PHY183-Lecture46 - Review: Yesterday ! Hooke's Force Law: F...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Review: Yesterday ! Hooke's Force Law: F = !kx d2x ! Equation of motion: m 2 = !kx dt Physics for Scientists & Engineers 1 Spring Semester 2006 Lecture 46 ! Solution: or: ! Amplitude: Phase: d2x k + x=0 dt 2 m x(t) = Asin(! 0t) + B cos(! 0t) x(t) = C sin(! 0t + " 0 ) C = A2 + B2 ! 0 = arctan(B / A) k m 2 ! Angular Frequency: ! 0 = April 14, 2006 Physics for Scientists&Engineers 1 1 April 14, 2006 Physics for Scientists&Engineers 1 Initial Conditions ! How does one determine constants A and B, or C and !0? ! One uses initial conditions to determine these constants Initial position Example: Initial Conditions Question 1: ! A spring with spring constant 56.0 N/m has a mass of 1.00 kg attached to its end. The mass is pulled +5.5 cm from its equilibrium position and pushed so that it receives an initial velocity of 0.32 m/s. What is the equation of motion for this oscillation? x0 = x(t = 0) Initial velocity v0 = v(t = 0) = (dx / dt) t = 0 ! Two equations for two unknowns ! April 14, 2006 Physics for Scientists&Engineers 1 3 April 14, 2006 Physics for Scientists&Engineers 1 4 Example: Initial Conditions (2) Answer 1: ! Equation of motion x(t) = Asin(! 0t) + B cos(! 0t) with ! 0 = k/m Example: Initial Conditions (3) Question 2: ! What is the amplitude of this oscillation? Answer 2: ! Although we have the constants A and B, they are not the amplitude. ! Have to use: ! We can already calculate the angular frequency ! Find equation for velocity by taking the time derivate x(t) = Asin(! 0t) + B cos(! 0t) " v(t) = ! 0 A cos(! 0t) # ! 0 Bsin(! 0t) ! 0 = k / m = (56.0 N/m)/(1.00 kg) = 7.48 s-1 C = A 2 + B 2 = 0.0432 + 0.055 2 m = 0.070 m ! Please note that amplitude (7 cm) is bigger than initial elongation from equilibrium (5.5 cm) ! Initial push added to amplitude, even though it was toward the equilibrium position 5 April 14, 2006 Physics for Scientists&Engineers 1 6 ! At time 0, because sin(0)=0 and cos(0)=1, we have x0 = x(t = 0) = B " ' B = x0 = 0.055 m #%( v0 = v(t = 0) = ! 0 A $ ) A = v0 / ! 0 = &0.043 m April 14, 2006 Physics for Scientists&Engineers 1 Biography: Robert Hooke ! Robert Hooke (1635-1703) ! Contemporary of Newton ! Curator of experiments for Royal Society ! Anticipated universal gravitation ! Finest mechanic of his time ! Invented spiral spring in watches ! Most important publication: Micrographia (1665) ! Coined the term "cell" as smallest parts of living organisms Position, Velocity, Acceleration ! Position: ! Velocity: ! Acceleration: x(t) = C sin(! 0t + " 0 ) v(t) = ! 0C cos(! 0t + " 0 ) 2 a(t) = !" 0 C sin(" 0t + # 0 ) Note: 1. Phase shifts of 90 between x & v, v & a 2. Acceleration always in opposite direction to displacement 3. Ratio of amplitudes (v/x, a/v) is !0 April 14, 2006 Physics for Scientists&Engineers 1 7 April 14, 2006 Physics for Scientists&Engineers 1 8 Period and Frequency ! Sinusoidal functions are periodic ... adding 2" to the argument does not change the value of the function Period and Frequency for Spring ! Now we use our result for the mass on a spring, ! This time interval over which function repeats itself is called period, T 2" sin(! t) = sin(2" + ! t) = sin(! ( + t)) ! !0 = ! Period: k m T= ! Frequency, f: 2! " T= ! Frequency: 2! = "0 f = 2! m = 2! k k/m k m 1 f = T !0 1 = 2" 2" (So far this is just the same as for circular motion) ! This relationship was used for time-keeping in antique clocks (since R. Hooke, and before quartz) 9 April 14, 2006 Physics for Scientists&Engineers 1 10 April 14, 2006 Physics for Scientists&Engineers 1 Tunnel Through the Moon ! Suppose we would drill a tunnel straight through the center of the moon, from one end to the other ! The fact that the moon is composed of solid rock makes this scenario slightly more realistic than drilling a tunnel through the center of Earth ! Question: ! If we go to one end of this tunnel and release a steel ball of mass 5.0 kg from rest, what can we say about the motion of this ball? ! Answer: ! In our chapter on gravity, we have found that the magnitude of the gravitational force inside a spherical mass distribution of constant density and radius R is Tunnel Through the Moon (2) ! The gravitational force inside a homogeneous spherical mass distribution follows Hooke's Law F(x) = !kx with a "spring constant" given by k = mg / R ! Here g = gM is the gravitational acceleration experienced at the surface of the moon ! We start by calculating the acceleration of gravity on the surface of the moon MM = 7.35#1022 kg, RM = 1735 km = 1.735#106 m Fg (r < R) = mgr / R ! This force points toward the center of the moon, in the opposite direction from the displacement April 14, 2006 Physics for Scientists&Engineers 1 11 6.67 !10 "11 m 3kg -1s "2 7.35 !10 22 kg GM M gM = = = 1.63 m/s 2 2 2 RM 1.735 !10 6 m ( )( ) ( ) April 14, 2006 Physics for Scientists&Engineers 1 12 Tunnel Through the Moon (3) ! We have found that the solution to the equation of motion with a spring force is the oscillatory motion x(t) = Asin(! 0t) + B cos(! 0t) ! Releasing it from rest from the surface of the moon implies that A=0 and B=x0. Thus our equation of motion is in this case: x(t) = RM cos(! 0t) ! The angular frequency of the oscillation is: k gM 1.63 m/s 2 !0 = = = = 9.69 "10 #4 Hz m RM 1.735 "10 6 m ! Please note the mass of the steel ball turned out to be irrelevant. The period of this oscillation is 2! T= = 6485 s "0 April 14, 2006 Physics for Scientists&Engineers 1 13 Tunnel Through the Moon (4) ! Our steel ball will come to the surface at the other side of the moon 3242 s after release, and then oscillate back ! To transit the entire moon in a little less than an hour would make this mode of transportation extremely efficient ! The velocity of the steel ball during this oscillation would be Especially because no power supply would be needed v(t) = ! The maximum velocity would be reached as the ball crosses the center of the moon and would have the numerical value of dx = !" 0 RM sin(" 0t) dt vmax = ! 0 RM = (9.69 "10 #4 s-1 )(1.735 "10 6 m) = 1680 m/s = 3760 mph ! If the tunnel were big enough, the same motion could be obtained for a vehicle holding people, providing very efficient transportation to the other side of the moon without the need for propulsion ! During the entire journey the people inside the vehicle would feel absolute weightlessness! April 14, 2006 Physics for Scientists&Engineers 1 14 Relationship with Circular Motion (1) Relationship with Circular Motion (2) ! Cartesian projections of circular motion perform simple harmonic oscillations during circular motion with constant angular velocity ! In circular motion with constant angular velocity: ! (t) = " t + ! 0 x(t) = r cos(! t + " 0 ) y(t) = r sin(! t + " 0 ) April 14, 2006 Physics for Scientists&Engineers 1 15 April 14, 2006 Physics for Scientists&Engineers 1 16 Quiz 46.1 ! The driver of a car decides to smoke a cigar. ! Which window should be opened so that the passenger, sitting beside the driver, does not have to breathe the secondhand smoke? Pendulum Motion ! There is a second important system that oscillates with which we are all familiar: the pendulum ! Ideal pendulum: a thin, massless string with a massive object attached to its end ! Let's look at the equation of motion first. ! The differential equation for a ball on a rope of length l at an angle ! relative to the vertical is A) The driver-side window B) The passenger-side window d 2! g + !=0 dt 2 l (for small angles ! ) April 14, 2006 Physics for Scientists&Engineers 1 17 April 14, 2006 Physics for Scientists&Engineers 1 18 Pendulum Motion (3) ! We recognize that this equation is similar to our equation of motion for a mass on a spring d2x k + x=0 dt 2 m d 2! g + !=0 dt 2 l with position replaced with angle x !" and the angular frequency replaced by ! This differential equation has the solution k /m ! g/l ! (t) = Asin(" 0t) + B cos(" 0t) with " 0 = g / l April 14, 2006 Physics for Scientists&Engineers 1 19 ...
View Full Document

This note was uploaded on 03/19/2008 for the course PHY 183 taught by Professor Wolf during the Spring '08 term at Michigan State University.

Ask a homework question - tutors are online