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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