ONLINE: MATHEMATICS EXTENSION 2 Topic 6 MECHANICS 6.3 HARMONIC MOTION

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Gwendolyn Harrell
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ONINE: MATHEMATICS EXTENSION Topic 6 MECHANICS 6.3 HARMONIC MOTION Vibrations or oscillations are motions that repeated more or less reularly in time.

propaation, electromanetic waves, AC currents and voltaes.

1 ONINE: MATHEMATICS EXTENSION Topic 6 MECHANICS 6.3 HARMONIC MOTION Vibrations or oscillations are motions that repeated more or less reularly in time. The topic is very broad and diverse and covers phenomena such as mechanical vibrations (swinin pendulums, motion of a piston in a cylinder and vibrations of strins, rods, plates), sound, wave propaation, electromanetic waves, AC currents and voltaes. Vibrations or oscillations are periodic if the values of physical quantities describin the motion are repeated in successive equal time intervals. The period T of vibration or oscillations is the minimum time interval in which all the physical quantities characterizin the motion are repeated. Thus, the period T is the time interval for one full vibration or cycle. The frequency f of periodic vibration is the number of vibration made per second. (1) f 1 1 T T f VIEW some reat animations on oscillation (UNSW Physclips) What do these fiures tell you about vibrations? physics.usyd.edu.au/teach_res/math/math.htm mec63 1

2 SIMPE HARMONIC MOTION The simplest type of periodic motion is called simple harmonic motion (SHM). The displacement of a particle eecutin SHM alon the X ais is iven by the sinusoidal function () cos( t ) is the manitude of the imum displacement from the equilibrium position ( = 0). is a positive number and is called the displacement amplitude t called the phase anle [radians] t is the time [s] is the anular frequency [rad.s -1 ] is the initial phase anle (value of the phase anle at t = 0) [rad]. Its value determines the initial displacement of the particle t = 0, cos( ) (3) f T Equation () can also be written as: A sine function sin( t ') A sine and cosine function Acos( t) Bsin( t) The values of the constants,,, A and B can be determined from the initial conditions ( and v at time t = 0). physics.usyd.edu.au/teach_res/math/math.htm mec63

= 1 m = 0 rad T = 0 s f = 0.05 Hz = 0.314 rad.

314 rad.s -1 = 1 m = / rad T = 0 s f = 0.

s -1 = 1 m = rad T = 0 s f = 0.s -1 = 0.

75 m = - / 4 rad T = 40 s f = 0.05 Hz = 0.

3 = 1 m = 0 rad T = 0 s f = 0.05 Hz = rad.s -1 = 1 m = - / rad T = 0 s f = 0.05 Hz = rad.s -1 = 1 m = / rad T = 0 s f = 0.05 Hz = rad.s -1 = 1 m = rad T = 0 s f = 0.05 Hz = rad.s -1 = 0.5 m = / 4 rad T = 0 s f = 0.05 Hz = rad.s -1 = 0.75 m = - / 4 rad T = 40 s f = 0.05 Hz = rad.s -1 physics.usyd.edu.au/teach_res/math/math.htm mec63 3

4 The velocity v is the time derivative of the displacement (4) d d v cos( t ) dt dt v sin( t ) v v sin( t ) v The displacement and velocity are / rad out of phase with each other 0 v v v 0 The velocity amplitude is v (always a positive number) The acceleration a is the time derivative of the velocity (5) dv d d dt dt dt a v sin( t ) a t cos( ) a a cos( t ) a a The acceleration amplitude is a (always a positive number) The displacement and acceleration are rad out of phase with each other 0 a 0 a a The acceleration is always in the opposite direction to the displacement ecept at the equilibrium position ( = 0 a = 0) and direction of the acceleration is directed towards the equilibrium position. physics.usyd.edu.au/teach_res/math/math.htm mec63 4

5 acceleration a velocity v position 10 SHM time t Since a the equation of the motion of the particle eecutin simple harmonic motion is (6) d 0 0 dt Another approach to the mathematical analysis of SHM is to start with the equation of motion. d dv a v dt d The equation of motion then becomes v dv d We can interate this equation 1 v dv d v C ' v C where C and C are constants which are determined from the initial conditions (t = 0). physics.usyd.edu.au/teach_res/math/math.htm mec63 5

6 Take the initial conditions to be t 0 v 0 0 C C Therefore, the equation for the velocity as a function of displacement is (7) v Accordin to Newton s Second aw, an acceleration results from a non-zero resultant force actin on an object (8) 1 a m i F i For SHM, the force actin on the particle is (9) F m The resultant force F is always in the same direction as the acceleration a. The force responsible for SHM is called the restorin force and is always directed towards the equilibrium position ( = 0) and is proportional to the displacement. physics.usyd.edu.au/teach_res/math/math.htm mec63 6

Eample (Syllabus) The deck of a ship was.4 m below the level of the wharf at low tide and 0.6 m above the level at hih tide. ow tide was at 8:30 am and hih tide was at.35 pm.

7 Eample (Syllabus) The deck of a ship was.4 m below the level of the wharf at low tide and 0.6 m above the level at hih tide. ow tide was at 8:30 am and hih tide was at.35 pm. Find when the deck was level with the wharf, if the motion was simple harmonic. Solution The most important part of answerin this question is constructin a ood scientific diaram of the physical situation. Take the initial conditions at the 8:30 am low tide t = 0 s v = 0 m.s -1 = - = -1.5 m The displacement as a function of time is cos( t ) At time t = 0 cos( ) cos( ) 1 rad Hence cos( t ) cos( t) physics.usyd.edu.au/teach_res/math/math.htm mec63 7

The time interval from low tide to hih tide is half-period T/ 8:30 am to :35 pm t = 6 hours 5 minutes = (6)(60)(60)+(5)(60) s = 1900 s period T = 43800 s anular velocity = 1.434510-4 rad.

8 The time interval from low tide to hih tide is half-period T/ 8:30 am to :35 pm t = 6 hours 5 minutes = (6)(60)(60)+(5)(60) s = 1900 s period T = s anular velocity = rad.s -1 T We want to find the time when the deck is level with the wharf t =? s = 0.9 m position of wharf above equilibrium position ( = 0) cos( t) cos( t) t acos acos t t = s = h = 4 h 17 m The deck will be level with the wharf at time 1:47 pm physics.usyd.edu.au/teach_res/math/math.htm mec63 8

SIMPE PENDUUM A simple pendulum is a particle of mass m suspended from a fied point by a weihtless, inetensible strin of lenth. It swins in a vertical plane.

9 SIMPE PENDUUM A simple pendulum is a particle of mass m suspended from a fied point by a weihtless, inetensible strin of lenth. It swins in a vertical plane. The forces actin on the particle are the ravitational force FG and the strin tension F T. For small anle deviations from the vertical, the motion of the particle is approimately SHM. Applyin Newton s Second aw to the particle of mass m (8) 1 a m i F i We assume that the amplitude of the oscillation is small such that the resultant force only acts in the X direction F F F F F 0 G T y The assumption that F y = 0 is only valid for small anles ( < ~15 o ) and the followin predictions do not ive ood areement with measurements for lare amplitude oscillations. Addin the components in the Y direction ives m FT cos m FT cos Addin the components in the X direction ives m F FT sin sin m tan tan cos m F physics.usyd.edu.au/teach_res/math/math.htm mec63 9

10 Therefore the acceleration a in the X direction is (10) a valid only for small values of But the acceleration a is opposite in direction to the displacement and proportional to the displacement, therefore, the motion of the particle is SHM. Equation (5) for the acceleration of a particle eecutin SHM is (5) a Comparin equations (10) and (5), the anular frequency must be (6) hence, the period T of vibration and frequency f are (7) T 1 f valid only for small values of The period T, frequency f and anular velocity only depend upon the lenth of the pendulum s strin and the acceleration due to ravity, they do not depend upon the mass m of the particle or the amplitude of oscillation. d Be careful not to think that as in rotational (circular motion). Here is the dt anle of the pendulum at any instant. We now use not as the rate at which the anle chanes, but rather as constant related to the period. T physics.usyd.edu.au/teach_res/math/math.htm mec63 10

11 Eample Consider a simple pendulum of lenth of the pendulum are t = 0 = 0 v =.. The initial conditions for the vibration (a) Find the first value of where v = 0 by solvin the equation of motion for the vibration of the pendulum. (b) The motion of the pendulum may more accurately be represented by the equation of motion 3 5 a 6 10 Use this equation to find a more accurate answer for than in part (a). Solution (a) Initial conditions t = 0 = 0 v = Final conditions =? v = 0 Equation of motion d dv dt d a v Rearranin the equation of motion v dv d Interatin this equation and usin the initial condition and final conditions ives 0 v dv d v 0 Note: the initial conditions ive the lower limits and the final values ive the upper limits for the interations physics.usyd.edu.au/teach_res/math/math.htm mec63 11

12 (b) dv a v d v dv d We need to find the value of. We can find by usin Newton s Method Newton s Method is a method for findin successively better approimations to the roots (or zeroes) of a real-valued function f(). =? f() = 0 We bein with a first uess 1 for a root of the function f(). Then a better estimate of the root is approimated by f ( ) d '( ) ( ) 1 1 f f f '( 1) d The process is repeated as f ( n) d n 1 n f '( ) f ( ) f '( ) d until a sufficiently accurate value is reached. n et f ( ) We can replace by a variable z where k z k z1 1 k k f ( z) k k k k f '( z) k physics.usyd.edu.au/teach_res/math/math.htm mec63 1

13 Take the first uess the solution iven in part (a) 1 z1 1 f( z ) z z z f '( z1) k k 30 k k f (1) k k k k 10 k 0 k k f '(1) z k k 0 k k k k k k 1 z k z k k k hopefully answer is correct but it seems rather complicated check carefully physics.usyd.edu.au/teach_res/math/math.htm mec63 13

Another look at the PENDUUM The displacement of the pendulum alon the arc is iven by must be in radians The restorin force F (force actin so that 0) is the component of the ravitational force (weiht

14 Another look at the PENDUUM The displacement of the pendulum alon the arc is iven by must be in radians The restorin force F (force actin so that 0) is the component of the ravitational force (weiht FG FG m ) tanent to the arc of the circle F m sin where the minus sin means that the restorin force F is in the direction opposite to the anular displacement. F is proportional to sin and not itself, hence the motion of the pendulum is not simple harmonic motion. However, for small anular displacements ( < 15 o ), the difference between the anle (in radians) and sin is less than 1%. < 15 o sin Therefore, for small anle oscillations of the pendulum F m the motion can be rearded as SHM. Usin the fact that the arc lenth is iven by acceleration can be epressed as m F F m a a, the restorin force and physics.usyd.edu.au/teach_res/math/math.htm mec63 14

15 Aain, we have the followin relationships Equation (5) for the acceleration of a particle eecutin SHM is (5) a Comparin equations (10) and (5), the anular frequency must be (6) hence, the period T of vibration and frequency f are (7) T 1 f For non-shm the acceleration a of the pendulum is a sin We can epand the function sin in terms of the variable sin 3! 5! 7! Therefore, we can epress the acceleration as a 3! 5! 7! It is now obvious that for small a SHM physics.usyd.edu.au/teach_res/math/math.htm mec63 15

16 physics.usyd.edu.au/teach_res/math/math.htm mec63 16

f 1. (8.1.1) This means that SI unit for frequency is going to be s 1 also known as Hertz d1hz

f 1. (8.1.1) This means that SI unit for frequency is going to be s 1 also known as Hertz d1hz ecture 8-1 Oscillations 1. Oscillations Simple Harmonic Motion So far we have considered two basic types of motion: translational motion and rotational motion. But these are not the only types of motion