PH 1-1D Spring 013 Oscillations Lectures 35-37 Chapter 15 (Halliday/Resnick/Walker, Fundaentals of Physics 9 th edition) 1
Chapter 15 Oscillations In this chapter we will cover the following topics: Displaceent, velocity and acceleration of a siple haronic oscillator Energy of a siple haronic oscillator Exaples of siple haronic oscillators: spring-ass syste, siple pendulu, physical pendulu, torsion pendulu Daped haronic oscillator Forced oscillations/resonance
Oscillations Periodic otion the otion of a particle or a syste of particles is periodic, or cyclic, if it repeats again and again at regular intervals of tie. Exaple: The orbital otion of a planet The unifor rotational otion of a phonograph turntable Back and forth otion of a piston in an autoobile engine Vibrations of a guitar string Oscillation back and forth or swinging periodic otion is called an oscillation Siple Haronic Motion Siple haronic otion is a special kind of one diensional periodic otion The particle oves back and forth along a straight line, repeating the sae otion again and again Siple haronic otion the particles position can be expressed as a cosine or a sine function of tie. Cosines and sines are called haronic functions => we call otion of the particle haronic. 3
When an object attached to a horizontal spring is oved fro its equilibriu position and released, the restoring force F = -kx leads to siple haronic otion Moving strip of paper at a steady rate Record position of the vibrating object vs. tie pen A = aplitude the axiu displaceent fro equilibriu x = 0 is the equilibriu position of the object. Position as a function of tie has the shape of trigonoetric sine or cosine function 4
Siple Haronic Motion ( SHM) In fig.a we show snapshots of a siple oscillating syste. The otion is periodic i.e. it repeats in tie. The tie needed to coplete one repetition is known as the period (sybol T, units: s ). The nuber of 1 repetitions per unit tie is called the frequency (sybol f, unit hertz, Hz) f T The displaceent of the particle is given by the equation: xt () x cos t Fig.b is a plot of xt ( ) versus t. The quantity x is called the aplitude of the otion. It gives the axiu possible displaceent of the oscillating object The quantity is called the angular frequency of the oscillator. It is given by the equation: f 5 T xt () x cost
xt () x cost The quantity is called the phase angle of the oscillator. The value of is deterined fro the displaceent x(0) and the velocity v(0) at t 0. In fig.a x( t) is plotted versus t for 0. x( t) x Velocity of SHM cost dx() t d vt () x cos t x sin t dt dt The quantity x is called the velocity aplitude v It expresses the axiu possible value of vt ( ) In fig.b the velocity vt ( ) is plotted versus t for 0. vt () x sint dv() t d Acceleration of SHM: at () xsin t xcos t x dt dt The quantity xis called the acceleration apli tue d a.it expresses the axiu possible value of a( t). In fig.c the acceleration a( t) is plotted versus t for 0. at x t 6 () cos
The Force Law for Siple Haronic Motion We saw that the acceleration of an object undergoing SHM is: If we apply Newton's second law we get: F a x x Siple haronic otion occurs when the force acting on an object is proportional to the disaplaceent but opposite in sign. The force can be written as: F Cx where C is a constant. If we copare the two expressions for F we have: C and T C a x Consider the otion of a ass attached to a spring of srping constant k than oves on a frictionless horizontal floor as shown in the figure. The net force F on is given by Hooke's law: F kx. If we copare this equation with the expression F Cx we identify the constant C with the sping constant k. We can then calculate the angular frequancy and the period T. C k and T C k k T k
Proble : When an object of ass 1 is hung on a vertical spring and set into vertical siple haronic otion, its frequency is 1.0 Hz. When another object of ass is hung on the spring along with 1, the frequency of the otion is 4.00 Hz. Find the ratio / 1 of the asses. 8
Three springs with force constants k 1 = 10.0 N/, k = 1.5 N/, and k 3 = 15.0 N/ are connected in parallel to a ass of 0.500 kg. The ass is then pulled to the right and released. Find the period of the otion. 9
A ass of 0.300 kg is placed on a vertical spring and the spring stretches by 10.0 c. It is then pulled down an additional 5.00 c and then released. Find: a) K; b) ω; c) frequency; d) T; e) ax velocity; f) a ax ; g) F ax (ax restoring force); h) V for x =.00 c; i) The equations for displaceent, velocity and acceleration at any tie 10
11
Proble 4 1
Energy in Siple Haronic Motion The echanical energy E of a SHM is the su of its potential and kinetic energies U and K. 1 1 Potential energy U kx kx cos t 1 1 1 k Kinetic energy K v xsin t xsin t 1 1 Mechanical energy E U K kx cos t sin t kx In the figure we plot the potential energy U (green line), the kinetic energy K (red line) and the echanical energy E (black line) versus tie t. While U and K vary with tie, the energy E is a constant. The energy of the oscillating object transfers back and forth between potential and kinetic energy, while the su of the two reains constant 13
Proble 35. A block of ass M=5.4 kg, at rest on a horizontal frictionless table, is attached to a rigid support by a string of constant k=6000 N/. A bullet of ass =9.5 g and velocity v of agnityde 630 /s strikes and is ebedded in the block. Assuing the copression of the spring is negligible untill the bullet is ebedded, deterine (a) the speed of the block iidiately after the collison, and (b) the aplitude of the resulting siple haronic otion The proble consists of two distinct parts: the copletely inelastic collision (which is assued to occur instantaneously, the bullet ebedding itself in the block before the block oves through significant distance) followed by siple haronic otion (of ass + M attached to a spring of spring constant k). (a) Moentu conservation readily yields v = v/( + M). With = 9.5 g, M = 5.4 kg and v = 630 /s, we obtain v ' 1.1 /s. (b) Since v occurs at the equilibriu position, then v = v for the siple haronic otion. The relation v = x can be used to solve for x, or we can pursue the alternate (though related) approach of energy conservation. Here we choose the latter: which siplifies to 1 1 1 v 1 M v kx M kx ' M x v k 3 (9.510 kg)(630 /s) 3.3 10. 3 M (6000 N/)(9.510 kg 5.4kg) 14
An Angular Siple Haronic Oscillator; Torsion Pendulu In the figure we show another type of oscillating syste It consists of a disc of rotational inertia I suspended fro a wire that twists as rotates by an angle. The wire exerts on the disc a restoring torque This is the angular for of Hooke's law. The constant is called the torsion constant of the wire. If we copare the expression for the torque with the force equation F Cx we realize that we identify the constant C with the torsion constant. We can thus readily deterine the angular frequency and the period T of the C I I oscillation. T I I C We note that I is the rotational inertia of the disc about an axis that coincides with the wire. The angle is given by the equation: () t cost 15
The Siple Pendulu A siple pendulu consists of a particle of ass suspended by a string of length L fro a pivot point. If the ass is disturbed fro its equilibriu position the net force acting on it is such that the syste exectutes siple haronic otion. There are two forces acting on : The gravitational force and the tension fro the string. The net torque of these forces is: rf Lg sin Here is the angle that the thread g akes with the vertical. If 1 (say less than 5 ) then we can ake the following approxiation: sin where is expressed in radians. With this approxiation the torque is: Lg If we copare the expression for with the force equation F Cx we realize that we identify the constant C with the ter Lg. We can thus readily deterine the angular frequency C gl I I and the period T of the oscillation. ; T I I C gl
In the sall angle approxiation we assued that << 1 and used the approxiation: sin We are now going to decide what is a sall angle i.e. up to what angle is the approxiation reasonably accurate? (degrees) (radians) sin 5 0.087 0.087 10 0.174 0.174 15 0.6 0.59 (1% off) 0 0.349 0.34 (% off) Conclusion: If we keep < 10 we ake less that 1 % error 17
The rotational inertia I about the pivot point is equal to L T L g T I gh I L Thus T gl gl Physical Pendulu A physical pendulu is an extened rigid body that is suspended fro a fixed point O and oscillates under the influence of gravity The net torque ghsin Here h is the distance between point O and the center of ass C of the suspended body. If we ake the sall angle approxiation 1, we have: gh. If we copare the torques with the force equation F Cx we realize that we identify the constant C with the ter hg. We can thus readily deterine period T of the oscillation. I I T Here I is the rotational C gh inertia about an axis through O. 18
Exaple 1: The pendulu can be used as a very siple device to easure the acceleration of gravity at a particular location. easure the length l of the pendulu and then set the pendulu into otion easure T by a clock For a unifor rod of length L 1 1 1 I I h L L L 1 3 co ( ) I I 1/3L L L T C gh gh 3 g( L /) 3g 8 L g 3T 19
Exaple : The Length of a Pendulu. A student is in an epty roo. He has a piece of rope, a sall bob, and a clock. Find the volue of the roo. 1. Fro the piece of rope and a bob we ake a siple pendulu. We set pendulu into otion 3. We easure period T by a clock 4. We calculate the length of the pendulu (rope) 5. With a help of the rope of the known length we easure the diensions of the roo a x b x h and its volue V = a x b x h 0
Pr oble 44. A physical pendulu consists of a eter stick that is pivoted at a sall hole drilled through the stick a distance d fro the 50 c ark. The period of oscillation is.5 s. Find d? T I gd We use Eq. 15-9 and the parallel-axis theore I = I c + h where h = d, the unknown. For a eter stick of ass, the rotational inertia about its center of ass is I c = L /1 where L = 1.0. Thus, for T =.5 s, we obtain T L / 1 d gd L 1gd d g. Squaring both sides and solving for d leads to the quadratic forula: d 4 g T/ g T/ L /3 =. Choosing the plus sign leads to an ipossible value for d (d = 1.5 L). If we choose the inus sign, we obtain a physically eaningful result: d = 0.056. 1
Siple Haronic Motion and Unifor Circular Motion Consider an object oving on a circular path of radius with a unifor speed v. If we project the position of the oving particle at point P on thex-axis we get point P. The coordinate of P is: xt ( ) x cos t. While point P executes unifro circular otion its projection P oves along the x-axis with siple haronic otion. The speed v of point P is equal to x. The direction of the velocity vector is along the tangent to the circular path. If we project the velocity v on the x-axis we get: vt ( ) xsin t The acceleration a points along the center O. If we project a along the x-axis we get: a( t) x cos t Conclusion: Whether we look at the displaceent,the velocity, or the acceleration, the projection of unifor circular otion on the x-axis is SHM x
Daped Siple Haronic Motion When the aplitude of an oscillating object is reduced due to the presence of an external force the otion is said to be daped. An exaple is given in the figure. A ass attached to a spring of spring constant k oscillates vertically. The oscillating ass is attached to a vane suberged in a liquid. The liquid exerts a daping force F d whose agnitude is given by the equation: bv The negative sign indicates that Fd opposes the otion of the oscillating ass. The pareter b is called the daping constant. The net force on is: F kxbv Fro Newton's second la w we have: net dx d x kx bv a We substitute v with and a with and get dt dt the following differential equation: d x dx b kx 0 dt dt Fd 3
Newton's second law for the daped haronic oscillator: d x dx b kx0 The solution has the for: dt dt bt / xt () xe cos t In the picture above we plot x( t) versus t. We can regard the above solution as a cosine function with a tie-dependent aplitude xe bt /. The angular frequency of the daped haronic oscillator is given by the equation: k b 1 = For an undaped haronic oscillator the energy E kx 4 If the oscillator is daped its energy is not constant but decreases with tie. If the Et bt / daping is sall we can replace x with x e By doing so we find that: 1 bt / ( ) kxe The echanical energy decreases exponentially with tie 4
Proble 59. The aplitude of a lightly daped oscillator decreases by 3.0% during each cycle. What percentage of the echanical energy of the oscillator is lost in each cycle? 1 E U K kx Since the energy is proportional to the aplitude squared (see Eq. 15-1), we find the fractional change (assued sall) is E E E de E dx = = xdx dx =. x x x Thus, if we approxiate the fractional change in x as dx /x, then the above calculation shows that ultiplying this by should give the fractional energy change. Therefore, if x decreases by 3%, then E ust decrease by 6.0 %. 5
Moving support Forced Oscillations and Resonance If an oscillating syste is disturbed and then allowed to oscillate freely the corresponding angular frequency is called the natural frequency. The sae syste can also be driven as shown in the figure by a oving support that oscillates at an arbitrary angular frequency. Such a forced oscillator oscillates at the angular frequency of the driving force. The displaceent is given by: xt () x co t The oscillation aplitude x varies s with the driving frequency as shown in the lower figure. The aplitude is approxiatetly greatest when This condition is called resonance. All echanical structures have one or ore natural frequencies and if a structure is subjected to a strong external driving force whose frequency atches one of the natural frequencies, the resulting oscillations ay daage the structure 6 d d d
Proble 63. A 100 kg car carrying four 8 kg people travels over a "washboard" dirt road with corrugations 4.0 apart. The car bounces with axiu aplitude when its speed is 16 k/h. When the car stops, and the people get out, how uch does the car body rise on its suspension? With M = 1000 kg and = 8 kg, we adapt Eq. 15-1 to this situation by writing k k T M 4. If d = 4.0 is the distance traveled (at constant car speed v) between ipulses, then we ay write T = d/v, in which case the above equation ay be solved for the spring constant: v k v = km 4. d M 4 d Before the people got out, the equilibriu copression is x i = (M + 4)g/k, and afterward it is x f = Mg/k. Therefore, with v = 16000/3600 = 4.44 /s, we find the rise of the car body on its suspension is = 4 = 4 g g d xi xf = 0.050. k M+4 v F H I K 7
Proble 6. Two springs are joined in series and connected to a block of ass 0.45 kg that is set oscillating over a frictionless floor. The springs each have spring constant k=6430 N/. What is the frequency of oscillations? We wish to find the effective spring constant for the cobination of springs. We do this by finding the agnitude F of the force exerted on the ass when the total elongation of the springs is x. Then k eff = F/x. Suppose the left-hand spring is elongated by x and the right-hand spring is elongated by x r. The left-hand spring exerts a force of agnitude kx on the right-hand spring and the right-hand spring exerts a force of agnitude kx r on the left-hand spring. By Newton s third law these ust be equal, so x x r. The two elongations ust be the sae and the total elongation is twice the elongation of either spring: x x. The left-hand spring exerts a force on the block and its agnitude is F kx. Thus keff kx / xr k /. The block behaves as if it were subject to the force of a single spring, with spring constant k/. To find the frequency of its otion replace k eff in f 1/ k / a f eff with k/ to obtain f = 1 k. With = 0.45 kg and k = 6430 N/, the frequency is f = 18. Hz. 8