Physics 207, Lecture 18, Nov. 3 Goals: Chapter 14 Interrelate the physics and mathematics of oscillations. Draw and interpret oscillatory graphs. Learn the concepts of phase and phase constant. Understand and use energy conservation in oscillatory systems. Understand the basic ideas of damping and resonance. Phase Contrast Microscopy Epithelial cell in brightfield (BF) using a 40x lens (NA 0.75) (left) and with phase contrast using a DL Plan Achromat 40x (NA 0.65) (right). A green interference filter is used for both images. 1

Physics 207, Lecture 18, Nov. 3 Assignment HW8, Due Wednesday, Nov. 12th Wednesday: Read through Chapter 15.4 1

Periodic Motion is everywhere Examples of periodic motion Earth around the sun Elastic ball bouncing up an down Quartz crystal in your watch, computer clock, iPod clock, etc.

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 What is the frequency? f = 1 / T = 1.17 Hz

1. A position of equilibrium A special kind of period oscillator: Harmonic oscillator What do all “harmonic oscillators” have in common? 1. A position of equilibrium 2. A restoring force, which must be linear [Hooke’s law spring F = -kx (In a pendulum the behavior only linear for small angles: sin θ where θ = s / L) ] In this limit we have: F = -ks with k = mg/L) 3. Inertia 4. The drag forces are reasonably small

Simple Harmonic Motion (SHM) In Simple Harmonic Motion the restoring force on the mass is linear, that is, exactly proportional to the displacement of the mass from rest position Hooke’s Law : F = -kx If k >> m  rapid oscillations <=> large frequency If k << m  slow oscillations <=> low frequency

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 -A A 0(≡Xeq)

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 …. 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 This oscillation is called Simple Harmonic Motion k m -A A 0(≡Xeq)

Simple Harmonic Motion (SHM) The time it takes the block to complete one cycle is called the period. Usually, the period is denoted 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 Note in the (x,t) graph that the vertical axis represents the x coordinate of the oscillating object, and the horizontal axis represents time. Which points on the x axis are located a displacement A from the equilibrium position ? A. R only B. Q only C. both R and Q Position time

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 time

Simple Harmonic Motion Now assume that the t coordinate of point K is 0.0050 s. What is the period T , in seconds? T = 0.02 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 = 0.01 s time

Simple Harmonic Motion Now assume that the x coordinate of point R is 0.12 m. What distance d does the object cover during one period of oscillation? d = 0.48 m What distance d does the object cover between the moments labeled K and N on the graph? d = 0.36 m time

At any given instant we know that F = ma must be true. SHM Dynamics At any given instant we know that F = ma must be true. But in this case F = -k x and ma = So: -k x = ma = k x m F = -k x a a differential equation for x(t) ! “Simple approach”, guess a solution and see if it works!

   T = 2/ SHM Solution... Try either cos (  t ) or sin (  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 ] Both sin and cosine work so need to include both    T = 2/ A

Combining sin and cosine solutions x(t) = B cos wt + C sin wt = A cos ( wt + f) = A (cos wt cos f – sin wt sin f ) =A cos f cos wt – A sin f sin wt) Notice that B = A cos f C = -A sin f  tan f= -C/B k x m    cos sin Use “initial conditions” to determine phase  !

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 = 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 = (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 + ) Kinetic energy is always K = ½ mv2 = ½ m(A)2 sin2(t+f) Potential energy of a spring is, U = ½ k x2 = ½ k A2 cos2(t + ) And w2 = k / m or k = m w2 U = ½ m w2 A2 cos2(t + )

Energy of the Spring-Mass System x(t) = A cos( t +  ) v(t) = -A sin( t +  ) a(t) = -2A cos(t + ) And the mechanical energy is K + U =½ m w2 A2 cos2(t + ) + ½ m w2 A2 sin2(t + ) K + U = ½ m w2 A2 [cos2(t + ) + sin2(t + )] K + U = ½ m w2 A2 = ½ k A2 which is constant

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 (or minimum) 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

What about Vertical Springs? For a vertical spring, if y is measured from the equilibrium position Recall: force of the spring is the negative derivative of this function: This will be just like the horizontal case: -ky = ma = j k y = 0 F= -ky m Which has solution y(t) = A cos( t + ) where

S Fy = mac = T – mg cos(q) = m v2/L S Fx = max = -mg sin(q) The Simple Pendulum (In class torques, t = Ia, were used instead but the results are the same) 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 = mac = 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  L d2q / dt2 - g q = 0 and q = q0 cos(wt + f) or q = q0 sin(wt + f) with w = (g/L)½ z y  L x T m mg

Velocity and Acceleration Position: x(t) = A cos(t + ) Velocity: v(t) = -A sin(t + ) Acceleration: a(t) = -2A cos(t + ) by taking derivatives, since: xmax = A vmax = A amax = 2A k x m

Physics 207, Lecture 18, Nov. 3 Assignment HW8, Due Wednesday, Nov. 12th Wednesday: Read through Chapter 15.4 The rest are for Wednesday, plus damping, resonance and part of Chapter 15. 1

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.

Exercise Simple Harmonic Motion A mass oscillates up & down on a spring. It’s position as a function of time is shown below. At which of the points shown does the mass have positive velocity and negative acceleration ? Remember: velocity is slope and acceleration is the curvature y(t) (a) (c) t (b)

Example A mass m = 2 kg on a spring oscillates with amplitude A = 10 cm. At t = 0 its speed is at a maximum, and is v=+2 m/s What is the angular frequency of oscillation  ? What is the spring constant k ? General relationships E = K + U = constant, w = (k/m)½ So at maximum speed U=0 and ½ mv2 = E = ½ kA2 thus k = mv2/A2 = 2 x (2) 2/(0.1)2 = 800 N/m, w = 20 rad/sec k x m

Home Exercise Simple Harmonic Motion You are sitting on a swing. A friend gives you a small push and you start swinging back & forth with period T1. Suppose you were standing on the swing rather than sitting. When given a small push you start swinging back & forth with period T2. Which of the following is true recalling that w = (g/L)½ (A) T1 = T2 (B) T1 > T2 (C) T1 < T2

Home Exercise Simple Harmonic Motion You are sitting on a swing. A friend gives you a small push and you start swinging back & forth with period T1. Suppose you were standing on the swing rather than sitting. When given a small push you start swinging back & forth with period T2. If you are standing, the center of mass moves towards the pivot point and so L is less, w is bigger, T2 is smaller (A) T1 = T2 (B) T1 > T2 (C) T1 < T2

BTW: The Rod Pendulum (not tested) A pendulum is made by suspending a thin rod of length L and mass M at one end. Find the frequency of oscillation for small displacements. S tz = I a = -| r x F | = (L/2) mg sin(q) (no torque from T) -[ mL2/12 + m (L/2)2 ] a  L/2 mg q -1/3 L d2q/dt2 = ½ g q z T  x CM L mg

BTW: General Physical Pendulum (not tested) Suppose we have some arbitrarily shaped solid of mass M hung on a fixed axis, that we know where the CM is located and what the moment of inertia I about the axis is. The torque about the rotation (z) axis for small  is (sin    )  = -MgR sinq  -MgR z-axis R    x CM Mg where  = 0 cos(t + )

Torsion Pendulum Consider an object suspended by a wire attached at its CM. The wire defines the rotation axis, and the moment of inertia I about this axis is known. The wire acts like a “rotational spring”. When the object is rotated, the wire is twisted. This produces a torque that opposes the rotation. In analogy with a spring, the torque produced is proportional to the displacement:  = - k  where k is the torsional spring constant w = (k/I)½ I wire  

Home Exercise Period All of the following torsional pendulum bobs have the same mass and w = (k/I)½ Which pendulum rotates the fastest, i.e. has the longest period? (The wires are identical) R R R R (A) (B) (C) (D)

Reviewing Simple Harmonic Oscillators k x m F = -kx a Spring-mass system Pendula General physical pendulum Torsion pendulum where z-axis x(t) = A cos( t + ) R   = 0 cos( t + ) x CM Mg I wire  

Energy in SHM For both the spring and the pendulum, we can derive the SHM solution using energy conservation. The total energy (K + U) of a system undergoing SMH will always be constant! This is not surprising since there are only conservative forces present, hence energy is conserved. -A A x U K E

SHM and quadratic potentials SHM will occur whenever the potential is quadratic. For small oscillations this will be true: For example, the potential between H atoms in an H2 molecule looks something like this: -A A x U K E U x