Chapter 13 Oscillations about Equilibrium
Periodic Motion Units of Chapter 13 Simple Harmonic Motion Connections between Uniform Circular Motion and Simple Harmonic Motion The Period of a Mass on a Spring Energy Conservation in Oscillatory Motion The Pendulum We will leave out Ch. 13.7 and Ch.13.8.
13-1 Periodic Motion Period: time required for one cycle of periodic motion Frequency: number of oscillations per unit time This unit is called the Hertz:
13- Simple Harmonic Motion A spring exerts a restoring force that is proportional to the displacement from equilibrium:
13- Simple Harmonic Motion A mass on a spring has a displacement as a function of time that is a sine or cosine curve: Here, A is called the amplitude of the motion.
13- Simple Harmonic Motion If we call the period of the motion T this is the time to complete one full cycle we can write the position as a function of time: It is then straightforward to show that the position at time t + T is the same as the position at time t, as we would expect.
13-3 Connections between Uniform Circular Motion and Simple Harmonic Motion An object in simple harmonic motion has the same motion as one component of an object in uniform circular motion:
13-3 Connections between Uniform Circular Motion and Simple Harmonic Motion Here, the object in circular motion has an angular speed of where T is the period of motion of the object in simple harmonic motion.
13-3 Connections between Uniform Circular Motion and Simple Harmonic Motion The position as a function of time: The angular frequency:
13-3 Connections between Uniform Circular Motion and Simple Harmonic Motion The velocity as a function of time: And the acceleration: Both of these are found by taking components of the circular motion quantities.
13-4 The Period of a Mass on a Spring Since the force on a mass on a spring is proportional to the displacement, and also to the acceleration, we find that. Substituting the time dependencies of a and x gives
13-4 The Period of a Mass on a Spring Therefore, the period is
13-5 Energy Conservation in Oscillatory Motion In an ideal system with no nonconservative forces, the total mechanical energy is conserved. For a mass on a spring: Since we know the position and velocity as functions of time, we can find the maximum kinetic and potential energies:
13-5 Energy Conservation in Oscillatory Motion As a function of time, So the total energy is constant; as the kinetic energy increases, the potential energy decreases, and vice versa.
13-5 Energy Conservation in Oscillatory Motion This diagram shows how the energy transforms from potential to kinetic and back, while the total energy remains the same.
Example: The period of oscillation of an object in an ideal mass-spring system is 0.50 sec and the amplitude is 5.0 cm. What is the speed at the equilibrium point? At equilibrium x 0: 1 1 E K + U mv + kx 1 mv Since E constant, at equilibrium (x 0) the KE must be a maximum. Here v v max Aω.
Example continued: The amplitude A is given, but ω is not. "! T! 0. 50 s 1. 6 rads/sec and v Aù ( 5. 0 cm)( 1. 6 rads/sec) 6. 8 cm/sec
Example: The diaphragm of a speaker has a mass of 50.0 g and responds to a signal of.0 khz by moving back and forth with an amplitude of 1.8 10 4 m at that frequency. (a) What is the maximum force acting on the diaphragm? # F F ( ) A" ma(! f )! maf max mamax m 4 The value is F max 1400 N.
Example continued: (b) What is the mechanical energy of the diaphragm? Since mechanical energy is conserved, E K max U max. U K max max 1 ka The value of k is unknown so use K max. 1 mv max 1 1 1 K max mvmax m! ( A" ) ma ( f ) The value is K max 0.13 J.
Example: The displacement of an object in SHM is given by: y ( t) ( 8.00 cm) sin[ ( 1.57 rads/sec) t] What is the frequency of the oscillations? Comparing to y(t) A sinωt gives A 8.00 cm and ω 1.57 rads/sec. The frequency is: f "! 1.57 rads/sec! 0.50 Hz
Example continued: Other quantities can also be determined: The period of the motion is T! "! 1.57 rads/sec 4.00 sec x v a max max max A A! A! 8. 00 cm ( 8. 00 cm)( 157. rads/sec) 1. 6 cm/sec ( 8. 00 cm)( 157. rads/sec) 19. 7 cm/sec
13-6 The Pendulum A simple pendulum consists of a mass m (of negligible size) suspended by a string or rod of length L (and negligible mass). The angle it makes with the vertical varies with time as a sine or cosine.
13-6 The Pendulum Looking at the forces on the pendulum bob, we see that the restoring force is proportional to sin θ, whereas the restoring force for a spring is proportional to the displacement (which is θ in this case).
13-6 The Pendulum However, for small angles, sin θ and θ are approximately equal.
13-6 The Pendulum Substituting θ for sin θ allows us to treat the pendulum in a mathematically identical way to the mass on a spring. Therefore, we find that the period of a pendulum depends only on the length of the string:
Example: A clock has a pendulum that performs one full swing every 1.0 sec. The object at the end of the string weighs 10.0 N. What is the length of the pendulum? T! L g Solving for L: L gt 4! ( 9. 8 m/s )( 10. s) 4! 0. 5 m
Example: The gravitational potential energy of a pendulum is U mgy. Taking y 0 at the lowest point of the swing, show that y L(1-cosθ). L θ L Lcosθ y0 y L( 1" cos! )