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CH15.Problems. JH. 132. a max = w 2 * x max = Pi^2 0.5 = 4.9m/s^2; General: Equation parameters Find also: Period, frequency, angular frequency Maximus speed, maximum displacement. If you are given mass then you could also find Maximum kinetic energy Maximum force.
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CH15.Problems JH.132
amax = w2 * xmax= Pi^2 0.5 = 4.9m/s^2; General: Equation parameters Find also: Period, frequency, angular frequency Maximus speed, maximum displacement. If you are given mass then you could also find Maximum kinetic energy Maximum force
Descriptions Parameters & Equation Zero velocity to next zero velocity: is 2 amplitudes Xmax = 40cm/2 = 20cm That trip is only T/2; so f= 1/T = 1/(2*0.25) = 2 Hz You could construct an equation: X = Xmax * Cos( 2Pi t/T + phi); let us assume zero phase factor X(t) = 20.0cm * Cos(Pi t)
Equation parameters at a specific position, velocity or time This is a special point x= 0 means also that v= vmax = 0.2 * 2Pi = 1.3, a = zero If the point selected is not special: X= 0 means that cos(2pi t) = 0, 2Pi t = Pi/2 or (3Pi/2) t = 0.25 sec then find v and a at that point
a = - w^2 x When x = -ve then a must be positive F = -k x
Simple Pendul: T = 2 Pi Sqrt(L/g) Physical: T = 2Pi sqrt (I/mgh), h = L/2 After made equal Lp/g = ML^2/3 /mg(L/2)= 2L/3g Lp = 2/3 L = 67cm
Parallel axis theorem: I = Icom +M h^2 = 1/12 ML^2 + M x^2 T = 2Pi Sqrt((Icom+Mh^2)/mg) dT/dx = 0 means d/dx[ Icom/x+ Mx)=0 x = L/sqrt(12) = 0.5m
From the given Etot = 4+8 = 12 J Etot = K @ xmax 12 = ½ k x^2 From x = 3cm , K = 8 J, find k of spring
Kmax = Etot Umax = ½ K x^2, we know Xmax
During each period 4 Xmax is travelled. In 1.9S there are 1.9/0.1 =19 oscillations Then, distance = 4* 19* 1.2 =91.cm.