2. Young’s Modulus
Spring F = -k x
What happens to “k” if cut spring in half?
1) decreases 2) same 3) increases
k is inversely proportional to length!
Define
Deformación unitaria = ∆L / L
Esfuerzo = F/A
Now
Esfuerzo = [Y] Deformación unitaria
F/A = Y ∆L/L
k = Y A/L from F = k x
Y (Young’s Modules) independent of L
Physics 101: Lecture 19, Pg 2
3. Simple Harmonic Motion
Vibrations
Vocal cords when singing/speaking
String/rubber band
Simple Harmonic Motion
Restoring force proportional to displacement
Springs F = -kx
Physics 101: Lecture 19, Pg 3
4. Ejemplos clásicos de osciladores armónicos
El péndulo El sistema
simple masa resorte
L
T
m
x
mg
5. Springs
Hooke’s Law: The force exerted by a spring is
proportional to the distance the spring is stretched
or compressed from its relaxed position.
FX = -k x Where x is the displacement from
the relaxed position and k is the
constant of proportionality.
relaxed position
FX = 0
x
x=0
Physics 101: Lecture 19, Pg 5
6. Springs
Hooke’s Law: The force exerted by a spring is
proportional to the distance the spring is stretched
or compressed from its relaxed position.
FX = -k x Where x is the displacement from
the relaxed position and k is the
constant of proportionality.
relaxed position
FX = -kx > 0
x
x<0
x=0
Physics 101: Lecture 19, Pg 6
7. Springs ACT
Hooke’s Law: The force exerted by a spring is proportional
to the distance the spring is stretched or compressed from its
relaxed position.
FX = -k x Where x is the displacement from
the relaxed position and k is the
constant of proportionality.
What is force of spring when it is stretched as shown
below.
A) F > 0 B) F = 0 C) F < 0
relaxed position
FX = - kx < 0
x
x>0
x=0
Physics 101: Lecture 19, Pg 7
8. Spring ACT II
A mass on a spring oscillates back & forth with simple
harmonic motion of amplitude A. A plot of displacement (x)
versus time (t) is shown below. At what points during its
oscillation is the magnitude of the acceleration of the block
biggest?
1. When x = +A or -A (i.e. maximum displacement)
2. When x = 0 (i.e. zero displacement)
3. The acceleration of the mass is constant
F=ma
La aceleración del bloque x
es máxima cuando la +A
fuerza que actúa sobre él
es máxima t
-A Physics 101: Lecture 19, Pg 8
9. Potential Energy in Spring
Force of spring is Conservative
F = -k x Force
W = Fx/2
F
W = -1/2 k x2
work
x x
Work done only depends on initial and final
position
Define Potential Energy Uspring = ½ k x2
El trabajo que se realiza para deformar el resorte
queda almacenado en él en forma de energía potencial.
Physics 101: Lecture 19, Pg 9
10. NOMENCLATURA
•Periodo (T) es el tiempo que un objeto tarda en dar una
vuelta o revolución.
•Frecuencia (f) es el número de vueltas o revoluciones que
un objeto da en un determinado intervalo de tiempo
•Frecuencia angular (ω)
1
f = ⇒( Hz )
T
Angulo 1Re volucion 2π
ω= = = = 2π f , (rad / s )
Tiempo 1Periodo T
2π
ω= = 2π f
T
11. En el movimiento armónico simple la fuerza que
acelera el objeto NO es constante y se la llama
fuerza restauradora
El movimiento armónico simple es un
movimiento con aceleración variable
12. En el M.A.S, sin fricción, la Energía Mecánica se
Conserva
La amplitud (A) es la máxima elongación (x)
Ver animación
13. El bloque es llevado a esta posición
inicial (x=A) y se lo suelta
Máxima elongación: Energía
cinética cero y Energía potencial
elástica máxima
Elongación cero: Energía
cinética máxima y Energía
potencial elástica cero
Máxima compresión:
Energía cinética cero y
Energía potencial elástica
máxima
14. La energía mecánica inicial se reparte entre
energía cinética y energía potencial elástica
Si no hay disipación de energía, la suma
es siempre una constante !
Ver animación
15. ***Energy in SHM***
A mass is attached to a spring and set to
motion. The maximum displacement is x=A
ΣWnc = ∆K + ∆U
0 = ∆K + ∆U or Energy U+K is
constant!
Energy = ½ k x2 + ½ m v2
At maximum displacement x=A, v = 0 PES
Energy = ½ k A2 + 0
At zero displacement x = 0
Energy = 0 + ½ mvm2
Since Total Energy is same
½ k A2 = ½ m vm2 x
0
k
vm = A m
m
x=0 x
Physics 101: Lecture 19, Pg 15
16. En cualquier instante, la suma de K y U es
una constante e igual al valor inicial de
energía que se le dio al sistema.
17. Preflight 3+4
A mass on a spring oscillates back & forth with simple
harmonic motion of amplitude A. A plot of displacement (x)
versus time (t) is shown below. At what points during its
oscillation is the total energy (K+U) of the mass and spring a
maximum? (Ignore gravity).
1. When x = +A or -A (i.e. maximum displacement)
2. When x = 0 (i.e. zero displacement)
3. The energy of the system is constant.
“When the kinetic energy is at a
minimum, the potential energy is at a
maximum and vice-versa.”
x
+A
CONSERVATION OF ENERGY
IS SACRED!!!!
t
-A Physics 101: Lecture 19, Pg 17
18. Preflight 1+2
A mass on a spring oscillates back & forth with simple
harmonic motion of amplitude A. A plot of displacement (x)
versus time (t) is shown below. At what points during its
oscillation is the speed of the block biggest?
1. When x = +A or -A (i.e. maximum displacement)
2. When x = 0 (i.e. zero displacement)
3. The speed of the mass is constant
“well it isn’t constant, and it is zero at the
maximums, so the zero position is the
only other choice”
x
+A
t
-A Physics 101: Lecture 19, Pg 18
19. SHM ACT
A spring oscillates back and forth on a frictionless horizontal
surface. A camera takes pictures of the position every 1/10th of a
second. Which plot best shows the positions of the mass.
1
EndPoint Equilibrium EndPoint
2
EndPoint Equilibrium EndPoint
3
EndPoint Equilibrium EndPoint
Physics 101: Lecture 19, Pg 19
20. Un bloque de 40 kg de masa se ata a un resorte
horizontal de constante elástica 500 N/m. Si la masa
descansa sobre una superficie horizontal sin fricción,
¿cuál es la energía total de este sistema cuando se
pone en movimiento armónico simple al desplazarlo
una distancia de 0.2 metros?
A) 10 J
B) 20 J
C) 50 J
D) 4 000 J
E) 100 000 J
21. Un resorte comprimido tiene una energía
potencial de 16 J. ¿Cuál es la máxima rapidez
que se puede impartir a un bloque de 2.0 kg de
masa ?
A. 2.8 m/s
B. 4.0 m/s
C. 5.6 m/s
D. 8.0 m/s
E. 16 m/s
24. What does moving in a circle have to do with
moving back & forth in a straight line ??
x = R cos θ = R cos (ωt)
since θ = ω t Movie
x x
1 1
2 8
R 2 8
θ
3 R 3 7
y 0 θ
7 π π 3π
2 2
4 6 -R 4 6
5 5
Physics 101: Lecture 19, Pg 24
27. La posición (x) en función del tiempo (t)
x = A cos θ
θ = ωt
x = A cos ωt
t =0 ⇒x = A
x = A cos ωt
T
t = ⇒x =0
4
T
t = ⇒ x = −A
2
28. La velocidad (v) en función del tiempo (t)
v =− o senθ
V
θ =ωt
v =− o senω
V t
Vo =vmax ima = Aω
t = 0 ⇒v = 0
v = −ωAsenωt
T
t = ⇒= Vo
v −
4
T
t = ⇒=
v 0
2
3
t = T ⇒v = Vo
4
29. La aceleración (a) en función del tiempo (t)
a = −ac cos θ
θ = ωt
a = −ac cos ωt
ac = amax ima = Aω2
2
t = 0 ⇒ a = −ω A
a = −ω A cos ωt
2
T
t = ⇒ =0
a
4
T 2
t = ⇒a = ω A
2
30. Gráficas que representan la posición (x), velocidad (v) y
aceleración (a) de un objeto en M.A.S. En función del tiempo
Observe el desfasamiento entre los valores de la posición,
velocidad y aceleración
x = A cos ωt v = −ωAsenωt
a = −ω 2 A cos ωt
31. Q13.3
To the right is an x-t graph for an object in
simple harmonic motion.
Which of the graphs below correctly shows
the velocity versus time for this object?
1. graph I
2. graph II
3. graph III
4. graph IV
32. Q13.2
This is an x-t graph for
an object in simple
harmonic motion.
At which of the following times does the object have
the most negative acceleration ax?
1. t = T/4
2. t = T/2
3. t = 3T/4
4. t = T
33. Q13.4
To the right is an x-t graph for an object in
simple harmonic motion.
Which of the graphs below correctly shows
the acceleration versus time for this object?
1. graph I
2. graph II
3. graph III
4. graph IV
34. Simple Harmonic Motion:
x(t) = [A]cos(ωt) x(t) = [A]sin(ωt)
v(t) = -[Aω]sin(ωt) OR v(t) = [Aω]cos(ωt)
a(t) = -[Aω2]cos(ωt) a(t) = -[Aω2]sin(ωt)
xmax = A Period = T (seconds per cycle)
Frequency = f = 1/T (cycles per second)
vmax = Aω
Angular frequency = ω = 2πf = 2π/T
amax = Aω 2
For spring: ω2 = k/m
Physics 101: Lecture 19, Pg 34
35. Example
A 3 kg mass is attached to a spring (k=24 N/m). It is
stretched 5 cm. At time t=0 it is released and oscillates.
Which equation describes the position as a function of
time x(t) =
A) 5 sin(ωt) B) 5 cos(ωt) C) 24 sin(ωt)
D) 24 cos(ωt) E) -24 cos(ωt)
We are told at t=0, x = +5 cm. x(t) = 5 cos(ωt) only
one that works.
Physics 101: Lecture 19, Pg 35
36. Example
A 3 kg mass is attached to a spring (k=24 N/m). It is
stretched 5 cm. At time t=0 it is released and oscillates.
What is the total energy of the block spring system?
A) 0.03 J B) .05 J C) .08 J
E=U+K
At t=0, x = 5 cm and v=0:
E = ½ k x2 + 0
= ½ (24 N/m) (5 cm)2
= 0.03 J
Physics 101: Lecture 19, Pg 36
37. Example
A 3 kg mass is attached to a spring (k=24 N/m). It is
stretched 5 cm. At time t=0 it is released and oscillates.
What is the maximum speed of the block?
A) .45 m/s B) .23 m/s C) .14 m/s
E=U+K
When x = 0, maximum speed:
E = ½ m v2 + 0
.03 = ½ 3 kg v2
v = .14 m/s
Physics 101: Lecture 19, Pg 37
38. Example
A 3 kg mass is attached to a spring (k=24 N/m). It is
stretched 5 cm. At time t=0 it is released and oscillates.
How long does it take for the block to return to x=+5cm?
A) 1.4 s B) 2.2 s C) 3.5 s
ω = sqrt(k/m)
= sqrt(24/3)
= 2.83 radians/sec
Returns to original position after 2 π radians
T = 2 π / ω = 6.28 / 2.83 = 2.2 seconds
Physics 101: Lecture 19, Pg 38
39. Summary
Springs
F = -kx
U = ½ k x2
ω = sqrt(k/m)
Simple Harmonic Motion
Occurs when have linear restoring force F= -kx
x(t) = [A] cos(ωt) or [A] sin(ωt)
v(t) = -[Aω] sin(ωt) or [Aω] cos(ωt)
a(t) = -[Aω2] cos(ωt) or -[Aω2] sin(ωt)
Physics 101: Lecture 19, Pg 39
40. Review Energy in SHM
A mass is attached to a spring and set to motion.
The maximum displacement is x=A
Energy = U + K = constant!
= ½ k x 2 + ½ m v2
At maximum displacement x=A, v = 0
PES
Energy = ½ k A2 + 0
At zero displacement x = 0
Energy = 0 + ½ mvm2
½ k A2 = ½ m vm2
x
0
vm = sqrt(k/m) A
Analogy w/ marble in bowl m
x=0 x
Physics 101: Lecture 19, Pg 40
41. Kinetic Energy ACT
In Case 1 a mass on a spring oscillates back and forth. In Case
2, the mass is doubled but the spring and the amplitude of the
oscillation is the same as in Case 1.
In which case is the maximum kinetic energy of the mass the
biggest?
A. Case 1
B. Case 2
C. Same
Physics 101: Lecture 19, Pg 41
42. Potential Energy ACT
In Case 1 a mass on a spring oscillates back and forth. In Case
2, the mass is doubled but the spring and the amplitude of the
oscillation is the same as in Case 1.
In which case is the maximum potential energy of the mass
and spring the biggest?
A. Case 1
B. Case 2
C. Same
Look at time of maximum displacement x = A
Energy = ½ k A2 + 0 Same for both!
Physics 101: Lecture 19, Pg 42
43. Kinetic Energy ACT
PE = 0
PE = /2kx
1 2
same KE = KEMAX
KE = 0 for both
same
for both
x=-A x=0 x=+A x=-A x=0 x=+A
A) Case 1
B) Case 2
C) Same
Physics 101: Lecture 19, Pg 43
44. Q13.6
This is an x-t graph for
an object connected to
a spring and moving in
simple harmonic
motion.
At which of the following times is the kinetic energy
of the object the greatest?
1. t = T/8
2. t = T/4
3. t = 3T/8
4. t = T/2
5. More than one of the above
45. Velocity ACT
In Case 1 a mass on a spring oscillates back and forth.
In Case 2, the mass is doubled but the spring and the
amplitude of the oscillation is the same as in Case 1.
Which case has the largest maximum velocity?
1. Case 1
2. Case 2
3. Same
Same maximum Kinetic Energy
K = ½ m v2 smaller mass requires larger v
Physics 101: Lecture 19, Pg 45
46. Period T of a Spring
Simple Harmonic Oscillator
ω=2πf =2π/T
x(t) = [A] cos(ωt)
v(t) = -[Aω] sin(ωt) m
a(t) = -[Aω2] cos(ωt) x=0 x
For a Spring F = -kx
amax = (k/m) A k m
ω= T = 2π
Aω2 = (k/m) A m k
Physics 101: Lecture 19, Pg 46
47. Period ACT
If the amplitude of the oscillation (same block and same spring)
is doubled, how would the period of the oscillation change? (The
period is the time it takes to make one complete oscillation)
A. The period of the oscillation would double.
B. The period of the oscillation would be halved
C. The period of the oscillation would stay the same
k m
ω= T = 2π
m k
x
+2A
t
-2A
Physics 101: Lecture 19, Pg 47
48. Vertical Mass and Spring
If we include gravity, there are two forces acting on mass. With
mass, new equilibrium position has spring stretched d
ΣF = kd – mg = 0
d = mg/k x
+A
Let this point be y = 0
ΣF = k(d-y) – mg t
-A
= -k y
Same as horizontal! SHO
New equilibrium position y=-d
Physics 101: Lecture 19, Pg 48
49. Vertical Spring ACT
If the springs were vertical, and stretched the same distance d from
their equilibrium position and then released, which would have the
largest maximum kinetic energy?
1) M 2) 2M 3) Same
PE = 1/2k y2
Y=0
PE = 1/2k y2
Just before being released, v=0 y=d
Etot = 0 + ½ k d2 Same total energy for both
When pass through equilibrium all of this energy will Y=0
be kinetic energy again same for both! Physics 101: Lecture 19, Pg 49
50. Q13.1
An object on the end of a spring is oscillating in simple harmonic
motion.
If the amplitude of oscillation is doubled,
1. the oscillation period and the object’s maximum speed
both double
2. the oscillation period remains the same and the
object’s maximum speed doubles
3. the oscillation period and the object’s maximum speed
both remain the same
4. the oscillation period doubles and the object’s
maximum speed remains the same
5. the oscillation period remains the same and the
object’s maximum speed increases by a factor of 21/2
51. Una masa m; atada a un resorte horizontal de constante
k; es puesto en movimiento armónico simple. El máximo
desplazamiento desde su posición de equilibrio es A.
¿Cuál es la magnitud de la velocidad de la masa en el
momento que pasa por su posición de equilibrio?
52. Un bloque de 1 kg se une a un resorte sobre una superficie sin
fricción como se indica en la figura. La posición del bloque en
el transcurso del tiempo está dada por la ecuación:
x = 0,05 sen 50π t (m). Determine:
a) La constante k del resorte
b) La rapidez máxima del bloque
c) La aceleración máxima del bloque
d) La posición del bloque a los dos segundos.
54. Energía del Oscilador Armónico Simple
1 2
Eo = kA
2
1 2 1 2
E (t ) = mv + kx
2 2
1 1
E (t ) = m( A ω sen ωt ) + k ( A2 cos 2 ωt )
2 2 2
2 2
1 2 k 1 2
E (t ) = mA ÷ sen ωt + kA cos 2 ωt
2
2 m 2
1 2
E (t ) = kA ( sen 2ωt + cos 2 ωt )
La energía mecánica 2
total del oscilador A.S. 1 2
E(t) = kA
se mantiene constante!! 2
55. Q13.5
This is an x-t graph for
an object connected to
a spring and moving in
simple harmonic
motion.
At which of the following times is the potential energy
of the spring the greatest?
1. t = T/8
2. t = T/4
3. t = 3T/8
4. t = T/2
5. More than one of the above
56. EL PÉNDULO SIMPLE
(frecuencia natural de oscilación)
Fmax ima = Mamax ima
Mgsenθ = M ω x 2
gsenθ = ω 2 x
Si, θ ≤ 10o
gθ = ω x 2
x =θL
g
ω= Period does not depend on x, or m!
L
Ver animación Ver animación
57. Pendulum Motion
For small angles
T = mg
Tx = -mg (x/L) Note: F proportional to x!
Σ F x = m ax
-mg (x/L) = m ax
ax = -(g/L) x L
Recall for SHO a = -ω2 x T
ω = sqrt(g/L)
m
T = 2 π sqrt(L/g) x
Period does not depend on A, or m! mg
Physics 101: Lecture 19, Pg 57
58. Preflight 1
Suppose a grandfather clock (a simple pendulum) runs
slow. In order to make it run on time you should:
1. Make the pendulum shorter
2. Make the pendulum longer
g
ω=
L
2π L
T= = 2π
ω g
Physics 101: Lecture 19, Pg 58
59. ACT
A pendulum is hanging vertically from the ceiling of an
elevator. Initially the elevator is at rest and the period of the
pendulum is T. Now the pendulum accelerates upward. The
period of the pendulum will now be. If you are accelerating
upward your weight is the same as if g had
1. increased
2. same
3. decreased
“Effective g” is larger when accelerating upward
(you feel heavier)
Physics 101: Lecture 19, Pg 59
60. Elevator ACT
A pendulum is hanging vertically from the ceiling of an
elevator. Initially the elevator is at rest and the period of the
pendulum is T. Now the pendulum accelerates upward. The
period of the pendulum will now be
1. greater than T
2. equal to T
g
3. less than T ω=
L
2π L
T= = 2π
ω g
“Effective g” is larger when accelerating upward
(you feel heavier)
Physics 101: Lecture 19, Pg 60
61. Preflight
Imagine you have been kidnapped by space invaders and are being held
prisoner in a room with no windows. All you have is a cheap digital
wristwatch and a pair of shoes (including shoelaces of known length).
Explain how you might figure out whether this room is on the earth or on
the moon
g
ω=
L
2π L
T= = 2π
ω g
L
g = ( 2π ) 2
2
T
Physics 101: Lecture 19, Pg 61
62. Q13.7
A simple pendulum consists of a point mass
suspended by a massless, unstretchable string.
If the mass is doubled while the length of the string
remains the same, the period of the pendulum
1. becomes 4 times greater
2. becomes twice as great
3. becomes 21/2 times greater
4. remains unchanged
5. decreases
63. Un péndulo simple de masa m y longitud L tiene un
periodo de oscilación T a una amplitud angular de θ
= 5o medida desde su posición de equilibrio. Si la
amplitud es cambiada a 10o y todos los demás
parámetros permanecen constantes, el nuevo valor del
periodo sería aproximadamente:
64. La energía de un péndulo simple de longitud
L y masa M que oscila con una amplitud A es
a) independiente de M
b) independiente de L
c) independiente de A
d) independiente de A, L, M
65. EN LAS SIGUIENTES 3 PREGUNTAS EL PÉNDULO
A TIENE UNA MASA MA Y LONGITUD LA. EL
PÉNDULO B TIENE UNA MASA MB Y UNA
LONGITUD LB
Si LA = LB y MA = 2MB, y las amplitudes de vibración son
iguales, entonces.
a) TA = TB y son iguales las energías de los péndulos.
b) TA = ½ TB y las energías de los péndulos son iguales.
c) TA = TB y A tiene mayor energía que B.
d) TA = TB y A tiene menor energía que B.
66. SI LA = 2LB, Y MA = MB,Y A DEMÁS LOS DOS
PÉNDULOS TIENEN IGUAL ENERGÍA DE
VIBRACIÓN, ENTONCES
a) sus amplitudes de movimiento angular son iguales
b) sus periodos de movimiento son iguales
c) B tiene mayor amplitud angular que A
d) ninguna de las afirmaciones anteriores es correcta
67. SI EL PÉNDULO A TIENE EL DOBLE DE PERIODO QUE
EL PÉNDULO B, ENTONCES
a) LA = 2LB y MA = 2 MB
b) LA = 2 LB, y las masas no cuentan
c) LA = 2 LB y MA = MB/2
d) Ninguna de las afirmaciones anteriores es correcta
68. PROBLEMA:
La aceleración de la gravedad varía ligeramente sobre la
superficie de nuestro planeta. Si un péndulo tiene un periodo de
3.000 s en un lugar donde g = 9.803 m/s2 y un periodo de
3.0024 s en otro lugar, ¿cuál es el valor de g en este último
lugar?
69. PROBLEMA
Un péndulo simple que
oscila con un periodo de
0.60 s en la Tierra es lleva a
la Luna. ¿Cuál será allí su
periodo?
70. Un péndulo simple tiene una masa de 0.25 kg y
una longitud de 1.00 m. Se desplaza por un ángulo
de 15o con la vertical y luego se suelta. Determine:
a) La rapidez máxima
b) La aceleración angular máxima
c) La fuerza restauradora máxima