Download CH15.Problems JH.132

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.