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Movimiento armónico Simple
 

Movimiento armónico Simple

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    Movimiento armónico Simple Movimiento armónico Simple Presentation Transcript

    • Elasticity and Oscillations Physics 101: Lecture 19, Pg 1
    • Young’s ModulusSpring F = -k x What happens to “k” if cut spring in half? 1) decreases 2) same 3) increasesk is inversely proportional to length!Define Deformación unitaria = ∆L / L Esfuerzo = F/ANow Esfuerzo = [Y] Deformación unitaria F/A = Y ∆L/L k = Y A/L from F = k xY (Young’s Modules) independent of L Physics 101: Lecture 19, Pg 2
    • Simple Harmonic MotionVibrations Vocal cords when singing/speaking String/rubber bandSimple Harmonic Motion Restoring force proportional to displacement Springs F = -kx Physics 101: Lecture 19, Pg 3
    • Ejemplos clásicos de osciladores armónicosEl péndulo El sistema simple masa resorte L T m x mg
    • SpringsHooke’s Law: The force exerted by a spring isproportional to the distance the spring is stretchedor 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
    • SpringsHooke’s Law: The force exerted by a spring isproportional to the distance the spring is stretchedor 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
    • 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
    • Spring ACT IIA mass on a spring oscillates back & forth with simpleharmonic motion of amplitude A. A plot of displacement (x)versus time (t) is shown below. At what points during itsoscillation is the magnitude of the acceleration of the blockbiggest?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 constantF=maLa aceleración del bloque xes máxima cuando la +Afuerza que actúa sobre éles máxima t -A Physics 101: Lecture 19, Pg 8
    • 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 x2El trabajo que se realiza para deformar el resortequeda almacenado en él en forma de energía potencial. Physics 101: Lecture 19, Pg 9
    • NOMENCLATURA•Periodo (T) es el tiempo que un objeto tarda en dar unavuelta o revolución.•Frecuencia (f) es el número de vueltas o revoluciones queun 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
    • 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
    • En el M.A.S, sin fricción, la Energía Mecánica se Conserva La amplitud (A) es la máxima elongación (x) Ver animación
    • El bloque es llevado a esta posición inicial (x=A) y se lo suelta Máxima elongación: Energíacinética cero y Energía potencial elástica máxima Elongación cero: Energíacinética máxima y Energía potencial elástica cero Máxima compresión: Energía cinética cero yEnergía potencial elástica máxima
    • La energía mecánica inicial se reparte entreenergía cinética y energía potencial elástica Si no hay disipación de energía, la suma es siempre una constante ! Ver animación
    • ***Energy in SHM***A mass is attached to a spring and set tomotion. 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
    • 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.
    • Preflight 3+4A mass on a spring oscillates back & forth with simpleharmonic motion of amplitude A. A plot of displacement (x)versus time (t) is shown below. At what points during itsoscillation is the total energy (K+U) of the mass and spring amaximum? (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 +ACONSERVATION OF ENERGY IS SACRED!!!! t -A Physics 101: Lecture 19, Pg 17
    • Preflight 1+2A mass on a spring oscillates back & forth with simpleharmonic motion of amplitude A. A plot of displacement (x)versus time (t) is shown below. At what points during itsoscillation 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
    • 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 EndPoint2 EndPoint Equilibrium EndPoint3 EndPoint Equilibrium EndPoint Physics 101: Lecture 19, Pg 19
    • Un bloque de 40 kg de masa se ata a un resortehorizontal de constante elástica 500 N/m. Si la masadescansa sobre una superficie horizontal sin fricción,¿cuál es la energía total de este sistema cuando sepone en movimiento armónico simple al desplazarlouna distancia de 0.2 metros?A) 10 JB) 20 JC) 50 JD) 4 000 JE) 100 000 J
    • 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/sB. 4.0 m/sC. 5.6 m/sD. 8.0 m/sE. 16 m/s
    • Springs and Simple Harmonic X=0 Motion X=A; v=0; a=-amax X=0; v=-vmax; a=0 X=-A; v=0; a=amax X=0; v=vmax; a=0 X=A; v=0; a=-amax X=-A X=A Physics 101: Lecture 19, Pg 22
    • RELACIÓN ENTRE EL MOVIMIENTO CIRCULAR Y EL MOVIMIENTO ARMÓNICO
    • 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 7y 0 θ 7 π π 3π 2 2 4 6 -R 4 6 5 5 Physics 101: Lecture 19, Pg 24
    • SHM and Circles Physics 101: Lecture 19, Pg 25
    • Modelo matemático deloscilador armónico simple
    • La posición (x) en función del tiempo (t)x = A cos θθ = ωtx = A cos ωtt =0 ⇒x = A x = A cos ωt Tt = ⇒x =0 4 Tt = ⇒ x = −A 2
    • La velocidad (v) en función del tiempo (t)v =− o senθ Vθ =ωtv =− o senω V tVo =vmax ima = Aωt = 0 ⇒v = 0 v = −ωAsenωt Tt = ⇒= Vo v − 4 Tt = ⇒= v 0 2 3t = T ⇒v = Vo 4
    • La aceleración (a) en función del tiempo (t)a = −ac cos θθ = ωta = −ac cos ωtac = amax ima = Aω2 2t = 0 ⇒ a = −ω A a = −ω A cos ωt 2 Tt = ⇒ =0 a 4 T 2t = ⇒a = ω A 2
    • Gráficas que representan la posición (x), velocidad (v) yaceleración (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
    • Q13.3To the right is an x-t graph for an object insimple harmonic motion.Which of the graphs below correctly showsthe velocity versus time for this object? 1. graph I 2. graph II 3. graph III 4. graph IV
    • 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
    • Q13.4To the right is an x-t graph for an object insimple harmonic motion.Which of the graphs below correctly showsthe acceleration versus time for this object? 1. graph I 2. graph II 3. graph III 4. graph IV
    • 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/Tamax = Aω 2 For spring: ω2 = k/m Physics 101: Lecture 19, Pg 34
    • ExampleA 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
    • ExampleA 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
    • ExampleA 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
    • ExampleA 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
    • SummarySprings 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
    • Review Energy in SHMA 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
    • Kinetic Energy ACTIn Case 1 a mass on a spring oscillates back and forth. In Case2, the mass is doubled but the spring and the amplitude of theoscillation is the same as in Case 1.In which case is the maximum kinetic energy of the mass thebiggest?A. Case 1B. Case 2C. Same Physics 101: Lecture 19, Pg 41
    • Potential Energy ACTIn Case 1 a mass on a spring oscillates back and forth. In Case2, the mass is doubled but the spring and the amplitude of theoscillation is the same as in Case 1.In which case is the maximum potential energy of the massand spring the biggest?A. Case 1B. Case 2C. SameLook at time of maximum displacement x = AEnergy = ½ k A2 + 0 Same for both! Physics 101: Lecture 19, Pg 42
    • Kinetic Energy ACT PE = 0 PE = /2kx 1 2 same KE = KEMAX KE = 0 for both same for bothx=-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
    • 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
    • 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. SameSame maximum Kinetic EnergyK = ½ m v2 smaller mass requires larger v Physics 101: Lecture 19, Pg 45
    • Period T of a SpringSimple Harmonic Oscillator ω=2πf =2π/T x(t) = [A] cos(ωt) v(t) = -[Aω] sin(ωt) m a(t) = -[Aω2] cos(ωt) x=0 xFor 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
    • Period ACTIf the amplitude of the oscillation (same block and same spring)is doubled, how would the period of the oscillation change? (Theperiod 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 halvedC. 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
    • Vertical Mass and SpringIf we include gravity, there are two forces acting on mass. Withmass, 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
    • Vertical Spring ACTIf the springs were vertical, and stretched the same distance d fromtheir equilibrium position and then released, which would have thelargest 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
    • 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
    • Una masa m; atada a un resorte horizontal de constantek; es puesto en movimiento armónico simple. El máximodesplazamiento desde su posición de equilibrio es A.¿Cuál es la magnitud de la velocidad de la masa en elmomento que pasa por su posición de equilibrio?
    • 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 resorteb) La rapidez máxima del bloquec) La aceleración máxima del bloqued) La posición del bloque a los dos segundos.
    • Energía potencial elástica del sistema masa resorte
    • 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 2total del oscilador A.S. 1 2 E(t) = kAse mantiene constante!! 2
    • 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
    • EL PÉNDULO SIMPLE (frecuencia natural de oscilación)Fmax ima = Mamax imaMgsenθ = M ω x 2 gsenθ = ω 2 x Si, θ ≤ 10o gθ = ω x 2 x =θL gω= Period does not depend on x, or m! L Ver animación Ver animación
    • Pendulum MotionFor 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
    • Preflight 1Suppose a grandfather clock (a simple pendulum) runsslow. In order to make it run on time you should:1. Make the pendulum shorter2. Make the pendulum longer g ω= L 2π L T= = 2π ω g Physics 101: Lecture 19, Pg 58
    • ACTA pendulum is hanging vertically from the ceiling of anelevator. Initially the elevator is at rest and the period of thependulum is T. Now the pendulum accelerates upward. Theperiod of the pendulum will now be. If you are acceleratingupward your weight is the same as if g had1. increased2. same3. decreased “Effective g” is larger when accelerating upward (you feel heavier) Physics 101: Lecture 19, Pg 59
    • Elevator ACTA pendulum is hanging vertically from the ceiling of anelevator. Initially the elevator is at rest and the period of thependulum is T. Now the pendulum accelerates upward. Theperiod of the pendulum will now be1. greater than T2. equal to T g3. 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
    • PreflightImagine you have been kidnapped by space invaders and are being heldprisoner in a room with no windows. All you have is a cheap digitalwristwatch and a pair of shoes (including shoelaces of known length).Explain how you might figure out whether this room is on the earth or onthe moon g ω= L 2π L T= = 2π ω g L g = ( 2π ) 2 2 T Physics 101: Lecture 19, Pg 61
    • 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
    • Un péndulo simple de masa m y longitud L tiene unperiodo de oscilación T a una amplitud angular de θ= 5o medida desde su posición de equilibrio. Si laamplitud es cambiada a 10o y todos los demásparámetros permanecen constantes, el nuevo valor delperiodo sería aproximadamente:
    • La energía de un péndulo simple de longitudL y masa M que oscila con una amplitud A esa) independiente de Mb) independiente de Lc) independiente de Ad) independiente de A, L, M
    • EN LAS SIGUIENTES 3 PREGUNTAS EL PÉNDULOA TIENE UNA MASA MA Y LONGITUD LA. ELPÉNDULO B TIENE UNA MASA MB Y UNALONGITUD LBSi LA = LB y MA = 2MB, y las amplitudes de vibración soniguales, 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.
    • SI LA = 2LB, Y MA = MB,Y A DEMÁS LOS DOSPÉNDULOS TIENEN IGUAL ENERGÍA DEVIBRACIÓN, ENTONCESa) sus amplitudes de movimiento angular son igualesb) sus periodos de movimiento son igualesc) B tiene mayor amplitud angular que Ad) ninguna de las afirmaciones anteriores es correcta
    • SI EL PÉNDULO A TIENE EL DOBLE DE PERIODO QUEEL PÉNDULO B, ENTONCESa) LA = 2LB y MA = 2 MBb) LA = 2 LB, y las masas no cuentanc) LA = 2 LB y MA = MB/2d) Ninguna de las afirmaciones anteriores es correcta
    • PROBLEMA:La aceleración de la gravedad varía ligeramente sobre lasuperficie de nuestro planeta. Si un péndulo tiene un periodo de3.000 s en un lugar donde g = 9.803 m/s2 y un periodo de3.0024 s en otro lugar, ¿cuál es el valor de g en este últimolugar?
    • PROBLEMAUn péndulo simple queoscila con un periodo de0.60 s en la Tierra es lleva ala Luna. ¿Cuál será allí superiodo?
    • Un péndulo simple tiene una masa de 0.25 kg yuna longitud de 1.00 m. Se desplaza por un ángulode 15o con la vertical y luego se suelta. Determine:a) La rapidez máximab) La aceleración angular máximac) La fuerza restauradora máxima