Chapter 14 - - Simple Harmonic Motion

Chapter 14 - SimpleHarmonic MotionA PowerPoint Presentation byPaul E. Tippens, Professor of PhysicsSouthern Polytechnic State University 2007

Photo by Mark TippensA TRAMPOLINE exerts a restoring force on thejumper that is directly proportional to the averageforce required to displace the mat. Such restoringforces provide the driving forces necessary forobjects that oscillate with simple harmonic motion.

Objectives: After finishing thisunit, you should be able to: Write and apply Hooke’s Law for objects movingwith simple harmonic motion. Write and apply formulas forfinding the frequency f, period T,velocity v, or acceleration a interms of displacement x or time t. Describe the motion of pendulumsand calculate the length requiredto produce a given frequency.

Periodic MotionSimple periodic motion is that motion in which abody moves back and forth over a fixed path,returning to each position and velocity after adefinite interval of time.1f rtzHertz(s(s-1))

Example 1: The suspended mass makes 30complete oscillations in 15 s. What is theperiod and frequency of the motion?15 sT 0.50 s30 cylcesxFPeriod:Period: TT 0.5000.500 ss11f T 0.500 sFrequency:Frequency: ff 2.002.00 HzHz

Simple Harmonic Motion, SHMSimple harmonic motion is periodic motion inthe absence of friction and produced by arestoring force that is directly proportional tothe displacement and oppositely directed.xFAA restoringrestoring force,force, F,F, actsacts ininthethe directiondirection oppositeopposite thethedisplacementdisplacement ofof thetheoscillatingoscillating body.body.FF -kx-kx

Hooke’s LawWhen a spring is stretched, there is a restoringforce that is proportional to the displacement.F -kxxmFThe spring constant k is aproperty of the spring given by:k F x

Work Done in Stretching a SpringWork done ON the spring is positive;work BY spring is negative.xFrom Hooke’s law the force F is:F (x) kxFFmTo stretch spring fromx1 to x2 , work is:Work ½ kx ½ kx22x1x221(Review module on work)

Example 2: A 4-kg mass suspended from aspring produces a displacement of 20 cm.What is the spring constant?The stretching force is the weight(W mg) of the 4-kg mass: 20 cmF (4 kg)(9.8 m/s2) 39.2 NFmNow, from Hooke’s law, the forceconstant k of the spring is:k F x 0.2 mkk 196196 N/mN/m

Example 2(cont.: The mass m is now stretcheda distance of 8 cm and held. What is thepotential energy? (k 196 N/m)The potential energy is equal tothe work done in stretching thespring:Work ½ kx ½ kx228 cm0m21U ½ kx ½(196 N/m)(0.08 m)2UU 0.6270.627 JJF2

Displacement in SHMxmx -Ax 0x A Displacement is positive when the position isto the right of the equilibrium position (x 0)and negative when located to the left. The maximum displacement is called theamplitude A.

Velocity in SHMv (- )v ( )mx -Ax 0x A Velocity is positive when moving to the rightand negative when moving to the left. It is zero at the end points and a maximumat the midpoint in either direction ( or -).

Acceleration in SHM a-x x-amx -Ax 0x A Acceleration is in the direction of therestoring force. (a is positive when x isnegative, and negative when x is positive.)F ma kx Acceleration is a maximum at the end pointsand it is zero at the center of oscillation.

Acceleration vs. Displacementavxmx -Ax 0x AGiven the spring constant, the displacement, andthe mass, the acceleration can be found from:F ma kx or kxa mNote: Acceleration is always opposite to displacement.

Example 3: A 2-kg mass hangs at the endof a spring whose constant is k 400 N/m.The mass is displaced a distance of 12 cmand released. What is the acceleration atthe instant the displacement is x 7 cm? kxa m (400 N/m)( 0.07 m)a 2 kg22aa -14.0m/s-14.0 m/samNote: When the displacement is 7 cm(downward), the acceleration is -14.0 m/s2(upward) independent of motion direction. x

Example 4: What is the maximum accelerationfor the 2-kg mass in the previous problem? (A 12 cm, k 400 N/m)The maximum acceleration occurswhen the restoring force is amaximum; i.e., when the stretch orcompression of the spring is largest.F ma -kxxmax A kA 400 N( 0.12 m)a 2 kgmMaximumAcceleration:m22aamax 24.0m/smax 24.0 m/s x

Conservation of EnergyThe total mechanical energy (U K) of avibrating system is constant; i.e., it is thesame at any point in the oscillating path.avxmx -Ax 0x AFor any two points A and B, we may write:22 ½kx 22 ½mv 22 ½kx 22½mv½mvAA ½kxAA ½mvBB ½kxBB

Energy of a Vibrating System:Axavmx -Ax 0Bx A At points A and B, the velocity is zero and theacceleration is a maximum. The total energy is:U K ½kA2 x A and v 0. At any other point: U K ½mv2 ½kx2

Velocity as Function of Position.xavmx -A12x 0mv kx kA212212vmax whenx 0:x Av 2v kAmkA2 x 2m

Example 5: A 2-kg mass hangs at the end ofa spring whose constant is k 800 N/m. Themass is displaced a distance of 10 cm andreleased. What is the velocity at the instantthe displacement is x 6 cm?½mv2 ½kx 2 ½kA2v kmA2 x 2800 N/mv (0.1 m) 2 (0.06 m) 22 kgvv 1.60 1.60 m/sm/sm x

Example 5 (Cont.): What is the maximumvelocity for the previous problem? (A 10cm, k 800 N/m, m 2 kg.)The velocity is maximum when x 0:0½mv2 ½kx 2 ½kA2v k800 N/mA (0.1 m)m2 kgvv 2.002.00 m/sm/sm x

The Reference CircleThe reference circle comparesthe circular motion of an objectwith its horizontal projection.x Acos tx A cos(2 ft )x Horizontal displacement.A Amplitude (xmax). Reference angle. 2 f

Velocity in SHMThe velocity (v) of anoscillating body at anyinstant is the horizontalcomponent of itstangential velocity (vT).vT R A; 2 fv -vT sin ; tv - A sin tvv -2 f-2 f AA sinsin 2 f2 f tt

Acceleration Reference CircleThe acceleration (a) of anoscillating body at any instant isthe horizontal component of itscentripetal acceleration (ac).a -ac cos -ac cos( t)v2 2 R2; ac 2 Rac RRa - cos( t)R Aa 4 2 f 2 A cos(2 ft )a 4 f x22

The Period and Frequency as aFunction of a and x.For any body undergoing simple harmonic motion:Since a -4 f2x1f 2 axandT 1/f xT 2 aTheThe frequencyfrequency andand thethe periodperiod cancan bebe foundfound ifif thethedisplacementdisplacement andand accelerationacceleration areare known.known. NoteNotethatthat thethe signssigns ofof aa andand xx willwill alwaysalways bebe opposite.opposite.

Period and Frequency as a Functionof Mass and Spring Constant.For a vibrating body with an elastic restoring force:Recall that F ma -kx:kx1f 2 kmmT 2 kTheThe frequencyfrequency ff andand thethe periodperiod TT cancan bebe foundfound ififthethe springspring constantconstant kk andand massmass mm ofof thethe vibratingvibratingbodybody areare known.known. UseUse consistentconsistent SISI units.units.

Example 6: The frictionless system shownbelow has a 2-kg mass attached to a spring(k 400 N/m). The mass is displaced adistance of 20 cm to the right and released.What is the frequency of the motion?xavmx -0.2 m1f 2 x 0k1 m 2 x 0.2 m400 N/m2 kgff 2.252.25 HzHz

Example 6 (Cont.): Suppose the 2-kg massof the previous problem is displaced 20 cmand released (k 400 N/m). What is themaximum acceleration? (f 2.25 Hz)xavmx 0x -0.2 mx 0.2 mAcceleration is a maximum when x Aa 4 f x 4 (2.25 Hz) ( 0.2 m)22222aa 40m/s40 m/s2

Example 6: The 2-kg mass of the previousexample is displaced initially at x 20 cmand released. What is the velocity 2.69 safter release? (Recall that f 2.25 Hz.)xavmvv -2 f-2 f AA sinsin 2 f2 f ttx -0.2 m x 0x 0.2 mv 2 (2.25 Hz)(0.2 m) sin 2 (2.25 Hz)(2.69 s) (Note: in rads)v 2 (2.25 Hz)(0.2 m)(0.324)vv -0.916-0.916 m/sm/sThe minus sign means itis moving to the left.

Example 7: At what time will the 2-kg massbe located 12 cm to the left of x 0?(A 20 cm, f 2.25 Hz) -0.12 mxavmx A cos(2 ft )x -0.2 m x 0x 0.12 mcos(2 ft ) ;A 0.20 m2 ft 2.214 rad;x 0.2 m(2 ft ) cos 1 ( 0.60)2.214 radt 2 (2.25 Hz)tt 0.1570.157 ss

The Simple PendulumThe period of a simplependulum is given by:LT 2 gLFor small angles 1f 2 gLmg

Example 8. What must be the length of asimple pendulum for a clock which has a periodof two seconds (tick-tock)?LT 2 g22 L;T 4 g(2 s) 2 (9.8 m/s 2 )L 24 LT 2gL 24 L 0.993 m

The Torsion PendulumThe period T of a torsionpendulum is given by:IT 2 k'WhereWhere k’k’ isis aa torsiontorsion constantconstant thatthat dependsdepends ononthethe materialmaterial fromfrom whichwhich thethe rodrod isis made;made; II isisthethe rotationalrotational inertiainertia ofof thethe vibratingvibrating system.system.

Example 9: A 160 g solid disk is attached tothe end of a wire, then twisted at 0.8 radand released. The torsion constant k’ is0.025 N m/rad. Find the period.(Neglect the torsion in the wire)For Disk:DiskI ½mR2I ½(0.16 kg)(0.12 m)2 0.00115 kg m2I0.00115 kg m 2 2 T 2 k'0.025 N m/radTT 1.351.35 ssNote: Period is independent of angular displacement.

SummarySimpleSimple harmonicharmonic motionmotion (SHM)(SHM) isis thatthat motionmotion ininwhichwhich aa bodybody movesmoves backback andand forthforth overover aa fixedfixedpath,path, returningreturning toto eacheach positionposition andand velocityvelocityafterafter aa definitedefinite intervalinterval ofof time.time.TheThe frequencyfrequency (rev/s)(rev/s) isis thethereciprocalreciprocal ofof thethe periodperiod (time(timeforfor oneone revolution).revolution).xmF1f T

Summary (Cont.)Hooke’sHooke’s Law:Law: InIn aa spring,spring, therethere isis aa restoringrestoringforceforce thatthat isis proportionalproportional toto thethe displacement.displacement.F kxxThe spring constant k is defined by:mF Fk x

Summary (SHM)xavmx -Ax 0F ma kxx A kxa mConservation of Energy:22 ½kx 22 ½mv 22 ½kx 22½mv½mvAA ½kxAA ½mvBB ½kxBB

Summary (SHM)12v mv kx kA2kA2 x 2mkAm212x A cos(2 ft )212v0 a 4 f x2v 2 fA sin(2 ft )2

Summary: Period andFrequency for VibratingSpring.avxmx -Ax 0x A1f 2 ax xT 2 a1f 2 kmmT 2 k

Summary: Simple Pendulumand Torsion Pendulum1f 2 gLIT 2 k'LLT 2 g

CONCLUSION: Chapter 14Simple Harmonic Motion

Simple Harmonic Motion, SHM Simple harmonic motion . Simple harmonic motion is periodic motion in the absence of friction and produced by a restoring force that is directly proportional to the displacement and oppositely directed. A restoring force, F, acts in the direction opposite the displacement of the oscillating body. F - kx. A .