Chapter 13: Oscillatory Motions

Transcription

1 Chapter 13: Oscillatory Motions Simple harmonic motion Spring and Hooe s law When a mass hanging from a spring and in equilibrium, the Newton s nd law says: Fy ma Fs Fg 0 Fs Fg This means the force due to the spring is equal to the force by gravity and opposite in direction when the spring is stretched. Hooe s law states that increasing the weight by equal amounts increases the stretch of the spring by equal amount. Therefore, the force due to the spring must be proportional to the stretch of the spring. x is deviation from the spring w/o weight r F s r x where is the spring constant. This is also true when the spring shrins y

2 Simple harmonic motion Simple harmonic motion (SHM) Let s study a motion of the mass m. When the mass is attached to the spring, the spring stretches by x 0. Then lift the mass by A and release it. Fy ma Fs Fg ma x x) mg ( 0 ma y y y The initial stretch is x 0 -x and from Hooe s law: Since in equilibrium F F 0 x0 mg, s g dv d dx d x x ma a ( / m) x and a dt dt dt dt d x x ω x Equation for SHM dt m

3 Simple harmonic motion Simple harmonic motion (SHM) (cont d) d x dt Solution: x ω x m x( Acos( ω t + φ) ω m phase constant : angular frequency (rad/s) x0 As x(0) Acosφ x0, φ arccos A Frequency : f ω / π, period : T 1/ Hz s f, amplitude: A dx( v( ωasin( ωt + φ), v( x) ± ω A x velocity dt dv( a( ω Acos( ωt + φ), a( x) ω x dt acceleration

4 Simple harmonic motion Simple harmonic motion (SHM) (cont d) Solution: x( Acos( ω t + φ) Acosφ fω/(π) What is SHM/SHO? t-φ/ω A simple harmonic motion is the motion of an oscillating system which satisfies the following condition: 1. Motion is about an equilibrium position at which point no net force acts on the system.. The restoring force is proportional to and oppositely directed to the displacement. 3. Motion is periodic. t0 ωω 0 ; ωω 0 ; ω3ω 0 By Dr. Dan Russell, Kettering University

5 Simple harmonic motion Connection between SHM and circular motion For an object in circular motion, the angular SHM! velocity is defined as, dθ ω θ ωt + φ dt The tangential velocity is related to the angular velocity : v rω The centripetal acceleration is also related to the angular velocity: Vector r is called a v ( ωr) a rω r r The position, velocity and acceleration of the object as a function of time are: x r cosθ x( v rω sinθ v( rω sin( ωt + φ) a ω r cosθ r cos( ωt + φ) a( ω r cos( ωt + φ) ω x( phasor

6 Simple harmonic motion Displacement, velocity and acceleration in SHM Displacement x( Acos( ω φ 0 Velocity dx( v( ωasin( ω dt Acceleration a( dv( dt ω Acos( ω Note: A x + ( 0) v (0) / ω

7 Energy conservation E K + U Energy in SHM Energy conservation in a SHM 1 1 K mv ; U x E BTW: 1 mv + x 1 const A mv + x v ± ω m ( A x ) Ch.7 E F s U x U x No friction du dx F dx s 0 1 x

8 Energy in SHM Energy conservation in a SHM (cont d) E 1 mv + x 1 const. E inetic energy energy energy distance from equilibrium point potential energy Time

9 Applications of SHM Simple pendulum The forces on the mass at the end are gravity and the tension. The tension, however, exerts no torque about the top of the string. g g τ Iα mglsinθ ml α α sinθ θ for small angle l l α dω dt d θ dt g θ l Angular frequency of a simple pendulum ω g ω 1 g 1, f, T π l π π l f l g mg

10 Applications of SHM Physical pendulum A simple pendulum has all its mass concentrated at a point and oscillates due to gravitational torques. Objects that do not have their mass concentrated at a point also oscillate due to gravitational torques. mgr τ Iα mgr sinθ Iα α sinθ I Angular frequency of ω mgr I a physical pendulum

11 Angular SHM An angular version of SHM is called torsion oscillation and shown on the right. A dis suspended by a wire experiences a restoring torsion when rotated by a small angle θ : τ κθ κθ c.f. Iα F x d θ κ θ dt I Angular frequency of Applications of SHM κ: torsion constant κ ω c.f. ω I an angular SHM : m

12 Oscillation with friction Damped oscillations In real world dissipative forces such as friction between a bloc and a table exist. Such a dissipative force will decrease the amplitude of an oscillation damped oscillation. The friction reduces the mechanical energy of the system as time passes, and the motion is said to be damped.

13 Damped oscillations A simple example of damped oscillation Consider a simple harmonic oscillation with a frictional damping force that is directly proportional to the velocity of the oscillating object. F x bv dx x b dt m ma d x dt If the damping force is relatively small, the motion is described by: x( Ae ( b / m) t cos( ω' t + φ) where ω' m b 4m

14 A simple example of damped oscillation Damped oscillations (cont d) ) / ( 4 ' where ) ' cos( ) ( m b m t Ae t x t m b + ω φ ω t m b Ae ) / ( By Dr. Dan Russell, Kettering University

15 Forced oscillations and resonance Driving force An example of resonantly driven damped harmonic oscillator Push Wait 1 period

16 Forced oscillations and resonance Driving force (cont d) The additional force that pushed by the person in the animation on the previous page is called a driving force. When a periodically varying driving force with angular frequency ω d is applied to a damped harmonic oscillator, the resulting motion is called a forced oscillation. driving force:cos( ωd ω d 0.4ω ω d 1.01ω By Dr. Dan Russell, Kettering University ω d 1.6ω

17 Forced oscillations and resonance Forced oscillation and resonance Damped SHM Forced damped SHM Fixed Moving/driving force F( Fmax cosω t d natural frequency ω ' m b 4m Damped

18 Forced oscillations and resonance Forced oscillation and resonance (cont d) Amplitude for a forced damped oscillation: F max A When mωd, A has a maximum ( mω d ) + b ωd near ω / m resonance: A d The fact that there is an amplitude pea at driving frequencies close to the natural frequency of the system is called resonance natural frequency angular freq. of driving force

19 Problem 1 Exercises a) The speed of the pan and the stea immediately after the collision (total inelastic collision): Initial speed of the meat just before the collision: v i gh 400 N/m Final speed of the meat-pan just after the collision: M. g M v f vi m + M. h0.40 m (9.80 m/s )(0.40 m).6 m/s.4 m0.0 g

20 Problem 1(cont d) Exercises b) The amplitude of the subsequent motion: When the stea hits the pan, the pan is Mg/ above the new equilibrium position. v f ± ω A f where x f Mg /, ω So the amplitude is: x A [ x f + v f /( m + M ) / ω ] 400 N/m M. g A Mg + ghm ( m + M ) 0.1 m. h0.40 m c) The period: m0.0 g T ( m + M ) π 0.49 s.

21 Problem Ma fr a I cm Rα f α, I Exercises x ( f : friction) Each:M/, R cm (1/ ) MR cylinders rolls w/o slipping stretched by x and then released a x x x ω x. α M + Icm / R (3/ ) M f T π ω π 3M.

22 Problem 3 Exercises Two identical, thin rods, each with 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 L- shaped object is deflected slightly, it oscillates. Find the frequency of the oscillation. Solution: The moment of inertia about the pivot: L (1/ 3) ml ( / 3) ml The center of gravity is located when balanced at a distance d L /( ) below the pivot. Thin the L-shaped object as a physical pendulum and is represented by the center of gravity. The period T is: T π I mgd L L d L /( L )

23 Problem 4 Exercises Find the effective spring constant. F x F x F eff x, x x + x 1 F F 1-1 x 1 F - x x F eff x 1 eff 1 + F F F 1 + F 1 + x F 1 F x, F x, F eff 1 1 x F 1-1 x eff 1 + F - x