Chapter 14 Oscillations Oscillations Introductory Terminology Simple Harmonic Motion: Kinematics Energy Examples of Simple Harmonic Oscillators Damped and Forced Oscillations. Resonance.
Periodic Motion Oscillations If an object oscillates back and forth over the same path, each cycle taking the same amount of time, the motion is called periodic The mass and spring system is a useful model for a periodic system performing the simplest type of oscillation: Physical situation: Consider a mass attached to a very light spring, and allowed to oscillate on a frictionless surface There is a point where the spring is neither stretched nor compressed; this is the equilibrium position We measure displacement x from that point The force exerted by the spring is proportional to the displacement as given by Hooke s law: F = kx where k is the spring constant. The minus sign on the force indicates that it is a restoring force because it is everywhere directed such that it attempts to restore the mass to its equilibrium position. Let s study the kinematics of this system equilibrium
Periodic Motion Terminology and Comments Simple oscillations can be described using concepts and physical quantities exemplified on the right using the mass-spring system: A cycle is a full to-and-fro motion; the figure shows a complete cycle Speed v and acceleration a of oscillation: time dependent kinematic quantities describing the oscillatory motion Displacement x, measured from the equilibrium point: when the displacement is zero (equilibrium point), the restoring force and acceleration are zero, a = 0, and the speed is maximum v = v max Amplitude A is the maximum displacement: when the amplitude is reached, the restoring force and acceleration are maximum, a = a max, and the speed is zero v = 0 Period T is the time required to complete one cycle Angular frequency ω is given by the number of cycles completed per second or frequency f via ω = 2πf F F F = 0 F = 0 F
Simple Harmonic Motion Definition Def: Any oscillatory system driven by a restoring force proportional to the negative of the displacement is in Simple Harmonic Motion (SHM), and is called a Simple Harmonic Oscillator (SHO): Comments: F = kx So, we see that the mass-spring system performs simple harmonic motion by the very nature of its restoring force: the elastic force in the spring However, simple harmonic oscillators do not necessarily have to oscillate in a line, so x in the definition is not necessarily a linear displacement. For instance, oscillators can swing along arcs of circle: in these cases the displacement is an angle and the force must be replaced by a torque Ex: a) the spiral spring of a mechanical watch b) various types of pendulums simple pendulum: physical pendulum: CM
Simple Harmonic Motion Kinematics using phasor projection In order to derive the time dependency of x, v and a in the case of SHM, consider a particle moving in a circle of radius A with constant angular speed ω: if the shadow of the particle is projected onto a screen perpendicular on the circle, the motion of the shadow is simple harmonic The frequency and period of the oscillation are the same as the frequency and period of rotation, and the amplitude is equal to the radius of the circle Therefore, we can find the position, velocity and acceleration of the SHO by writing the components along the diameter for the position (here called phasor), velocity and acceleration of the revolving particle Displacement Velocity v Acceleration r θ A A θ x= Acosθ vx = v sinθ ω ω ω x Acos( ωt ϕ) = + v = Aω sin( ωt+ ϕ) a 2 cos x = Aω ( ωt+ ϕ) x v Aω = 2 A θ a a = a cosθ Since the particle revolves uniformly, we have θ = ωt, so, assuming that the phasor had an initial angle φ called initial phase, the kinematics of the simple harmonic motion along an x-axis is given by: a = Aω x
Simple Harmonic Motion Sinusoidal kinematics Alternatively, if a particle performs SHM, its kinematics can be derived from the definition of the force driving the motion and Newton s 2 nd Law: nd SHM ewton' s 2 2 k 2 d x denote ω= m d x 2 x = = x = + ω = 0 2 2 F kx ma m x dt dt The resulting equation is the SHO equation of motion which has solutions of the form ( ) = cos( ω + ) x t A t ϕ The initial phase is given by the position of the SHO at moment t = 0: 1 x ( 0 ) ϕ = cos A Hence, the other kinematic quantities can be derived immediately: v max dx vx t A t dt ( ) = = ω sin( ω + ϕ) a max dv ax t A t dt x ( ) = = ω 2 cos( ω + ϕ) x A A v max v max a max a max v φ = π/2 ( ω ) π x= Acos t+ T/2 3T/2 t T 2T φ = 0 x= Acos( ωt) t T 2T a t T 2T 2
Simple Harmonic Motion Isochrony of SHM On the previous slide we found an expression for the angular frequency ω of the SHO which can be combined with the generic definition of frequency to find the period: ω= k m 2π ω= T = 2π T Notice that the period depends only on the configuration of the oscillator, and it does not depend on the amplitude of oscillation: this property is called isochrony In the case of SHO, the mass and the force constant characteristics of the system completely determine the frequency and period of the system Particularly, notice that the frequency and period do not depend on the amplitude of the oscillator This means that, if the system mass-spring oscillates, oscillations with large amplitudes will be faster than oscillations with small amplitudes Ex: Isochrony was one of the first properties of simple harmonic oscillators that has been ever observed (we owe it to Galileo Galilei). It was subsequently used to build mechanical clocks, such as the clocks built by the famous physicist Christiaan Huygens m k
Simple Harmonic Motion Energy conservation A SHO is a conservative machine based on the cyclic reversible conversion of energy from one form into another. Let s look at the spring-mass oscillator: The total mechanical energy at an arbitrary displacement is given by: E= K+ U = mv + kx 1 2 1 2 2 2 At the two extremes of the motion (x = ±A) the energy is exclusively potential, while at the equilibrium point (x = 0), the energy is exclusively kinetic. Then, using the conservation of mechanical energy we can derive a relationship between the instantaneous SHO velocity and displacement: E= mv + kx = ka 1 2 1 2 1 2 2 2 2 We find thence a speed-position relation: v 2 k =± v where max v 2 max = A = A m 1 x A ω
Problems: 1. Mass-Spring Simple Harmonic Oscillator: A mass-spring SHO has mass m = 0.500 kg and an ideal spring with force constant k = 140 N/m. Find a) the period, b) the frequency, c) the angular frequency. 2. SHM solution check: Substitute the following equations, in which A, ω and β are constants, into the equation for SHM equation to see if they describe a SHM. a) x= Asin( ωt+ β) b) x 2 ω β = A t + 3. SHM given by an equation of motion: A particle of mass 1.50 kg is connected to a spring and oscillates with a time dependent displacement given by the equation Find a) the frequency and period 1 ( ) = ( 7.40 cm) cos( 4.16 s ) x t b) the force constant and maximum force c) the maximum speed and maximum energy e) the position, speed and acceleration of the mass at t = 1.00 s f) the kinetic energy and potential energy at t = 1.00 s t
Simple Harmonic Oscillators Types The specific nature of the constant k and the displacement x may be different, depending on the type of SHO considered: spring and oscillating mass: k stiffness constant, x linear coil spring: k torsion constant, x θ angular pendulums: k weight per length, x θ angular cm molecule vibrations: k energy per distance squared, x linear Steps in analyzing the oscillation: 1. identify the restoring cause (force or torque) and see if it can arranged to be proportional to the displacement 2. identify the force constant and the parameters of the oscillation: T, ω, f etc.
Pendulums The simple pendulum Periodic motion is not necessarily linear: the motion of a pendulum is also periodic, but it is better described using angular displacements For instance, a simple pendulum consists of a pointlike mass at the end of a unstretchable lightweight cord The mass is allowed to oscillate in the vertical plane, as described by the varying angular displacement θ We already know how to describe a SHO, so it makes sense to ask: Is this oscillator simple harmonic? We can easily identify the restoring force to be the tangential component of the weight F = mg sinθ which is proportional to sinθ and not to the displacement θ, as expected for a SHO. We conclude that, in general, the simple pendulum is not a simple harmonic oscillator However, note that it can be approximated as a SHO, when the angle of oscillation is small since sinθ ~ θ for small angles. Then the restoring force becomes F mgθ L θ restoring force mgsinθ T θ m mg
Pendulums The period and frequency of simple pendulum Therefore, for small angles, the simple pendulum can be approximated as SHO since the restoring force is approximately proportional to the displacement: F mg θ F mg x = k x x Lθ L denote k Hence, we can use the results that we obtained for a generic SHO to characterize the period of the respective oscillation: T m m = 2π = 2π k mg L Comments: T = 2π The relationship suggests that the simple pendulum is indeed isochrone The period of the same device will depend on the local gravitational acceleration g: it will be different on the Moon, in the water, or in an accelerated vehicle L g Ex: Already in the sixteenth century Galileo noticed that a chandelier oscillates with constant period. Later on he suggested using pendulums as time measuring devices
Pendulums The physical pendulum What if the mass of the pendulum arm cannot be neglected? In this case, we are dealing with a physical pendulum: a rigid body oscillating about a pivot In this case, the restoring cause is the torque of the weight acting in the center of mass We see again that, in general, the physical pendulum is not simple harmonic, since the torque is not proportional to sinθ not to the angular displacement θ : τ = mgd sinθ However, when the oscillation is small, it is almost harmonic: τ mgdθ From this expression for the restoring force and Newton s 2 nd law, one can derive the period of the physical pendulum in terms of the moment of inertia I of the rigid body with respect to the pivot, the distance L between the pivot and the cm, the mass m of body and the local gravitational acceleration: Ex: The expression reproduces the period of a simple pendulum when the rigid body reduces to a point mass located in the center of mass: T Pivot θ d = 2π cm θ mg I mgl I ml 2 L T = 2π = 2π = 2π mgd mgl g
Pendulums Angular SHM kinematics In the case of pendulums, it makes sense to consider oscillatory angular quantities rather than linear quantities In this case, the SHM equation takes an angular form. For instance, using the torque derived on the previous slide and Newton s 2 nd law for rotational motion, we get d θ d θ τ θ = α = + ω θ = 2 2 mg d I z I dt 2 dt 2 2 0 Ω θ d cm where the angular frequency is consistent with the period of the physical pendulum: ω= mgd I T = The solutions of this angular SHM equation are the angular position θ, velocity Ω, and acceleration α of the swinging pendulum: angular amplitude When the oscillations are small, we θ = Θ cos( ωt+ ϕ) can use a linear approximation which reduces the angular equations to the Ω= dθ dt= ωθ sin( ωt+ ϕ) linear SHM: ( ) 2 = dω dt= Θ cos t+ α ω ω ϕ 2π I mgd x θd A Θd
Problems: 4. Simple pendulum: You pull a simple pendulum of length L = 0.240 m to the side through an angle Θ = 3.50 and release it. a) How long does it take the pendulum to reach its highest speed? How long does it take if the pendulum is released at an angle of Θ/2? b) Write out the angular position, velocity and acceleration of this pendulum in terms of known quantities. 5. Finding g using a simple pendulum: After landing on an unfamiliar planet, a space explorer constructs a simple pendulum of length 50.0 cm. She finds that the pendulum makes 100 complete swings in 136 s. What is the value of g on this planet? 6. Physical pendulum: Two identical thin rods, each of mass m and length L, are joined at right angles to form an L-shaped object. This object is balanced on top of a sharp edge. If the object is displaced slightly, it oscillates. Find, the angular frequency of oscillation of the object. (The moment of inertia of a rod about one of its ends is ml 2 /3.) 7. Finding I using a physical pendulum: A 1.80-kg connecting rod from a car engine is pivoted about a horizontal knife edge as in the figure. The center of gravity of the rod was located by balancing and is 0.200 m from the pivot. When it is set into small amplitude oscillation, the rod makes 100 complete swings in 120 s. Calculate the moment of inertia of the rod about the rotation axis through the pivot.
Damped Oscillations Recall that the simple harmonic oscillator is an idealized model: in reality, oscillators are affected by dissipative forces slowing them down A SHO in the presence of a frictions performs a damped harmonic motion So, typical mechanical oscillation can be maintained only if they are helped, or driven by external forces. However, in many cases controlled damping can be useful Ex: automobile shock absorbers and earthquake protection for buildings If the damping is small, we can treat it as an envelope that modifies the un-damped oscillation, but if it is large, the motion no longer resembles an oscillation at all Accordingly, the damped oscillations can be classified based on the strength of the damping force: A: underdamping: there are a few small oscillations before the oscillator comes to rest. B: critical damping: this is the fastest way to get to equilibrium. C: overdamping: the system is slowed so much that it takes a long time to get to equilibrium.