Examples of oscillating motion are all around us… a child on a swing, a boat bobbing up and down in the ocean, a piston in the engine of a car, a vibrating.

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Presentation transcript:

Examples of oscillating motion are all around us… a child on a swing, a boat bobbing up and down in the ocean, a piston in the engine of a car, a vibrating guitar string, etc… If the “periodic” motion, when graphed, is a sinusoidal function of time, we call it ___________________________________. SIMPLE HARMONIC MOTION (SHM)

For example, in this series of “snapshots” taken at equal time intervals, it is evident that the particle is oscillating. A graph of the displacement as a function of time produces a sinusoidal function.

Definitions: Cycle – one complete oscillation (“round trip”) __________ (__) – number of cycles or oscillations completed per second UNITS: ____ or ______ (___) ________ (__) – the time for one cycle _________ (__) – the maximum displacement from equilibrium (labeled x m in the diagram below) T A frequency f 1/sHertz Hz Period T Amplitude A

SHM- Video Clip hysics.unsw.edu.au/mec hanics/chapter4_simpleh armonicmotion.html

A proof that the oscillating motion of an ideal spring on a frictionless surface is SHM… Note that when x (displacement from equilibrium) is +, a is ___, and vice-versa. Define new variable: -k x -

A sinusoidal function is the solution to the differential equation that had resulted from the analysis of Newton’s Laws, which proves, then, that the oscillating motion of an ideal spring is SHM! Phew… The solution is not limited to just the sin and cos functions. Any cosine function with a phase shift, , also works! (Cos function) (Sin function) So, in general…

Graphs of SHM: x max = ___v max = ____a max = ____ When displacement is a maximum, velocity is _____, and acceleration is ________ (most negative). When displacement is zero, speed is _________, and acceleration is _______. A AA AA zero minimum maximum zero

An alternative way to arrive at V max using energy conservation: Recall… So… An alternative way to arrive at a max using the relationship discovered earlier from F = ma:

How is  related to the period of the motion, T? Since one period later, the motion repeats itself… Since a cosine function repeats itself every 2  radians…  is known as the angular frequency in rad/s; f is the frequency in Hz (cycles/sec) Same relationship as between angular velocity, , and period for circular motion….

SHM- Video Clip s.physics.unsw.edu.a u/mechanics/chapter 4_simpleharmonicmo tion.htmlhttp:// s.physics.unsw.edu.a u/mechanics/chapter 4_simpleharmonicmo tion.html

Hmm… it seems, then, that simple harmonic motion is related to uniform circular motion… To see how, recall that Galileo observed the moons of Jupiter with his telescope. The moons seemed to be be moving back and forth with the center of the planet at the midpoint of the motion. The graph represents actual data obtained by Galileo in 1610 for the measured angle between Jupiter and its moon Calisto. The “best-fit curve” suggests simple harmonic motion!

Of course, we know that Callisto was actually moving in a circular orbit around Jupiter. So… it seems that UNIFORM CIRCULAR MOTION viewed “edge-on” is simply SIMPLE HARMONIC MOTION. “The projection of an object in uniform circular motion onto the diameter of the circle results in simple harmonic motion.” As P‘ moves in a circle with angular velocity, , the point, P moves in simple harmonic motion with angular frequency, .

Again, recall… The period is proportional to the ________, … Recall that the best-fit curve for T vs. m was _______ T = # m 0.5 How is the period related to the mass for a mass oscillating on a spring?

The graph, then, of _________ will be linear. In the lab analysis, you will use the slope of this graph to calculate the spring constant, k of the spring. HOW?.... Compare to… T vs. m 0.5 T m 0.5

SHM EXAMPLES: 1. An object oscillates in SHM with an amplitude of 2 m and a frequency of 1 Hz. 2 m- 2 mx = 0 (a) How long does it take to move from its center position to 1 m away? How do we know whether to use x = Acos(  t) or x = Asin(  t)? The sine function “begins timing” at x = __________, while the cosine function “begins timing” at the _______________________. So, in this case we should choose _____ function. 0 (center) maximum displacement sine

…. where  = 2  f = _________ Take the sin -1 of both sides….. Don’t forget to be in rad mode! 2  (1) = 2 

(b) How long does it take to move from 1 m away from center to 2 m away? 2 m- 2 mx = 0 Should we use x = Acos(  t) or x = Asin(  t)? Since the time to move from x = 1 m to 2 m is the ______ time as it would take to move from 2 m away to 1 m, we should choose the ______ function (it begins “timing” at the maximum displacement.) Take the ____ of both sides….. SAME cosine Cos -1

A couple alternative methods to solve the problem: Using …. If we plug in x = 2 into the equation, we will find the time to travel from x = ___ to x = ____. Since we know the time from x = 0 to x = 1, we can then _________ to find the answer. Using the definition of a period….. So, the time to go from 0 to 2 m, would be ____. Again… 0 2 subtract ¼ s 2 m- 2 mx = 0

a max = ____________________________ Example 2: In a linear electric shaver, the cutting blades move back and forth over a total distance of 2 mm (from one end to the other). The motion is SHM with a frequency of 60 Hz (AC voltage in the U.S.) (a) Find the blade’s maximum speed (at the __________ of its motion). A = ____________  = _________________ V max = _________________________ (b) Find the blade’s maximum acceleration (at the ______ of its motion) midpoint 1 mm =.001 m2  f = 2  (60) = 120  A  = (.001)(120  ) = 0.38 m/s ends A  2 = (.001)(120  ) 2 = 142 m/s 2

(c) Find the time for the blades to travel from one end to the other (a total distance of just 2 mm). METHOD 1: Using the definition of the period: The time that we want is _____ of the period. METHOD 2: Using the appropriate SHM position function: Remember that x is NOT the distance moved!! x=0x=1x=-1 half

IF  is SMALL  ….. _________ (in radians) For example, 5°=0.873 rad and sin 5° = So… the “restoring force”, F, can be approximated as…. The arc length that the pendulum “sweeps out” on its way to its lowest point is S = _____= ______. Solving for  …. So… the “restoring force”, F, is then… This is, in essence, Hooke’s Law…F= -Kx. A pendulum is like a “gravitational spring”! RR LL

We have already derived that for an oscillator that obeys Hooke’s Law... For the pendulum (“gravitational spring”), the “spring” constant, k, is now So, now…. The Period of a Pendulum:

If the angle is NOT small, the departure from SHM can be substantial. In that case, the period can be expressed as a nasty “infinite series”…. So, how small is “small”? < 15°…If  = 15°, the true period is longer than that given by by less than 0.5%! Precise measurements of a pendulum’s period are often made in order to find the acceleration of gravity, g, at a location. This information can be a predictor of oil or coal deposits (different densities will slightly affect the “local” value of g!)

Instead of measuring the length and period of just one pendulum, more precise results are obtained if length and period data is obtained for a number of different length pendulums. What graph could then be made that would lead to a calculation of g? T Or, a graph of T 2 vs L would work!