Physics 1501: Lecture 25, Pg 1 Physics 1501: Lecture 25 Todays Agenda l Homework #9 (due Friday Nov. 4) l Midterm 2: Nov. 16 l Topics Review of static equilibrium Oscillation Simple Harmonic Motion masses on springs Energy of the SHO Physics 1501: Lecture 25, Pg 2 Approach to Statics: l In general, we can use the two equations to solve any statics problems. l When choosing axes about which to calculate torque, we can be clever and make the problem easy.... Physics 1501: Lecture 25, Pg 3 Lecture 25, Act 1 Statics l A box is placed on a ramp in the configurations shown below. Friction prevents it from sliding. The center of mass of the box is indicated by a white dot in each case. In which cases does the box tip over ? (a) all (b) 2 & 3 (c) 3 only 1 2 3 Physics 1501: Lecture 25, Pg 4 Lecture 25, Act 1 Solution l We have seen that the torque due to gravity acts as though all the mass of an object is concentrated at the center of mass. l Consider the bottom right corner of the box to be a pivot point. l If the box can rotate in such a way that the center of mass is lowered, it will ! Physics 1501: Lecture 25, Pg 5 Lecture 25, ACT 1 Solution l We have seen that the torque due to gravity acts as though all the mass of an object is concentrated at the center of mass. l Consider the bottom right corner of the box to be a pivot point. l If the box can rotate in such a way that the center of mass is lowered, it will ! Physics 1501: Lecture 25, Pg 6 Lecture 25, Act 1 Addendum l What are the torques ?? (where do the forces act ?) rGrG mgmg gg switches sign at critical point rfrf f f always zero goes to zero at critical point rNrN N NN Physics 1501: Lecture 25, Pg 7 New topic (Ch. 13) Simple Harmonic Motion (SHM) l We know that if we stretch a spring with a mass on the end and let it go the mass will oscillate back and forth (if there is no friction). l This oscillation is called Simple Harmonic Motion, and is actually very easy to understand... k m k m k m Physics 1501: Lecture 25, Pg 8 SHM Dynamics Fa l At any given instant we know that F = ma must be true. l But in this case F = -kx and ma = l So: -kx = ma = k x m F F = -kxa a differential equation for x(t) ! Physics 1501: Lecture 25, Pg 9 SHM Dynamics... Try the solution x = Acos( t) this works, so it must be a solution ! define Physics 1501: Lecture 25, Pg 10 SHM Solution l We just showed that (which came from F=ma) has the solution x = Acos( t). This is not a unique solution, though. x = Asin( t) is also a solution. l The most general solution is a linear combination of these two solutions! x = Bsin( t)+ Ccos( t) ok Physics 1501: Lecture 25, Pg 11 Derivation: x = Acos( t+ ) is equivalent to x = Bsin( t)+ Ccos( t) x = Acos( t+ ) = Acos( t) cos - Asin( t) sin where C = Acos( ) and B = Asin( ) It works! = Ccos( t) + Bsin( t) We want to use the most general solution: So we can use x = Acos( t+ ) as the most general solution! Physics 1501: Lecture 25, Pg 12 SHM Solution... Drawing of Acos( t ) l A = amplitude of oscillation T = 2 / A A Physics 1501: Lecture 25, Pg 13 SHM Solution... Drawing of Acos( t + ) Physics 1501: Lecture 25, Pg 14 SHM Solution... Drawing of Acos( t - /2) A = Asin( t) ! Physics 1501: Lecture 25, Pg 15 What about Vertical Springs? l We already know that for a vertical spring if y is measured from the equilibrium position l The force of the spring is the negative derivative of this function: l So this will be just like the horizontal case: -ky = ma = j k m F= -ky y = 0 Which has solution y = Acos( t + ) where Physics 1501: Lecture 25, Pg 16 Velocity and Acceleration k x m 0 Position: x(t) = Acos( t + ) Velocity: v(t) = - Asin( t + ) Acceleration: a(t) = - 2 Acos( t + ) by taking derivatives, since: x MAX = A v MAX = A a MAX = 2 A Physics 1501: Lecture 25, Pg 17 Lecture 25, Act 2 Simple Harmonic Motion l A mass oscillates up & down on a spring. Its position as a function of time is shown below. At which of the points shown does the mass have positive velocity and negative acceleration ? t y(t) (a) (b) (c) Physics 1501: Lecture 25, Pg 18 Example l A mass m = 2kg on a spring oscillates with amplitude A = 10cm. At t=0 its speed is maximum, and is v = +2 m/s. What is the angular frequency of oscillation ? What is the spring constant k ? k x m = Also: k = m 2 So k = (2 kg) x (20 s -1 ) 2 = 800 kg/s 2 = 800 N/m v MAX = A Physics 1501: Lecture 25, Pg 19 Initial Conditions k x m 0 Use initial conditions to determine phase ! Suppose we are told x(0) = 0, and x is initially increasing (i.e. v(0) = positive): x(t) = Acos( t + ) v(t) = - Asin( t + ) a(t) = - 2 Acos( t + ) sin cos x(0) = 0 = Acos( ) = /2 or - /2 v(0) > 0 = - Asin( ) < 0 = - /2 So Physics 1501: Lecture 25, Pg 20 Initial Conditions... k x m 0 x(t) = Acos( t - /2 ) v(t) = - Asin( t - /2 ) a(t) = - 2 Acos( t - /2 ) So we find = - /2 !! x(t) = Asin( t) v(t) = Acos( t) a(t) = - 2 Asin( t) tt x(t)A -A Physics 1501: Lecture 25, Pg 21 Lecture 25, Act 3 Initial Conditions l A mass hanging from a vertical spring is lifted a distance d above equilibrium and released at t = 0. Which of the following describe its velocity and acceleration as a function of time (upwards is positive y direction): k m y 0 d (a) v(t) = - v max sin( t) a(t) = -a max cos( t) (b) v(t) = v max sin( t) a(t) = a max cos( t) (c) v(t) = v max cos( t) a(t) = -a max cos( t) (both v max and a max are positive numbers) t = 0 Physics 1501: Lecture 25, Pg 22 Energy of the Spring-Mass System We know enough to discuss the mechanical energy of the oscillating mass on a spring. Kinetic energy is always K = 1/2 mv 2 K = 1/2 m (- Asin( t + )) 2 We also know what the potential energy of a spring is, U = 1/2 k x 2 U = 1/2 k (Acos( t + )) 2 x(t) = Acos( t + ) v(t) = - Asin( t + ) a(t) = - 2 Acos( t + ) Remember, Physics 1501: Lecture 25, Pg 23 Energy of the Spring-Mass System Add to get E = K + U 1/2 m ( A) 2 sin 2 ( t + ) + 1/2 k (Acos( t + )) 2 Remember that U~cos 2 K~sin 2 E = 1/2 kA 2 so, E = 1/2 kA 2 sin 2 ( t + ) + 1/2 kA 2 cos 2 ( t + ) = 1/2 kA 2 [ sin 2 ( t + ) + cos 2 ( t + )] = 1/2 kA 2 Physics 1501: Lecture 25, Pg 24 SHM So Far The most general solution is x = Acos( t + ) where A = amplitude = frequency = phase constant l For a mass on a spring The frequency does not depend on the amplitude !!! We will see that this is true of all simple harmonic motion ! l The oscillation occurs around the equilibrium point where the force is zero! Physics 1501: Lecture 25, Pg 25 The Simple Pendulum l 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. L m mg z Physics 1501: Lecture 25, Pg 26 The Simple Pendulum... Recall that the torque due to gravity about the rotation (z) axis is = -mgd. d = Lsin L for small so = -mg L L d m mg z where Differential equation for simple harmonic motion ! = 0 cos( t + ) But = I I = mL 2 Physics 1501: Lecture 25, Pg 27 Lecture 25, Act 4 Simple Harmonic Motion l You are sitting on a swing. A friend gives you a small push and you start swinging back & forth with period T 1. l Suppose you were standing on the swing rather than sitting. When given a small push you start swinging back & forth with period T 2. Which of the following is true: (a) T 1 = T 2 (b) T 1 > T 2 (c) T 1 < T 2 Physics 1501: Lecture 25, Pg 28 The Rod Pendulum l 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. L mg z x CM Physics 1501: Lecture 25, Pg 29 The Rod Pendulum... The torque about the rotation (z) axis is = -mgd = -mg{L/2}sin -mg{L/2} for small l In this case L d mg z L/2 x CM where d I So = I becomes Physics 1501: Lecture 25, Pg 30 Lecture 25, Act 25 Period (a)(b)(c) l What length do we make the simple pendulum so that it has the same period as the rod pendulum? LRLR LSLS