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General Physics I Spring 2011 Oscillations 1 Oscillations • A quantity is said to exhibit oscillations if it varies with time about an equilibrium or reference value in a repetitive fashion. • Oscillations are periodic when the time for one complete cycle of the oscillations is constant. This repeat time is called the period (T) of the oscillation. • Let us assume that for a given oscillation, f cycles are completed in one second. The number of cycles per second is called the frequency of the oscillation. Since f cycles are completed in one second, it follows that one cycle is completed in 1/f seconds. The time taken to complete one cycle is the period. Thus, T=1 f or f = 1. T The unit of frequency is the hertz (Hz). 1 Hz = 1 cycle/sec. 2 Simple Harmonic Motion • Consider an object attached to a spring at one end. The other end of the spring is fixed. The object rests on a frictionless horizontal surface. The object is in equilibrium when the spring is not stretched or compressed. If the object is displaced from the equilibrium position and released, it will oscillate about this position. • If the displacement of the object from its equilibrium position is graphed as a function of time, one obtains sinusoidal behavior (sine or cosine function). f =1/T . 3 Simple Harmonic Motion Oscillatory motion for which the position is a sinusoidal function of time is called simple harmonic motion. 4 Restoring Force • An oscillation must have a restoring force. This is the net force that always acts in a direction toward the equilibrium position, and thus tends to “restore” the object back to the equilibrium position. The oscillation is sustained because when the object reaches the equilibrium position (where the net force is zero), it has a non-zero velocity and so its inertia carries it past the equilibrium position. The restoring force then slows it down, it stops momentarily and then moves back toward the equilibrium position. Marble in a bowl Object attached to spring 5 Linear Restoring Force • The spring force is a linear restoring force because the force is proportional to the displacement from the equilibrium position: (Fsp )x =−k ∆x. • The equation above is called Hooke’s law. Note that the force is in the opposite direction to the displacement of the free end of the spring from its equilibrium position. Thus, the force always points toward the equilibrium position. • If we take the equilibrium position to be x = 0, then ∆x = x. So, (Fsp )x =−kx. • A linear restoring is required for simple harmonic motion. Slope is –k, where k is the spring constant. 6 Workbook: Chapter 14, Questions 1, 2 7 Describing Simple Harmonic Motion • The graph to the right shows the position (or displacement) x versus time t for a particle undergoing simple harmonic motion (SHM). The position as time progresses is described by the cosine function: 2π t x = Acos . T In the equation above, A is the magnitude of the maximum displacement from equilibrium, which is the amplitude. When x = A or –A, the particle is at its maximum distance from equilibrium and so the velocity is zero. The particle changes directions at these points, which are called turning points. 8 Describing Simple Harmonic Motion • • • • • x = Acos 2π t . T Note that the argument of the cosine function (the quantity inside the parentheses) must be in radians. At t = 0, the argument of the cosine function is zero. The cosine of zero radians = 1. Thus, x = A. The object starts at the maximum positive displacement from equilibrium. At t = T/4 (one-fourth of one period), the argument of the cosine function is π/2 radians. The cosine of π/2 radians is zero. Thus, at this time, the object is at the equilibrium position. At t = T/2 the argument of the cosine function is π radians. The cosine of π radians is -1. Thus, at this time, the particle is at x = -A and is about to change directions. At t = 3T/4, the argument of the cosine function is 3π/2 radians. The cosine of 3π/4 radians is zero. Thus, the particle is at the equilibrium position again. 9 Describing Simple Harmonic Motion x = Acos 2π t . T • At t = T, the argument of the cosine function is 2π radians. The cosine of 2π radians = 1. Thus, x = A. The object is back to its starting point. One cycle has been completed. Velocity • To find the velocity of the object at any instant, we can find the slope of the position graph. It is easy to see that the slope is zero when x = ±A, so the velocity is zero at these times. These are the turning points as mentioned before. Also, the slope has maximum magnitude when x = 0, i.e., at the equilibrium. Thus, the speed of the object is greatest at the equilibrium position. 10 Describing Simple Harmonic Motion • From the slope of the position graph, we find that the velocity of the object undergoing SHM is given by 2π t vx =−vmax sin , T where vmax = 2π A. T • Like the cosine function, the sine function will be positive, zero, or negative at different times. Thus, the velocity will negative, zero, or positive at these times. A negative velocity simply indicates motion in the negative x direction (leftward). 11 Describing Simple Harmonic Motion Acceleration • One can find the acceleration versus time by finding the slope of the velocity vs. time graph. However, another way to obtain the acceleration vs. time behavior is to use Newton’s second law. The restoring force is the only force acting along the direction of motion and so is the net force in this direction. The acceleration is given by ax =(Fnet)x/m. But, (Fnet)x = −kx. Thus, k x. ax =− m • We see that the acceleration is proportional to the position but in the opposite direction to it. Thus, the acceleration vs. time graph must also be a cosine graph, but an inverted one because of the minus sign. Using our previous expression for x as a function of t, we find that k x =− kA cos 2π t . ax =− m m T 12 Describing Simple Harmonic Motion • We can rewrite the acceleration as a function of time as 2π t ax =−amax cos , T where 2A kA 4 π amax = m = 2 . T 13 Describing Simple Harmonic Motion 14 A mass attached to a spring oscillates back and forth as indicated in the position vs. time plot below. At point P, the mass has 1. 2. 3. 4. 5. 6. 7. positive velocity and positive acceleration. positive velocity and negative acceleration. positive velocity and zero acceleration. negative velocity and positive acceleration. negative velocity and negative acceleration. negative velocity and zero acceleration. zero velocity but is accelerating (positively or negatively). 8. zero velocity and zero acceleration. 15 Vertical Springs • If an object is mounted on a vertical spring, it will execute simple harmonic motion in exactly the same way as an object that is attached to a horizontal spring. • The vertical orientation causes gravity to lower the equilibrium position. That is all. 16 Workbook: Chapter 14, Question 7 Textbook: Chapter 14, Question 23, Problem 9 17