motion of medium perpendicular to the velocity of the wave.

-5 -2.5 0 2.5 5 7.5 10 12.5 15 17.50

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

1.8

2

Transverse wave

x

y

v y

Longitudinal Waves – motion of medium parallel to the velocity of the wave.

v

x

0 2 4 6 8 10 12 14 16 18-3

-2.5

-2

-1.5

-1

-0.5

0

0.5

1

1.5

2

2.5

3

Y vs. X

x(m)

y(m

)

wavelength

Amplitude A

0 2.5 5 7.5 10 12.5 15 17.5 20-3

-2.5

-2

-1.5

-1

-0.5

0

0.5

1

1.5

2

2.5

3

Y vs. Time

t(s)

y(m

)

Period T

Amplitude A

Pulse y x = 4

x21

Pulse y x = 4

x21

-5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 150

0.250.5

0.751

1.251.5

1.752

2.252.5

2.753

3.253.5

3.754

Y vs. X

x(m)

y(m

)

Pulse y x = 4

x21

-5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 150

0.250.5

0.751

1.251.5

1.752

2.252.5

2.753

3.253.5

3.754

Y vs. X

x(m)

y(m

)

Now have pulse move at a constant velocity of 3 m/s.

Pulse y x = 4

x21Now have pulse move at a constant velocity of 3 m/s in the positive x direction.

If moving in positive x direction:

Replace x with x-vt

If moving in negative x direction:

Replace x with x+vt

Pulse y x = 4

x21Now have pulse move at a constant velocity of 3 m/s in the positive x direction.

If moving in positive x direction:

Replace x with x-vt

y x , t = 4

x−vt 21= 4

x−3t21

Pulse

Now have pulse move at a constant velocity of 3 m/s in the positive x direction.

y x , t = 4

x−vt 21= 4

x−3t 21

-5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 150

0.250.5

0.751

1.251.5

1.752

2.252.5

2.753

3.253.5

3.754

Y(x,t) vs X

x(m)

y(m

)

t= 0.0 s

Pulse

Now have pulse move at a constant velocity of 3 m/s in the positive x direction.

y x , t = 4

x−vt 21= 4

x−3t 21

t= 2.0 s

-5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 150

0.250.5

0.751

1.251.5

1.752

2.252.5

2.753

3.253.5

3.754

Y(x,t) vs X

x(m)

y(m

)

∆x=6m

Pulse

Now have pulse move at a constant velocity of 3 m/s in the positive x direction.

y x , t = 4

x−vt 21= 4

x−3t 21

t= 3.0 s

-5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 150

0.250.5

0.751

1.251.5

1.752

2.252.5

2.753

3.253.5

3.754

Y(x,t) vs X

x(m)

y(m

)

∆x=9m