Chapter 12. Recall that when a spring is stretched a distance x, it will pull back with a force given by: F = kx


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1 Chapter 1 Lecture Notes Chapter 1 Oscillatory Motion Recall that when a spring is stretched a distance x, it will pull back with a force given by: F = kx When the mass is released, the spring will pull it back to the left. If the surface is frictionless, the mass will move through the equilibrium position (x = 0) and continue to move to the left until the spring has been compressed a distance x (energy is conserved). Now the spring will exert a force kx on the mass causing it to move back to the right. If we assume that the spring never changes and that there is no friction, the mass will move between +x and x forever. This is known as simple harmonic motion. If we look at the sum of forces on the mass, we have: x k F = kx = ma a = x m Physics 10 Page 1
2 Chapter 1 Lecture Notes This means that the acceleration of a simple harmonic oscillator is proportional to the position of the oscillator. Recall that the acceleration is defined as: dx a = dt Therefore we can set up the following differential equation: dx dt k = x m (1) For simplicity, let s define: k ω = m () This means that our differential equation is: dx dt = ω x (3) If we solve this differential equation, we have: x(t) = Acos( ω t +φ) (4) We can find the velocity and acceleration functions for the oscillator by taking the appropriate derivatives: dx v(t) = = Aωsin( ω t +φ) dt dv a(t) = = Aω cos( ω t + φ ) = ω x(t) dt (5) Physics 10 Page
3 Chapter 1 Lecture Notes If we plot the position of our oscillator, it might look like: A is the amplitude of our oscillation (this is equal to the original distance that we stretch the spring). T is known as the period (in seconds) and it is the time it takes for the mass to go from A to A and then back to A. This time is related to the angular frequency ( ω ) by: π T = (6) ω We can also define the frequency which is the number of oscillations which occur each second. The frequency is: f = 1 = ω T π (7) The units of frequency are Hertz (Hz) which is equivalent to 1 cycle/second. Physics 10 Page 3
4 Chapter 1 Lecture Notes Example 1: A fisherman s scale stretches.8 cm when a 3.7 kg fish hangs from it A. What is the spring constant? B. What will be the amplitude and frequency of vibration if the fish is pulled down.5 cm more and released so that the fish vibrates up and down? Physics 10 Page 4
5 Chapter 1 Lecture Notes Example : At what displacement from equilibrium is the speed of a SHO half the maximum value? At what displacement from equilibrium is the acceleration of a SHO half the maximum value? Physics 10 Page 5
6 Chapter 1 Lecture Notes Example 3: The position of a SHO as a function of time is given by 7π π x = 3.8cos t + where t is in seconds and x is in meters. Find 4 6 A. the period and the frequency, B. the position and velocity at t = 0, and C. the velocity and acceleration at t =.0 s. Physics 10 Page 6
7 Chapter 1 Lecture Notes Energy in Harmonic Motion We know that the kinetic energy of an object is: 1 K = mv This means that the kinetic energy (as a function of time) for a simple harmonic oscillator is: 1 K = mω A sin ω t +φ ( ) (8) We have previously seen that the elastic potential energy for the spring is: 1 U = kx If we substitute our position function into the above, we have: 1 U = ka cos ω t +φ ( ) (9) The total energy is E = K + U, therefore: 1 1 E = mω A sin ω t +φ + ka cos ω t +φ 1 = ka sin ( ω t +φ ) + cos ( ω t +φ) 1 = ka ( ) ( ) So we can show how energy converts during the oscillation of the mass: Physics 10 Page 7
8 Chapter 1 Lecture Notes Example 4: It takes a force of 95.0 N to compress the spring of a popgun m to load a 0.00 kg ball. With what speed will the ball leave the gun? Physics 10 Page 8
9 Chapter 1 Lecture Notes Example 5: A kg bullet strikes a kg block attached to a fixed horizontal spring whose spring constant is.5 10 and sets it into m vibration with amplitude of 1.4 cm. What was the speed of the bullet if the two objects move together after impact? 3 N Physics 10 Page 9
10 Chapter 1 Lecture Notes The Simple Pendulum A simple pendulum consists of a point mass (no size) attached to the end of a massless string. As we swing the mass out, the component of the weight tangent to the circular path, will cause the mass to swing. If there is no friction, the mass will swing back and forth in simple harmonic motion just like the mass attached to a spring. If we look at the sum of tangential forces on the mass, we have: ds d Ft = mat mgsinθ= m = m Lθ dt dt ( ) d θ g g = sin θ θ dt L L (10) We can now say that the angular position of the swinging mass as a function of time is: ( ) θ (t) =θmax cos ω t +φ (11) This means that the angular frequency of the simple pendulum is: ω= g L (1) Physics 10 Page 10
11 Chapter 1 Lecture Notes Example 6: Determine the length of a simple pendulum whose period is 1.00 s. What would the period of a 1.00 m long simple pendulum be? Physics 10 Page 11
12 Chapter 1 Lecture Notes Example 7: What is the period of a 73 cm long simple pendulum on the Earth? What is it inside of a freely falling elevator? Physics 10 Page 1
13 Chapter 1 Lecture Notes The Physical Pendulum The term pendulum refers to any object which oscillates back and forth. So far we have only looked at a pendulum where the mass is concentrated at a single point, but we are not limited to this type of pendulum. When the swinging object cannot be modeled as a particle, we have a physical pendulum. From Chapter 10, we know that the net torque on any object is given by Newton s nd law for rotations: d θ τ= Iα= I dt For the object shown on the previous page, the applied torque (due to the weight of the object) will cause a clockwise rotation. So the net torque is: So our differential equation is: τ= mgdsin θ d θ mgd mgd = sinθ θ dt I I This means that the angular frequency for this type of pendulum is: Physics 10 Page 13
14 Chapter 1 Lecture Notes ω= mgd (13) I Example 8: An easy way to determine the moment of inertia of an object about any axis is to measure the period of oscillation about that axis. Suppose a nonuniform 1.0 kg stick can be balanced at a point 4 cm from one end. If it is pivoted about that end it oscillates with a period of 3.0 s. What is the moment of inertia about the end? What is the moment of inertia about an axis perpendicular to the stick through its center of mass? Physics 10 Page 14
15 Chapter 1 Lecture Notes Example 9: A plywood disk of radius 0.0 cm and mass 3.00 kg has a small hole drilled through it.00 cm from its edge. The disk is hung from the wall by means of a metal pin through the hole and is used as a pendulum. What is the period of this pendulum for small oscillations? Physics 10 Page 15
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