1 / 17

CH15.Problems

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.

erelah
Download Presentation

CH15.Problems

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. CH15.Problems JH.132

  2. 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

  3. 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)

  4. 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

  5. a = - w^2 x When x = -ve then a must be positive F = -k x

  6. 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

  7. 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

  8. 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

  9. Kmax = Etot Umax = ½ K x^2, we know Xmax

  10. 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.

More Related