In PHYS 111/121 developed equation of motion for constant acceleration πβ π‘ πβ (remember vector so acceleration constant in all direction) π₯ π₯ π£ π‘ π£ π£ ππ‘ π£ π£ 1 ππ‘ 2 2π π₯ π₯ What happens if πβ not constant => the equations do not apply Many real objects behave this way ο· ο· ο· ο· leaf blowing in the wind wind up cars teacher (me!) pendulum (velocity keeps changing direction) Let’s discuss a special case of nonβconstant acceleration: Simple Harmonic motion – special kind of periodic motion What is periodic motion ο· ο· motion that repeats itself at regular intervals Example o tapping finger to music o steady walk Periodic motion that moves back and forth over the same path is oscillatory motion – has 2 precise end points Example – piston in a car engine ο· ο· ο· Good windshield wipers Periodic motion is hard to get in realβlife friction can damp the motion o brings end points close together Called a damped oscillation – will not deal with this in class “least complicated periodic motion” Simple harmonic motion (SHM) Simple harmonic oscillation (SHO) Period T is the repeat time π₯ π‘ π΄πππ ωπ‘ describes SHM A β amplitude of motion (m) π (omega) angular frequency of motion (rad/sec) π 2ππ ο· ο· f frequency of motion (cycle/sec or 1/sec or Hz (hertz)) How many times a second the motion repeats T period of motion = how long it takes one full cycle to complete (sec) π 1 π So ωπ ω 2π π 2π 2ππ What type of force gives SHM? ο· Linear restoring force Consider mass spring system on frictionless surface Unstretched spring Mass at equilibrium position ο· ο· x=0 F=0 Unstretched spring Pull mass a distance to the right Spring force pulls mass to the left towards equilibrium ο· ο· x>0 F<0 Pull mass to the left Spring force pushes mass to the right towards equilibrium ο· ο· x<0 F>0 This is a restoring force – wants to restore position to equilibrium Hooke’s law πΉ ππ₯ negative sign means it’s a restoring force Linear in position (not x or √x) Today we want position as a function of time i.e. what is the position at any given time? 2 ways of doing this 1. Cheating (like the book does with the circle) see appendix at the end 2. The true honest method Use calculus to go answer Consider mass spring system Force not constant so can’t use equations of motion like last semester πΉ mπ πΉ mπ ππ₯ π ⇒ N ππ mg m 0 so N mg ππ π π₯ π Get acceleration as a function of position What have we done so far Stated SHM goes as a cosine wave Stated SHM requires linear restoring force really want to prove this Non constant acceleration so need to try other method Remember (from calculus) If position is π₯ π‘ π₯ Velocity π£ Acceleration a Can write from force equations (see above) ππ π ππ₯ π π π Define ω π/π π π₯ π π π₯ ππ‘ ω π₯ Differential equation which as solution π₯ π‘ π΄ πππ ωπ‘ θ π π ω ν π‘ ππ₯ ππ‘ π΄ω sin ωπ‘ π π‘ πν ππ‘ π΄ω cos ωπ‘ θ θ A amplitude of motion x cannot be greater than A or smaller than – A ω is angular frequency of motion (how many radius per sec the motion goes) ω π π π 1 π 2π π π 1 π 2 ππ 2π π π θ phase constant depends on initial conditions (when the stopwatch starts) 0) θ Do you pull the mass and release (θ Do you give the mass a kick at equilibrium θ Do you pull the mass to the left (βA) and release θ For this class x π‘ Asin ωt or x π‘ What do these functions look like π Acos ωt will not consider different initial conditions X = A max ν π΄ ππ π₯ π΄ω 0 πππ’πππππππ’π ν π΄ 0 π΄ωππ 0 π πππππ ππ πππ’πππππππ’π π₯ π΄ πππ₯ ν 0 π π΄ω πππ₯ π₯ 0 ν π π΄ω πππ₯ 0 π πππππ ππ‘ πππ’πππππππ’π π₯ π΄ πππ₯ ν 0 π π΄ω (max, spring fully compressed) π₯ 0 ν π΄ω (max) π΄ 0 π πππππ ππ‘ πππ’πππππππ’π How about vertical mass spring Drawing on the left: Unstretched spring Middle drawing: Equilibrium with mass hanging at equilibrium πΉ mg mg kΔπΏ 0 k ΔπΏ Drawing at the right, spring stretched some arbitrary amount πΉ ππ πΉ ππ mg k ΔπΏ π₯ ma mg kΔπΏ ππ₯ ma π π π₯ π Just like before ω π π Just new equilibrium position Now consider energy Kinetic energy is ½ππ£ πΎπΈ ππΈ Potential energy stored in spring ½ππ₯ If we had vertical mass spring also need to consider gravitational potential energy ππΈ (using as equilibrium position the position of the mass hanging on spring with no motion) Initial energy = final energy = total energy KπΈ ½ππ£ PπΈ KπΈ ½ππ₯ PπΈ E ½ππ£ ½ππ₯ πΈ πππ₯ What is total energy? When mass is at the end point of motion a distance A from equilibrium So total energy πΈ πΎπΈ ππΈ ½ππ΄ ½π 0 ½ππ΄ and this is conserved So π‘ππ‘ππ ππππππ¦ ½ππ΄ ½ππ£ πΎπΈ ½ππ₯ Textbook solves for ν from forces get π ππΈ π΄ –π₯ π₯ and the other equations of motion Example 1 A spring has an unstretched length of 20cm. It stretches 5cm after a 200g weight is placed hanging vertically from the spring. The spring is then used in a horizontal springβmass oscillator with mass 300 g. The mass is pulled 3 cm from equilibrium and released. Calculate π, f, T, A. What equations describe π π , π π , π π ? What is total energy of the system? What is the maximum velocity? What is the displacement at π ππ seconds? Consider forces acting on 200 gm mass. ππ₯ π π π π π π₯ π π 0.2 ⋅ 9.8 0.05 0 39.2 π/π Mass pulled 3 cm = 0.03 m and released, so furthest displacement is 0.03 m. So amplitude is 0.03 m. A = 0.03 m . Angular frequency, π . Frequency π 1 π ππππππ π . 1 1.82 11.4 rad/s 1.82π 0.55 π Equations defining motion π₯ π‘ π΄ cos ππ‘ 0.03 cos 11.4π‘ π π‘ π΄π π ππ ππ‘ 0.03 ⋅ 11.4π ππ11.4π‘ π π‘ π΄ω πππ ω π‘ 0.03 11.4 πππ 11.4π‘ 0.342π ππ 11.4π‘ 3.9 πππ 11.4π‘ total energy E = KE + PE (no gravity so ignore mgh) 1 ππ£ 2 E 1 ππ₯ 2 Easy to calculate if one of v or x is 0. Say v =0, this occurs when mass is at an extreme end of the motion, ie x = A So πΈ π 0 ½ 3.92 0.03 0.0176 π½ Remember m, kg, s ππ΄ Maximum velocity occurs when displacement = 0 (i.e. x = 0) 1 ππ£ 2 πΈ 1 ππ₯ 2 Get max when x = 0 ππ£ So πΈ 0.0176 ⋅ 0.3 ν 0.342π/π Solving gives ν OR From equations of motion ωπ΄ ν 11.4 ⋅ 0.03 0.342 π/π Displacement a t = 10 sec π₯ π‘ 0.03πππ 11.4π‘ 0.03πππ 11.4 ⋅ 10 0.0186 π If you get β0.012, remember deg vs rad. Example 2 The motion of a 1 kg mass attached to a spring is described by x(t) = 0.3 cos(5t) What is the total energy of the system? Remember general form of equation of motion π₯ π‘ π΄πππ ππ‘ so A = 0.3 m and π 5 rad/sec and remember m=1kg 2 ways to solve this this πΈ πΎπΈ ππΈ 1 ππ₯ 2 1 πν 2 Easy to calculate when ν max (x = 0) E ν 1 m ν 2 ωπ΄ So πΈ π ωπ΄ 1 ⋅ 1 5 ⋅ 0.3 2 = 1.125 J OR If at xmax ( π πΈ 0 ππ π₯ π΄ 1 π π₯ 2 1 ππ΄ 2 ω π π π π π πΈ ω ππ΄ (same equation as before ω π = 1.125 J Example 3: A bullet of mass 10 g travels at 600 m/s before striking (and embedding itself) into a 1 kg mass attached to a horizontal spring with spring constant 25 N/m. What is the amplitude of the resulting simple harmonic motion? Calculate frequency of oscillation. What is frequency if the bullet has mass 1 gm? A little different as the simple harmonic motion starts with velocity at the equilibrium position, NOT by pulling spring and releasing From drawings 1 to 2: Conservation of momentum Initial momentum = final momentum π ν π 0 0.01 600 so π£ π ν π 0.01 . 1 π£ 5.94π/π . total energy of the SHO (i.e. at 2) is ½ππ₯ We can calculate the energy just after the collision and then the πΈ πΎπΈ ππΈ ½ππ£ energy at the amplitude position to calculate A Or ωπ΄ We can use ν ω π π π£ π£ . ω 2π 4.975 πππ/π ππ 5.94 . π΄ f 25 1.01 4.976 2π 1.19π 0.79π ππ If mass changes to 1 g, what happens? Amplitude changes and frequency changes to (no need to calculate all the momentum stuff as the frequency does not depend on the amplitude) F= = . = 0.795 secβ1
