Lecture 19 Goals: Chapter 14 Assignment Periodic motion. No HW this week. Wednesday: Read through Chapter 15.4 1

Periodic Motion is everywhere Examples of periodic motion Earth around the sun Elastic ball bouncing up and down Quartz crystal in your watch, computer clock, iPod clock, etc. Earth revolving around the sun is a very good example of periodic motion, motion that is repeating itself over and over again. Earth has been revolving around the sun for billions of years without too much change in its orbit. There are some variations that happen at very long time-scales, at about every 100,000 years but this variation is not very much. An Elastic ball bouncing up and down is another example of periodic motion. When you drop a ball, if the ball is sufficiently elastic, it will bounce up and down many times before the motion dies out. Unlike earth revolving around the sun though, this is not going to happen a few billion times, even if the ball is very good, this motion will die out after maybe 100 bounces or so. So after each bounce, the ball will reach a height that is slightly smaller than the its previous height. The reason for this is that there are dissipative processes that suck energy out of the system, each collision with the floor will not be completely elastic, sot there will be some energy loss there, also there may also be significant energy loss due to air resistance. Now the key reason why periodic motion is so important for us is that we always use a type of periodic motion to keep track of time. As we know, we use earth’s rotation around the sun to keep track of year, so every one rotation is one year. But the same idea carries to much more precise tracking of time. For example, every digital watch or computer has a crystal that generates a precise voltage, and the device keeps track of time by counting how many times that voltage goes up and down. In fact, currently, our definition of one second relies on a a bunch of Cs atoms, that emit a very precise electromagnetic radiation at the microwave, and we count how many times the Electric field of that EM radiation changes its sign, that is infact how we define one second. So the central concept is that periodic motion is very important because it allows us to keep track of time.

Periodic Motion is everywhere Examples of periodic motion Heart beat In taking your pulse, you count 70.0 heartbeats in 1 min. What is the period, in seconds, of your heart's oscillations? Period is the time for one oscillation T= 60 sec/ 70.0 = 0.86 s Now, whenever we have periodic motion, the most important quantity that relates to that motion, is the period, that is the time it takes to complete one cycle of that motion. For example, for earth revolving around the sun, the period is one year And here in this slide, I have another example, the beating of your heart. Beating of the heart is also an example of periodic motion. The heart beats about two billion times during a lifetime. Now, let’s say that you take your pulse, and you find that it is 70 beats per minute. Then the question is what is the period of your heart’s motion, of your heart’s oscillations, in seconds? We usually denote the period with capital T, since we have counted 70 beats every minute, every 60 seconds, the duration for one beat, one cycle, or one oscillation is 60 sec / 70 = 0.86 seconds. If you look at the EKG from the heart, it looks something quite complicated, but the duration between the two beats would be 0.86 seconds in this case. And if you started running for example, your heart will start to beat faster, and the period will decrease.

Simple Harmonic Motion (SHM) We know that if we stretch a spring with a mass on the end and let it go the mass will, if there is no friction, ….do something 1. Pull block to the right until x = A 2. After the block is released from x = A, it will A: remain at rest B: move to the left until it reaches equilibrium and stop there C: move to the left until it reaches x = -A and stop there D: move to the left until it reaches x = -A and then begin to move to the right k m The periodic motion that we will analyze in detail in this course is where we connect a mass to spring. And the motion in this case is called simple harmonic motion, SHM. Harmonic is the term that is usually reserved for the case when the periodic motion is sinusoidal, that is you can describe the motion with sines and cosines. We will say more about that later. So let’s say that we have a spring whose equilibrium position is here. We stretch the spring by an amount A, we connect a mass m, and then let it go. If the system is frictionless, what will happen? The correct answer is D. We will analyze this system quantitatively in a moment, but first let’s discuss qualitatively why this has to be the case. When we let it go, since the spring is stretched, there will be a force towards of magnitude kA to the left, so the mass will start to accelerate. And this acceleration towards left will continue until the mass reaches the equilibrium position. Now how would we find the velocity of the mass when it passes through the equilibrium position. Well, because there is no friction, energy is conserved, so we start with the potential energy that is stored in the spring, 1/2kA^2. and at equilbrium position, all that potential energy is converted into the kinetic energy of the mass, 1/2mV^2. Now after the equilibrium position, now the spring will start to be compressed, and will start to apply a force to the right. So, the mass will start to decelerate, will start to lose its speed. Now, where will the mass lose all of its spee, Well, now we will be converting all the mechanical energy back to potential energy, and since the energy that we started with is 1/2kA^2, when the mass comes a stop, the stored energy in the spring will have to exactly equal that, so it will reach –A. But once it reaches –A, then there is a force to the right because the spring is compressed, so the mass will start moving to the right. So the mass will oscillate back and forth between A and –A The position vs time graph will look something like this. It will start at A, it will reach to –A, and then it will come to A again, and as we will show later on, this function will turn out to be a cosine function. Here we call A the amplitude of oscillation. -A 0(≡Xeq) A

Simple Harmonic Motion (SHM) The time it takes the block to complete one cycle is the period T and is measured in seconds. The frequency, denoted f, is the number of cycles that are completed per unit of time: f = 1 / T. In SI units, f is measured in inverse seconds, or hertz (Hz). If the period is doubled, the frequency is A. unchanged B. doubled C. halved

Simple Harmonic Motion (SHM) An oscillating object takes 0.10 s to complete one cycle; that is, its period is 0.10 s. What is its frequency f ? Express your answer in hertz. f = 1/ T = 10 Hz

Simple Harmonic Motion Suppose that the period is T. Which of the following points on the t axis are separated by the time interval T? A. K and L B. K and M C. K and P D. L and N E. M and P Now we will go through several problems to make sure that we understand how to read these oscillatory graphs. time

Simple Harmonic Motion Now assume that the t coordinate of point K is 0.25 s. What is the period T , in seconds? How much time t does the block take to travel from the point of maximum displacement to the opposite point of maximum displacement? time

Simple Harmonic Motion Now assume that the t coordinate of point K is 0.25 s. What is the period T , in seconds? T = 1 s How much time t does the block take to travel from the point of maximum displacement to the opposite point of maximum displacement? time

Simple Harmonic Motion Now assume that the t coordinate of point K is 0.25 s. What is the period T , in seconds? T = 1 s How much time t does the block take to travel from the point of maximum displacement to the opposite point of maximum displacement? t = 0.5 s time

Simple Harmonic Motion Now assume that the x coordinate of point R is 1 m. What total distance d does the object cover during one period of oscillation? What distance d does the object cover between the moments labeled K and N on the graph? time

Simple Harmonic Motion Now assume that the x coordinate of point R is 1 m. What total distance d does the object cover during one period of oscillation? d = 4 m What distance d does the object cover between the moments labeled K and N on the graph? time

Simple Harmonic Motion Now assume that the x coordinate of point R is 1 m. What total distance d does the object cover during one period of oscillation? d = 4 m What distance d does the object cover between the moments labeled K and N on the graph? d = 3 m time

T = 1 s T is: T > 1 s T < 1 s T=1 s k m -A 0(≡Xeq) A k 2m -A

T = 1 s T is: T > 1 s T < 1 s T=1 s k m -A 0(≡Xeq) A 2k m -A

SHM Dynamics: Newton’s Laws still apply At any given instant we know that F = ma must be true. But in this case F = -k x and ma = F So: -k x = ma =m d2x/dt2 k x m F = -k x a 1) The acceleration is not constant, so this is different than what we have done in the course so far. d2x/dt2=-(k/m)x a differential equation for x(t) ! “Simple approach”, guess a solution and see if it works!

   T = 2/ SHM Solution... Try cos (  t ) Below is a drawing of A cos (  t ) where A = amplitude of oscillation [with w = (k/m)½ and w = 2p f = 2p /T ] T=2 (m/k)½    T = 2/ A

The general solution is: x(t) = A cos ( wt + f) SHM Solution... The general solution is: x(t) = A cos ( wt + f) Use “initial conditions” to determine phase  ! k m -A A 0(≡Xeq) x(t) = A cos ( wt + 0) x(t) = A cos ( wt + π) at t=0 It is easy to see that adding a phi to the argument of cosine does not really change anything and we can go through exactly the same argument and we would find that instead of A cos(wt), A cos(wt+phi) would still satisfy the same differential equation, so A cos(wt+phi) would also be a solution. But that phi is critical because we need that to satisfy the initial condition. To motivate this idea, consider these two different systems here, where in one of them, at t=0, when we start the system, the mass is at x=A, so the spring is stretched, whereas in the second one, at t=0, the spring is compressed, so the system starts at x=-A. So I would have written exactly the same differential equation for both systems, because Newton’s law reads exactly the same, and I would have found the same equation. Now, for my solution to be a satisfactory answer for both of these problems, the solution has to satisfy the initial conditions, that is at t=0, the first one must be at x=A, and the second system must be at x=-A. To guarantee that, I need this phi, and phi would be zero for the first one, and at t=0, x=A, for the second one, phi would be pi, so that at t=0, x=-A.

Energy of the Spring-Mass System We know enough to discuss the mechanical energy of the oscillating mass on a spring. x(t) = A cos ( wt + f) If x(t) is displacement from equilibrium, then potential energy is U(t) = ½ k x(t)2 = ½ k A2 cos2 ( wt + f) v(t) = dx/dt  v(t) = A w (-sin ( wt + f)) And so the kinetic energy is just ½ m v(t)2 K(t) = ½ m v(t)2 = ½ m (Aw)2 sin2 ( wt + f) Finally, a(t) = dv/dt = -2A cos(t + )

Energy of the Spring-Mass System x(t) = A cos( t +  ) v(t) = -A sin( t +  ) a(t) = -2A cos( t +  ) Potential energy of the spring is U = ½ k x2 = ½ k A2 cos2(t + ) The Kinetic energy is K = ½ mv2 = ½ m(A)2 sin2(t+f) And w2 = k / m or k = m w2 K = ½ k A2 sin2(t+f)

Energy of the Spring-Mass System So E = K + U = constant =½ k A2 At maximum displacement K = 0 and U = ½ k A2 and acceleration has it maximum At the equilibrium position K = ½ k A2 = ½ m v2 and U = 0 E = ½ kA2 U~cos2 K~sin2

SHM So Far The most general solution is x = A cos(t + ) where A = amplitude  = (angular) frequency  = phase constant For SHM without friction, The frequency does not depend on the amplitude ! This is true of all simple harmonic motion! The oscillation occurs around the equilibrium point where the force is zero! Energy is a constant, it transfers between potential and kinetic

The “Simple” Pendulum A pendulum is made by suspending a mass m at the end of a string of length L. Find the frequency of oscillation for small displacements. S Fy = may = T – mg cos(q) ≈ m v2/L S Fx = max = -mg sin(q) If q small then x  L q and sin(q)  q dx/dt = L dq/dt ax = d2x/dt2 = L d2q/dt2 so ax = -g q = L d2q / dt2 and q = q0 cos(wt + f) with w = (g/L)½ z y  L x T m mg

The shaker cart You stand inside a small cart attached to a heavy-duty spring, the spring is compressed and released, and you shake back and forth, attempting to maintain your balance. Note that there is also a sandbag in the cart with you. At the instant you pass through the equilibrium position of the spring, you drop the sandbag out of the cart onto the ground. What effect does jettisoning the sandbag at the equilibrium position have on the amplitude of your oscillation? It increases the amplitude. It decreases the amplitude. It has no effect on the amplitude. Hint: At equilibrium, both the cart and the bag are moving at their maximum speed. By dropping the bag at this point, energy (specifically the kinetic energy of the bag) is lost from the spring-cart system. Thus, both the elastic potential energy at maximum displacement and the kinetic energy at equilibrium must decrease

The shaker cart Instead of dropping the sandbag as you pass through equilibrium, you decide to drop the sandbag when the cart is at its maximum distance from equilibrium. What effect does jettisoning the sandbag at the cart’s maximum distance from equilibrium have on the amplitude of your oscillation? It increases the amplitude. It decreases the amplitude. It has no effect on the amplitude. Hint: Dropping the bag at maximum distance from equilibrium, both the cart and the bag are at rest. By dropping the bag at this point, no energy is lost from the spring-cart system. Therefore, both the elastic potential energy at maximum displacement and the kinetic energy at equilibrium must remain constant.

The shaker cart What effect does jettisoning the sandbag at the cart’s maximum distance from equilibrium have on the maximum speed of the cart? It increases the maximum speed. It decreases the maximum speed. It has no effect on the maximum speed. Hint: Dropping the bag at maximum distance from equilibrium, both the cart and the bag are at rest. By dropping the bag at this point, no energy is lost from the spring-cart system. Therefore, both the elastic potential energy at maximum displacement and the kinetic energy at equilibrium must remain constant.