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Physics

Physics. Oscillations About Equilibrium Teacher: Luiz Izola. Chapter Preview. Periodic Motion 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.

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Physics

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  1. Physics Oscillations About Equilibrium Teacher: Luiz Izola

  2. Chapter Preview Periodic Motion 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

  3. Introduction The basic cause of oscillations is the fact that when an object is displaced from a position of stable equilibrium it experiences a restoring force that is directed towards the equilibrium position. The most familiar oscillating system is the PENDULUM

  4. Periodic Motion A motion that repeats itself over and over. For example: The heart beating, the clock ticking, the child on a swing. One of the key features of a periodic system is the time required for the completion of one cycle of its repetitive motion. This time is called PERIOD (T). SI unit is seconds. For example, the pendulum on a clock may take one second to swing from maximum displacement in one direction to maximum displacement in the opposite direction. This will take 2 seconds to complete 1 cycle, then T = 2secs.

  5. Another important definition related to periodic motion is FREQUENCY (f). The frequency of an oscillation is the number of oscillations per unit of time. Frequency is the inverse of period. • f = 1/T SI unit = s-1 • The frequency is also measured in Hz (Hertz). Ex: If the processing speed of a computer is 1.80Ghz, how much time is required for one processing cycle? Ex: A tennis ball is hit back and forth between 2 players. If it takes 2.31secs for the ball to go from 1 player to another, what are T and f? Periodic Motion

  6. Simple Harmonic Motion It is a special type of periodic motion. A classic example is provided by the oscillation of a mass attached to a spring. Consider a cart of mass m attached to a spring of force constant k. We know, by Hooke’s Law, that if the cart is displaced from equilibrium a distance x, the spring exerts a force of magnitude: F = -kx. Mathematical analysis, using calculus, shows that the position of the cart as function of time can be represented by the sine or cosine function.

  7. Simple Harmonic Motion The extreme displacement on either side of the equilibrium position is called AMPLITUDE (A). The amplitude is ½ of the total range of motion. Position x Time is SHM X = A cos(2πt/T) SI unit: meters (m) Remember that the cart’s motion repeats with a period T. Therefore the position of the cart must be the same at time t and time t+T.

  8. A Mass Attached to a Spring Undergoes Simple Harmonic Motion About x = 0

  9. Simple Harmonic Motion as a Sine or a Cosine

  10. Practice Problems Ex1: An air track attached to a spring completes one oscillation every 2.4 secs. At t=0, the cart is released from rest at a distance of 0.10 meters from its equilibrium position. What is the position of the cart at (a) 0.30 secs, (b) 0.60 secs? Ex2: A mass attached to a spring oscillates with a period of 3.15 secs. (a) If the mass starts from rest at x = 0.0440 meters and time t = 0, where is it at time t = 6.37 secs? (b) Is the mass moving on the positive or negative direction at t = 6.37s?

  11. The Relationship Between Uniform Circular Motion and Simple Harmonic Motion

  12. Uniform Circular Motion and SHM Lets assume that the turntable rotates with a constant angular speed of ω = 2π/T, taking the time T to complete 1 revolution. Also, lets assume that the radius of the turntable is A. We can see from the previous slide that, if we adjust the period of the mass connected to the spring to be the same as the turntable to complete 1 revolution, the mass and the peg’s shadow move as one. Let us use Circular Motion in order to obtain detailed results for the position, velocity, and acceleration of a particle undergoing SHM.

  13. Position Versus Time in SHM

  14. Position Versus Time in SHM In the previous picture, we show the peg at the angular position Θ, where Θ is measured relative to the x-axis. If the peg starts at Θ = 0 and t = 0, and the turntable rotates with a constant angular speed of ω, we know that Θ = ωt. This means, the angular position increases with time. Notice in the previous picture that the x-location in the shadow is at location x = A cos(Θ) , which is the x component of the radius vector. Therefore, the position as a function of time is: X = Acos(Θ) = Acos(ωt) = Acos(2πt/T)

  15. Position Versus Time in SHM When talking about SHM, ω means angular frequency. It definition is: ω = 2π/T Ex: A ball rolls on a circular track of radius 0.50 m with a constant angular speed of 1.3rad/sec. If the angular position of the ball at t=0 is Θ = 0, find the component of the ball’s position at times 2.50 secs, 5.00 secs, and 7.50 secs.

  16. Velocity Versus Time in SHM

  17. Velocity Versus Time in SHM From the previous picture, we can define the x-component velocity as –Vsin(Θ) at the peg position Θ. Knowing that tangential circular motion velocity is equals to rω, we have: Vx = -Vsin(Θ) = -rωsin(ωt) = -Aωsin(ωt) Note that when the displacement from equilibriumis a maximum, the velocity is zero. This is to be expected since at x+A and x-A the object is momentarily at rest (turning points of the motion). The maximim velocity is Vmax = Aω

  18. Acceleration Versus Time in SHM

  19. Acceleration Versus Time in SHM Acceleration in SHM is: a = -Aω2cos(ωt) Since x = Acos(ωt), we can rewrite the acceleration formula as: a = ω2x The maximum acceleration is: a = Aω2

  20. Practice Problems Ex3: An air track attached to a spring completes one oscillation every 2.40secs. At t=0 the cart is released from rest with the spring stretched 0.10m from its equilibrium position. What are the velocity and acceleration of the cart at (a) 0.30secs and (b) 0.60secs? Ex4: On 12/29/1997, airplane was hit with severe turbulence 31 minutes after takeoff. Data from the black box indicated that the plane moved up and down with an amplitude of 30.0meters and a maximum acceleration of 1.8g. Treating the up/down as SHM, find (a) time to complete 1 oscillation (b) plane’s maximum vertical speed

  21. Practice Problems Ex5: A red apple floats in a barrel of water. If you lift the apple 2.00cm above its floating level and release it, it bobs up and down with a period of T= 0.75secs. Assuming the motion is a SHM, find the position, velocity and acceleration of the apple at (a) T/4 and (b) T/2. Ex6: A child rocks back and forth on a swing with an amplitude of 0.204meters and a period of 2.80secs. Assuming the motion is approximately SHM, find the child’s maximum velocity.

  22. The Period of a Mass on a Spring We will show how the period of a mass on a spring is related to the mass m, the force constant k, and the amplitude of motion A. First remember from Hooke’s Law that at position x, the force acting on the mass is F = -kx Since F = ma, we have: ma = -kx We know that a = -Aω2cos(ωt) and x = Acos(ωt) Replacing in the equation above, we have: ω2 = k/m ω = (k/m)1/2 T = 2π(m/k)1/2

  23. Practice Problems Ex7: When a 0.22kg air track cart is attached to a spring, it oscillates with a period of 0.84secs. What is the force constant of this spring? Ex8: A 0.12kg mass attached to a spring oscillates with an amplitude of 0.075 meters and a maximum speed of 0.524 m/s. Find k and T Ex9: When a 0.42kg mass is attached to a spring, it oscillates with a period of 0.350secs. If a different mass m2, is attached to the same spring, it oscillates with a period of 0.70secs. Find (a) the force constant of the spring and (b) the mass m2.

  24. Energy Conservation on Oscillatory Motion The total energy (E) of a mass on a horizontal spring is the sum of its kinetic energy (K = 1/2mv2) and its potential energy (U = 1/2kx2). E = 1/2mv2 + 1/2kx2 Solving the equation with SHM, we conclude that: E = ½ kA2

  25. Energy as a Function of Position in Simple Harmonic Motion

  26. Practice Problems Ex10: A 0.980kg block slides on a frictionless horizontal surface with a speed of 1.32m/s. The block encounters an un-stretched spring with a force constant of 245N/m. (a) How far is the spring compressed before the block comes to rest? (b) How long is the block in contact with the spring before it comes to rest? Ex11: A bullet of mass m embeds itself in a block of mass M, which is attached to a spring of force constant k. If the initial speed of the bullet is vo, find (a) maximum spring compression. (b) the time for the bullet-block system to come to rest.

  27. The Pendulum A simple pendulum consists of a mass m suspended by a light string or rod of length L. The pendulum has a stable equilibrium when the mass is directly below the suspension point and oscillates about this position if displaced from it. As a pendulum swings away from equilibrium, it rises a vertical distance equals ( L – Lcos(Θ) ). The angle of the pendulum with the vertical varies with time like a sine or cosine

  28. Motion of a Pendulum

  29. The Potential Energy of a Simple Pendulum

  30. Forces Acting on a Pendulum Bob

  31. Forces Acting on a Pendulum Bob The net tangential force acting on m is the tangential component of its weight: F = mgsinΘ For small angles measured in radians, the sine of Θ is almost equals to Θ itself. Also, we know that the arc length S = LΘ. Therefore, Θ = S/L. Replacing those assumptions above, we have: F = (mg/L)S Comparing with the to a mass on a spring: F = kx. If we let x = s: k = mg/L. Then the period is: T = 2π(L/g)1/2

  32. Practice Problems Ex12: A pendulum in a grandfather clock is designed to take one second to swing in each direction; that is, 2.00secs for a complete period. Find the length of the pendulum with a period of 2.00secs. Ex13: A pendulum is constructed from a string 0.627meters long attached to a mass of 0.25kg. When set in motion, the pendulum completes one oscillation every 1.59secs. If the pendulum is held at rest and the string is cut, how long does it take for the mass to fall through a distance of 1 meters?

  33. Formulas Summary Period (T) : time required for one motion cycle. Frequency(f): # of oscillations per unit time. f = 1/T Angular Frequency(ω): ω = 2π/T = 2πf Amplitude (A): max. displacement from equilibrium SHM Formulas Position (x): x = Acos(Θ) = Acos(ωt) = Acos(2πt/T) Velocity (v): Vx = -Asin(Θ) = -rωsin(ωt) = -Aωsin(ωt) Acceleration (a): a = -Aω2cos(ωt) = a = ω2x

  34. Formulas Summary Period of a Mass on a Spring (T): T = 2π(m/k)1/2 Energy Conservation In Oscillatory Motion: Total Energy on a Spring E = 1/2kA2 Potential Energy on a Spring Varying with Time U = 1/2kA2cos2(ωt) Kinetic Energy on a Spring Varying with Time K = 1/2kA2sin2(ωt)

  35. Formulas Summary The Pendulum A pendulum oscillating with small amplitude also exhibits SHM. A simple or ideal pendulum is one in which all the mass is concentrated at a single point a distance L from the suspension point. Period of a Simple Pendulum of length L (T) T = 2π(L/g)1/2

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