CHAPTER 15: Vibratory Motion

CHAPTER 15: Vibratory Motion courtesy of Richard White courtesy of Richard White 2.) 1.)

Two glaring observations can be ade fro the graphic on the previous slide: 1.) The PROJECTION of a point on a circle oving with constant angular velocity ω follows the sae path as a ass attached to an ideal hanging spring as the spring oscillates up and down. And: 2.) If you trac the oscillation in tie, it traces out a sinusoidal path. Both of these observations fall out fro the ath if we start with Newton s Second Law applied to a ass attached to an ideal spring oscillating over a frictionless, horizontal surface. x That analysis follows: x = 0 2.)

Keeping the sign of the acceleration ebedded (it will be either positive or negative, depending upon the point in tie, so we ll leave it iplicit), and noting that the spring force in the x- direction on a ass attached to the x = 0 spring will be F spring = ( x) î (Hooe s Law), Newton Second Law yields: F x : x = a x = d2 x dt 2 Pulling the x to the right side and dividing by yields the relationship: x d 2 x dt + 2 x = 0 This is the characteristic equation of SIMPLE HARMONIC MOTION. 3.)

This relationship: d 2 x dt + 2 x = 0 x essentially ass us to find a function x such that when you tae its second derivative and add it to a constant ties itself, the su will always add to zero. x = 0 The function that does this is either a cosine or a sine (I usually use a sine, but your boo for no particularly good reason uses a cosine, so we ll use that). There are restrictions on the cosine function we need. In fact, we want the ost general for possible. Specifically: 1.) We need the cosine s angle to be tie dependent, so instead of using an angle θ, we will use a constant ties t, where the constants units have to be radians/second. As we have already run into a variable with those units ( ω), we will use that sybol. (Interesting note: If the angular velocity of the rotating point-on-the-circle shown in the first slide had been ω, the constant in question for the vibratory otion s cosine function would have been that sae nuber ω.) 4.)

2.) We need the ability to start the cloc when we want. A siple cosine function sets the position to be at a positive axiu at t = 0 (see setch). We need to be able to shift the axis by soe phase shift aount φ, essentially starting the cloc (i.e., setting t = 0) when the body is at any chosen x-coordinate. 3.) Lastly, we need to be able to accoodate otion whose axiu displaceent is other than one. cos( ωt) at t=0, x=ax pos. displ. cos( ωt + φ) φ at t=0, x-coord. shifted new axis The function that does all of this for us is: x = Acos( ωt + φ) 5.)

So bac to the proble at hand. Does x = Acos( ωt + φ) x satisfy d 2 x dt + 2 x = 0? x = 0 The only way to tell is to try it out: dx dt d( Acos ( ωt + φ ) ) = dt = ωasin( ωt + φ) This, by the way, is the velocity function. And as a sine function can never be larger than one, this eans the agnitude of the axiu velocity for this oscillatory otion will be: v ax = ωa This will happen when the force is copletely spent, or at equilibriu. 6.)

Continuing: d 2 x dt 2 d( ωasin ( ωt + φ ) ) = dt = ω 2 Acos( ωt + φ) x = 0 x Another side point: This eans the agnitude of the axiu acceleration, which happens at the extrees where the spring force is axiu, will be: a ax = ω 2 A Plugging everything in: d 2 x dt + 2 x = 0 ω 2 Acos( ωt + φ) + Acos( ωt + φ) = 0 ω 2 + = 0 ω = 1 2 7.)

In other words, the differential equation d 2 x dt + 2 x = 0 x is satisfied by the position function x = Acos( ωt + φ) x = 0 as long as the angular frequency ω satisfies: ω = 1 2 Big Note: Notice that in concluding that ω =, we are saying that the angular frequency of our oscillating syste is equal to the square root of the constant that sits in front of the position ter in the Newton s Second Law equation! Put a little differently, if you can get a N.S.L. evaluation into the for: acceleration + constant ( )( position) = 0 you will now the oscillation is siple haronic in nature AND you will now that the angular frequency of the syste is ω = constant. ( ) 1 2 1 2 8.)

Minor point: So what is the angular frequency ω really doing for us? It is siply another way to identify how quicly the syste is oscillating bac and forth. But instead of telling us the frequency ν in cycles per second, it is telling us how any radians being swept through per cycle. Noting that there are 2π radians per cycle, the relationship between frequency and angular frequency is: ω = 2πν x = 0 x And as the frequency ν in cycles per second is the inverse of the nuber of seconds required to traverse one cycle (or the period T in seconds per cycle), we can also write: T = 1 ν In short, if we can derive an expression for ω for a syste, we also now ν and T. 9.)

Energy in a Spring Syste x At the extrees where the velocity is zero and acceleration a axiu: F ax = A, where A is the axiu displaceent (the aplitude), so: E total = 1 2 v2 + 1 2 A2 0 x = 0 = 1 2 A2 At the equilibriu where the acceleration is zero and velocity a axiu: E total = 1 2 v 2 ax + 1 2 x2 = 1 2 ( ωa)2 0 10.)

Suary of Relationships Relationships always true: d 2 x dt + ( κ )x = 0 or α + ( κ )θ = 0 2 ω = κ ω = 2πν characteristic equation for siple haronic otion ( ) 1 2 angular frequency fro characteristic equation angular frequency and frequency related T = 1 ν x = Acos( ωt + φ) period inversely related to frequency position function for s.h.. v = dx dt v ax = ωa and a = dv dt = d2 x dt 2 velocity and acceleration functions axiu velocity (happens at equilibriu) a ax = ω 2 A axiu acceleration (happens at extrees) 11.)

Suary of Relationships For a spring: d 2 x dt + 2 ω = x = 0 1 2 angular frequency fro characteristic equation characteristic equation for siple haronic otion E tot = 1 2 A2 total echanical energy in syste The period of oscillation for a spring is constant no atter what the spring s aplitude. How so? A larger displaceent will require ore distance traveled to execute a single cycle, but because force is a function of displaceent, it will also generate a larger axiu force, so the period will stay the sae no atter what! 12.)

REALLY SIMPLE Exaple 1: A spring with spring constant is hung fro the ceiling. A ass =.4 g is attached and allowed to gently elongate the spring until it coes to rest at a point.6 eters below its free-hanging position. The ass is then pulled an additional.2 eters down and released. a.) What is the spring s spring constant? The spring constant identifies how uch force is required to displaceent the spring one eter (units: N/). It easures, essentially, stiffness. We ve been told that a ass, whose weight is g, displaced the spring a distance d =.6 eters, so we now: = F x = g d = ( ).6 (.4 g) 9.8 /s2 ( ) = 6.53 N/ 13.)

b.) What is the aplitude of the resulting oscillation? Being elongated an additional.2 eters fro equilibriu eans will oscillate about the equilibriu point.2 eters above and below, which eans the otion s aplitude is.2 eters. c.) What is the otion s angular frequency? Knowing and, we can write: d.) What is the otion s frequency? ω = = ω = 2πν ( 6.53 N/).4 g ( ) = 4.04 rad/sec ν = ω 4.04 rad/sec = 2π 2π =.643 cycles/sec (or Hertz) 14.)

e.) What is the period of the otion? T = 1 ν = 1.643 cycles/sec = 1.56 sec/cycle f.) What is the axiu velocity? v ax = ωa = 4.04 rad/sec =.81 /s ( )(.2 ) g.) Where is the velocity a axiu? at equilibriu f.) What is the axiu acceleration? a ax = ω 2 A ( ) 2 (.2 ) = 4.04 rad/sec = 3.26 /s 2 g.) Where is the acceleration a axiu? at the extrees 15.)

h.) How uch echanical energy is in the syste? E = 1 2 A2 = 1 2 ( 6.53 N/ )(.2 )2 =.13 joules i.) How would this proble have changed if the spring had been inverted? it wouldn t 16.)

Physical Characteristics of Springs Consider a spring of spring constant that has a ass attached to it. What happens to the spring constant if: a.) You cut the spring in half? It doubles for each new spring, but how so? Assue the ass elongates the spring a distance d down to its equilibriu position. In that case, g = d If we put a ar halfway down the spring and thin of it as two springs, one on top of the other, with each having its own new spring constant 1, and if we notice that the force on the upper spring has to be the sae as the force on the lower spring (note that this is called a series cobination and is associated with situations in which the stress is the sae for all parts involved) we can write: d 1 d/2 1 d/2 g = 1 y 1 d = 1 2 = d = 2 1 17.)

a.) What happens if you cut the spring in half? (con t.) Another way to loo at it is through energy. The spring potential energy for an elongated spring stretched fro its equilibriu position is: U = 1 2 d2 If we thin of the spring at two springs attached one on top of the other, the energy content won t change and we can write: d U = 1 2 d2 = 1 2 d 1 2 + 1 2 1 d 2 d 2 = 2 1 4 1 = 2 2 d 2 2 1 d/2 b.) What happens if you double the spring s length? 1 d/2 Playing off the reasoning fro above, the spring constant would halve. 18.)

a.) What happens if you put two springs side by side? This is a parallel cobination where the stress is distributed between the ebers, and in it the effective spring constant is the su of the individual spring constants involved. How so? 1 2 d F net = F 1 +F 2 = 1 y + 2 y g = ( 1 + 2 )d new = 1 + 2 19.)

Exaple 2: A siple pendulu of length L is observed as shown to the right. What is its period of otion? If we can show that this syste s N.S.L. expression confors to siple haronic otion, we have it. As the otion is rotational, we need to su torques about the pivot point. The torque due to the tension is zero. Noting that r-perpendicular for gravity is Lsinθ, we can write: τ pin : g ( ) Lsinθ ( ) = I piin α d 2 θ dt + 2 g L = ( L 2 ) d2 θ dt 2 sinθ = 0 This isn t quite the right for, but if we tae a sall angle approxiation, we find that for θ <<, sinθ θ and we can write: d 2 θ dt + 2 g L θ = 0 pin i L g θ T L r = Lsinθ g 20.)

With d 2 θ dt + 2 g L θ = 0 We can see we are dealing with siple haronic otion, which eans: ω = g 1 2 L and ω = 2πν θ L and ν = ω 2π ν = 1 2π g L 1 2 T = 1 ν T = 2π L 1 2 g And wasn t that fun... 21.)

Exaple 3: A physical pendulu of length L taes the for of a bea pinned at one end as shown to the right. What is its period of otion? Again, starting with N.S.L.: τ pin : ( ) g L 2 sinθ = I piin α d 2 θ dt 2 + = 1 3 L2 d2 θ dt 2 3g 2L sinθ = 0 pin i L 2 r = L 2 sinθ center of ass g θ g L Again, this isn t quite the right for, but if we tae a sall angle approxiation, we find that for θ <<, sinθ θ and we can write: d 2 θ dt 2 + 3g 2L θ = 0 22.)

With d 2 θ dt 2 + 3g 2L θ = 0 We can see we are dealing with siple haronic otion, which eans: ω = 3g 1 2 2L and ω = 2πν θ L and ν = ω 2π ν = 1 2π 3g 2L 1 2 T = 1 ν T = 2π 2L 1 2 3g And wasn t THAT fun... 23.)

Exaple 4: An oddball physical pendulu consists of a dis of ass and radius R with a wad of ass.2 attached at its edge as shown. What is its period of otion? Using the parallel axis theore on the dis and the definition of oent of inertia for a point ass on the wad, we can deterine the net oent of inertia about the pin as: I pin = ( I c/dis + M dis R 2 )+ wad ( 2R ) 2 = 1 2 R2 + R 2 +.2 = 2.3R 2 ( )( 2R ) 2 pin i ω θ d g wad w g 24.)

Taing the torque about the pin, yields: pin i ω τ pin : ( ) Rsinθ g ( ).2g ( )( 2Rsinθ ) = I pin α 1.4gRsinθ = ( 2.3R 2 ) d2 θ dt 2 d 2 θ dt +.61g 2 R sinθ = 0 pin i θ d g wad w g With θ <<, sinθ θ ω =.61g R and d 2 θ dt +.61g 2 R θ = 0 1 2 ν = ω 2π = 1 2π.61g R 1 2 R r = Rsinθ g T = 1 ν = 2π R.61g 1 2 25.)

Exaple 5--an old AP question: A spring whose force agnitude is! equal to F = x 3 is displaced a distance A and released whereupon its period is deterined to be T. If it is then displaced a distance 2A, will its period go up, go down or stay the sae? This is a very cool proble because it aes you thin about what you now, then extrapolate to this new situation. What you now! is that for a noral, Hooe s Law spring where the force agnitude is F = x, the period will be constant. That is, if the displaceent is sall, the force will be sall and it will tae soe aount of tie to cover one cycle. And if the displaceent is large, the proportionally larger force will allow the additional distance to be covered in the sae tie... hence a constant period. In this scenario, increasing the displaceent eans that the force will be larger, but not proportionally larger it will be GREATER than proportionally larger. With the larger than expected force, the ass should cover the ground in less tie than usual, eaning the period should diinish. Bizarre, but that the case. 26.)

Exaple 6: A great proble! A id jups into a tunnel drilled through the earth fro pole to pole. He accelerates toward the center of the earth, reaches that point whereupon he begins to negatively accelerate as he travels bac toward the surface on the other side of the planet. Once at the surface on the other side, he stops, then start bac toward the center of the earth again. In other words, the id oscillates bac and forth. His father, wanting to see his id occasionally, coisions a satellite to orbit in just the right circular path so as to be overhead every other tie the id s head eerges fro the! hole. Reebering that the force due to gravity inside a solid sphere is F g = G M R 3 (this was derived in the last chapter), what altitude above the earth s surface ust the satellite be to effect this feat? The ey to this proble is in finding the frequency (or period) of otion of the id s oscillation between the poles. This will be the sae as the period of the satellite s otion. So how do you deterine that period? If we could show that the id was executing siple haronic otion, coplete with its characteristic equation, we d be set. So let s try that! r ( ) r 27.)

Writing out Newton s Second Law on the id yields: F r : ( ) r = d 2 r dt 2 ( ) r = 0 G e R 3 d 2 r dt 2 + G e R 3 r This is the characteristic equation of S.H.M. (an acceleration plus a constant ties a position equals zero). That eans the angular frequency of the oscillation is: But: ( ω = G )1 2 e R 3 ω = 2πν = 2π T T = 2π ω = 2π ( G M ) 1 2 R 3 28.)

Knowing that: 2π T = ( G )1 2 e R 3 r we can deterine the satellite s velocity with twice that period (it sees the id every other pass) as: v = 2π ( R + r orbit ) 2T ( = R + r orbit ) 2 = 2 2π ( G e R 3 2π( R + r orbit ) )1 2 ( G e R 3 )1 2 29.)

We can use Newton s Second Law on the satellite: F r : G e s ( R + r orbit ) = 2 s v 2 ( R + r orbit ) G e s ( R + r orbit ) = 2 s G e s R + r orbit ( ) = 2 s ( ) 3 R 3 = R + r orbit 4 ( ) 1 3 R = R + r orbit r orbit = 0.59R 4 ( R + r orbit ) G 1 2 e 2 R 3 ( R + r orbit ) 2 ( R + r orbit ) G e 2 R 3 ( R + r orbit ) 2 r orbit 30.)