Survey

* Your assessment is very important for improving the work of artificial intelligence, which forms the content of this project

Survey

Document related concepts

no text concepts found

Transcript

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); X(t) = 20.0cm * Cos(Pi t) let us assume zero phase factor 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.

Related documents