Vibrating cubeExamplel: A woody cube can float with height hunder water. If it is pushed down and released, therewill be a SHM. Determine the angular frequencySolution: Assume a mass m and length lmg = pgl’hxWith displacement xx0ma = pgl2 (h-x) -mghPg/?gUa=XXhm=0=/g / h
Vibrating cube Example1: A woody cube can float with height h under water. If it is pushed down and released, there will be a SHM. Determine the angular frequency. Solution: Assume a mass m and length l 2 mg gl h = h o x x With displacement x ( ) 2 ma gl h x mg = − − 2 gl g a x x m h = − = − = g h/
ThinkingquestionThinking: A mass M is connected to two springs (k)k). What is the angular freguency?FFFk,k,k:x = x,+xkkik2k, +kzkXiX2QMFki1 kzxM
Thinking question Thinking: A mass M is connected to two springs (k1 , k2 ). What is the angular frequency? M k1 k2 1 2 1 2 F F F x x x k k k = + = + = F x x1 x2 1 2 1 2 k k k k k = + k M =
Motional equation of SHMd?x+x=0dt?= x= Acos(t+motional equationA and @ : integral constants → initial conditions1) Amplitude A: the maximum magnitude ofdisplacement from equilibrium2) Angular Frequency : rate of the vibrationの=/k /mdetermined by the system
Motional equation of SHM 2 2 2 0 d x x dt + = = + cos x A t ( ) motional equation 1) Amplitude A: the maximum magnitude of displacement from equilibrium 2) Angular Frequency ω: rate of the vibration = k m/ determined by the system A and : integral constants → initial conditions
Quantities in SHMPeriod T: time required for one complete cycle2元2元 =→ Acos(ot +p)= Acos[o(t+T)+β10Frequency f: number of complete cycles per sec10natural frequencyT2元3) のt +@: the phase of SHM at time t, note as 0β: phase at t=O, initial phase or phase anglePhase: key point in oscillations or waves
Quantities in SHM Period T: time required for one complete cycle 2 T = + = + + A t A t T cos cos ( ) ( ) Frequency f: number of complete cycles per sec 1 2 f T = = Phase: key point in oscillations or waves natural frequency 3) t + : the phase of SHM at time t, note as θ : phase at t=0, initial phase or phase angle
Phasedifference1) For one SHM at different time0=(ot, +β)-(ot, +P) =の(t -t)= 0t2) For two SHM with same o at same time=(t +P2)-(ot+) = P2 - = △p>0: x2 is in front of xi=O: xi and x2 are in phase△0= P2 -P1<0: x2 is behind of xi=π: xi and x2 are inverse phase
Phase difference 1) For one SHM at different time = + − + ( t t 2 1 ) ( ) = − = (t t t 2 1 ) 2) For two SHM with same ω at same time = + − + ( t t 2 1 ) ( ) = − = 2 1 >0: x2 is in front of x1 =0: x1 and x2 are in phase <0: x2 is behind of x1 =π: x1 and x2 are inverse phase = − 2 1