Vibrations and Waves

m Vibrations and Waves Physics 2053 Lecture Notes Vibrations and Waves

Vibrations and Waves Topics 13.01 Hooke’s Law 13.02 Elastic Potential Energy 13.03 Comparing SHM with Uniform Circular Motion 13.04 Position, Velocity and Acceleration as a Function of Time 13.05 Motion of a Pendulum Vibrations and Waves (3 of 33)

Hooke’s Law x = 0 We assume that the surface is frictionless. There is a point where the spring is neither stretched nor compressed; this is the equilibrium position. We measure displacement from that point (x = 0 ). m If an object vibrates or oscillates back and forth over the same path, each cycle taking the same amount of time, the motion is called periodic (T). Vibrations and Waves (4 of 33)

Hooke’s Law F m x x = 0 The restoring force exerted by the spring depends on the displacement: m The minus sign on the force indicates that it is a restoring force – it is directed to restore the mass to its equilibrium position. Vibrations and Waves (5 of 33)

Hooke’s Law F x m (a) (k) is the spring constant (b) Displacement (x) is measured from the equilibrium point (c) Amplitude (A) is the maximum displacement (d) A cycle is a full to-and-fro motion (e) Period (T) is the time required to complete one cycle (f) Frequency (f) is the number of cycles completed per second Vibrations and Waves (6 of 33)

Hooke’s Law mg If the spring is hung vertically, the only change is in the equilibrium position, which is at the point where the spring force equals the gravitational force. xo m Equilibrium Position Vibrations and Waves (7 of 33)

Hooke’s Law Any vibrating system where the restoring force is proportional to the negative of the displacement moves with simple harmonic motion (SHM), and is often called a simple harmonic oscillator. Vibrations and Waves (8 of 33)

Elastic Potential Energy Potential energy of a spring is given by: The total mechanical energy is then: The total mechanical energy will be conserved Vibrations and Waves (9 of 33)

Elastic Potential Energy x = 0 If the mass is at the limits of its motion, the energy is all potential. vmax A If the mass is at the equilibrium point, the energy is all kinetic. m m Vibrations and Waves (10 of 33)

Elastic Potential Energy where The total energy is, therefore And we can write: This can be solved for the velocity as a function of position: Vibrations and Waves (11 of 33)

Elastic Potential Energy F x m The acceleration can be calculated as function of displacement Vibrations and Waves (12 of 33)

Comparing Simple Harmonic Motion with Circular Motion v A q x vmax If we look at the projection onto the x axis of an object moving in a circle of radius A at a constant speed vmax, we find that the x component of its velocity varies as: This is identical to SHM. Vibrations and Waves (13 of 33)

Comparing Simple Harmonic Motion with Circular Motion Therefore, we can use the period and frequency of a particle moving in a circle to find the period and frequency of SHM: Vibrations and Waves (14 of 33)

Comparing SHM with Uniform Circular Motion (Problem) A mass m at the end of a spring vibrates with a frequency of 0.88 Hz. When an additional 680 g mass is added to m, the frequency is 0.60 Hz. What is the value of m? Vibrations and Waves (15 of 33)

Vibrations and Waves A mass on a spring undergoes SHM. When the mass passes through the equilibrium position, its instantaneous velocity A) is maximum. B) is less than maximum, but not zero. C) is zero. D) cannot be determined from the information given. Vibrations and Waves

Vibrations and Waves A mass is attached to a vertical spring and bobs up and down between points A and B. Where is the mass located when its kinetic energy is a minimum? A) at either A or B B) midway between A and B C) one-fourth of the way between A and B D) none of the above Vibrations and Waves

Comparing SHM with Uniform Circular Motion (Problem) A 0.60 kg mass at the end of a spring vibrates 3.0 times per second with an amplitude of 0.13 m. Determine the velocity when it passes the equilibrium point, Vibrations and Waves (18 of 33)

Comparing SHM with Uniform Circular Motion (Problem) con’t A 0.60 kg mass at the end of a spring vibrates 3.0 times per second with an amplitude of 0.13 m. Determine the velocity when it is 0.10 m from equilibrium, Vibrations and Waves (19 of 33)

Comparing SHM with Uniform Circular Motion (Problem) con’t A 0.60 kg mass at the end of a spring vibrates 3.0 times per second with an amplitude of 0.13 m. Determine the total energy of the system, Vibrations and Waves (20 of 33)

Vibrations and Waves A mass is attached to a vertical spring and bobs up and down between points A and B. Where is the mass located when its potential energy is a minimum? A) at either A or B B) midway between A and B C) one-fourth of the way between A and B D) none of the above Vibrations and Waves

Vibrations and Waves Doubling only the amplitude of a vibrating mass-and-spring system produces what effect on the system's mechanical energy? A) increases the energy by a factor of two B) increases the energy by a factor of three C) increases the energy by a factor of four D) produces no change Vibrations and Waves

Comparing SHM with Uniform Circular Motion (Problem) A mass of 2.62 kg stretches a vertical spring 0.315 m. If the spring is stretched an additional 0.130 m and released, how long does it take to reach the (new) equilibrium position again? Vibrations and Waves (23 of 33)

Vibrations and Waves Doubling only the spring constant of a vibrating mass-and-spring system produces what effect on the system's mechanical energy? A) increases the energy by a factor of three B) increases he energy by a factor of four C) produces no change D) increases the energy by a factor of two Vibrations and Waves

The Simple Pendulum A simple pendulum consists of a mass at the end of a lightweight cord. We assume that the cord does not stretch, and that its mass is negligible. Vibrations and Waves (25 of 33)

The Simple Pendulum q L x F q mg s k for SHM Small angles x  s m Vibrations and Waves (26 of 33)

Vibrations and Waves A simple pendulum consists of a mass M attached to a weightless string of length L. For this system, when undergoing small oscillations A) the frequency is proportional to the amplitude. B) the period is proportional to the amplitude. C) the frequency is independent of the length L. D) the frequency is independent of the mass M. Vibrations and Waves

Comparing SHM with Uniform Circular Motion (Problem) The length of a simple pendulum is 0.760 m, the pendulum bob has a mass of 365 grams, and it is released at an angle of 12.0° to the vertical. With what frequency does it vibrate? Assume SHM. Vibrations and Waves (28 of 33)

Comparing SHM with Uniform Circular Motion (Problem) con’t The length of a simple pendulum is 0.760 m, the pendulum bob has a mass of 365 grams, and it is released at an angle of 12.0° to the vertical. What is the pendulum bob’s speed when it passes through the lowest point of the swing? Vibrations and Waves (29 of 33)

Comparing SHM with Uniform Circular Motion (Problem) con’t The length of a simple pendulum is 0.760 m, the pendulum bob has a mass of 365 grams, and it is released at an angle of 12.0° to the vertical. Assume SHM.What is the total energy stored in this oscillation, assuming no losses? Vibrations and Waves (30 of 33)

Summary of Chapter 11 For SHM, the restoring force is proportional to the displacement. Period for a mass on a spring: During SHM, the total energy is continually changing from kinetic to potential and back. The period is the time required for one cycle, and the frequency is the number of cycles per second. Vibrations and Waves (31 of 33)

Summary of Chapter 11 A simple pendulum approximates SHM if its amplitude is not large. Its period in that case is: Velocity Acceleration The kinematics of a mass/spring system: Vibrations and Waves (32 of 33)

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Vibrations and Waves