Physics-Oscillatory Problem!?
919976
Apparently we are ignoring any pitching motion due to their getting in the car and due to the bump, since you'd need to know a lot more about mass distribution if it was pitching. So we're only considering vertical movement and the distance between axles is irrelevant. I'm assuming "a total of 8.9 cm" means the sum of the 4 spring deflections, so their average deflection is 2.225 cm. If it actually means the average deflection, then k and m are 1/4 what you'll get with the following calculations. Stiffness k (of the 4 springs in parallel) in n/M = 146*g/.02225. Natural radian frequency w = sqrt(k/m), so m = k/w^2 = k/(2*pi/1.7)^2. (That's the total m; mass of car = m-146.) \nThis was not a well-phrased question, because "a total of 8.9 cm" also could mean 8.9 cm from their no-load length (and thus due to the weight of the car and the students). However this would result in an insoluble problem. \nBTW, why are they geeky? Did they figure all this out before me?
136454
I will give you a direction to solve the problem...\n\nFirst I will do the problem from a rectilinear formulation(using force, acceleration, velocity) then transfer it to a curvilinear formulation(using Torque, angular accelaration, angular velocity).\n\ngiven F=ma (or F=m(v/t) or more specifically F=m(dv/dt) or even more specifically F=d(mv)/dt).\nWhere F=force, m=mass, v=velocity, and t=time.\n\ngives you Ft=mv (conservation of energy=>conservation of momentum or the impulse-momentum equation).\n\nThis states that it will take a certain amount of force for a certian period of time to produce a certain change in velocity of an object assuming that the mass remains constant.\n\nTransmitting to curvilinear motion force becomes torque, v becomes rv(radius time velocity or w), mass becomes moment of inertia (or I) and t stays the same.
550813
I believe it is called a phase difference. They will go in and out of sync periodically if there is a slight frequency difference.
883480
this seems an easy question but most of the student get it wrong since it is so easy to mess up your mind. it was a part of an exercise for my University exams. it makes you realise what is going on with spring ...just consider it as a opposing force. the body will drop and will expirience g --> it will accelerate. at the same time it will also expirience the sping;s opposing force.\n\nit will accelerate till the force become equal. this is the importnat factor. weight will keep on accelerating the body till it is completely cancelled by the spring force.\n\nso when \nFs = Weight then Velocity is maximum. \nso kx =mg =>\n\nx = mg / k.  (at this displacement the velocity  will be maximum)\n\nat this specigic point due to the theorem of conservation of energy we get that.\n\nkinetic energy + sping stored energy = potential energy at the start.\n\nso \n1/2mu^2 = (m*g)^2 / k => (be carefull mass = 76.7/1000kg)\nU = \n\nU = sqrt (( m*g^2)/k)) => u = 0.361m/s 9\n\nIF THE ALGBRA IS CORRECT\n\nNOW AFTER THAT POINT the body will still travel due to inertia and the springs force  larger now than gravity wll slow it down and bring it at rest \nat that point all the potential energy (due to height difference) will be transfornmed to springs stored energy therefore,\n\n1/2 k xmax^2 = mgXmax =>\n\nXmax = 2*m*g / k =>\n\nXmax = 0.02658m \n\n\n( i am at work and i believe the algebra is correct. the main point is to realise where certain things happen.)
415527
First of all, the sum is incompleted.  Is it for maximum height, or for range?  Secondly, let us assume you are asking for maximum height.  Then differentiating the given equation of motion, d = -5(t+1)^2 + 258 with respect to 't', we get dt/dt = -10(t+1).  At maximum height, the velocity is zero. \nTherefore, -10(t+1)=0, or t = -1.  This is absurd time can not become negative once the motion is initiated.  Check up there is something wrong in the problem.
227728
Very simple. start out asking yourself what information do i know about the problem. Anything from masses and forces to symmetries of the problem. Next Ask yourself what additional information you need to solve the problem, values, formulas, etc. from there it should all fall out with ease.
969486
I just helped someone else with their physics question, so I'm alittle tired of it.  But, I suggest you make an equation sheet...physics is all about equations.  So, if you have a piece of paper with your equations, all you have to do is look at the sheet and plug things in...
617970
I am terribly afraid of the equilibrium problem.. the first one on the free response!!! well, good luck!
919525
Vibration:  ω = √(k/m) = √(100/4) = 5.0 rad/sec\n\nV at neutral point = Vo = Aω = 2*5.0 = 10 m/sec\n\nx-direction momentum: This momentum will be the same before and after dropping the 6 on the 4 →\n6*0 + 4*5 = (6+4)*v →  v = 20/10 = 2 m/s\n\nThe new KE = .5mv² = .5*10*2² = 20 Nm ; This will equal the new spring energy:\n\n20Nm = .5kA² → A = √(40/100) = .6324 m, so the change in amplitude is\na)  2-.6324 = 1.3676 m\n\nb) New period: T = 2π√(m/k) = 2π√(10/100) = 1.987 sec\n....Old period:  To = 2π√(4/100) = 1.2566 sec\n....Difference = .73 sec greater\n\nc)  KE1 = .5*4*10² = 200 Nm.  From (above) KE2 = 20 Nm\n....so, energy change is -180 Nm\n\nd) The energy loss is released as heat between the (momentarily) sliding surfaces of the 2 objects when they contact.
827544
1) use conservation of energy:\n\n1/2*m*vb^2=1/2*m*va^2+m*g*h\n\nm=2\nva=4\ng=9.8\nh=6\nv=sqrt(16+2*9.8*6)\n=11.56 m/s\n\n2.\nSame concept\nm*g*(bounce)=1/2*m*va^2+m*g*(drop)\nmass divides out\nbounce=va^2/(2*g)+drop\nva=6\ng=9.8\ndrop=4\nbounce=36/(2*9.8) + 4\n=5.84 m\n\n3)  since the mass isn't moving it's offset the 300N force\n\n4)\nAgain, use conservation of energy\nNote that the change in potential energy will subtract from the kinetic energy of the start \n\n1/2*m*vc^2=1/2*m*va^2-m*g*delta-h\nthe mass divides out\nva=8\ndelta-h=2\ng=9.8\nvc=sqrt(64-2*9.8*2)\n=4.98\n\nj
288780
From the following link, look for "Chapt22, Problem-24".\n\nThis problem is similar, except the speed is 10% instead of 11%. I'm sure you can use the same formulae though.\n\nhttp://www.jca.umbc.edu/~george/html/courses/phys112/2004fall/homework/04fall_phys112_hk8.pdf
579183
Yes. If there is no dampening the energy just keeps going back and forth between maximum potential energy (U) and maximum kinetic energy (K). If there is nothing opposing the restoring force it will remain in SHM (simple harmonic motion) and oscillate continuosly.
1059521
This involves second order differential equations.\n\nAn oscillator is damped when the equation looks like:\n          mx''+bx'+kx = 0 \nwhere b is the damping coefficient.  The solutions go to 0 as time goes on.\n\nA driven oscillator takes the equation similiar to:\n         mx''+kx=F(t) \nwhere F(t) is some function, typically a sine or cosine type function.  Here, the oscillator changes in how it moves, its frequency and amplitude.\n\nRemember for a non-driven undamped oscillator mx'' + kx = 0\n\nNote: x' is the first time derivative of x, and x'' is the second
35590
I concur with previous answerer. The acceleration is 1.5 m/s^2. Rearrange Newton's Second Law, to solve for the acceleration. Then, substitute what you know.
881328
this requires way more than anyone can do here, but, i know of a cool java program that will do it for you: \n\nhttp://www.math.rutgers.edu/courses/ODE/sherod/phase-local.html\n\nyou can even set your initial conditions and print it out!
820986
1.\nPhases:\nf1 = (t/T1)*2*pi, where T1 = 1 sec\nf2 = (t/T2)*2*pi, T2 = 1 sec\nIn phase: f1 = f2 + 2*pi*n, where n is integer (0, +-1, +-2, ...)\n(t/T1)*2*pi = (t/T2)*2*pi + 2*pi*n\nt(1/T1 + 1/T2) = n\nT2 = 0.98 = 98/100 = 49/50, 1/T2 = 50/49\nt(49/49 - 50/49) = n\n-t/49 = n\nt = 49 seconds, n = -1\nNumber of oscillations if second, shorter pendulum (shorter pendulum has shorter period): N = t/T2 = 49/(49/50) = 50\n\n2.\nCoordinate: y(t) = 3*sin(20*pi*t)\nSpeed: v(t) = y'(t) = 20*pi*3*cos(20*pi*t)\nMean position: y(t) = 0. Take t0 = 0.\nMaximum displacement: y(t1) = amplitude = 3, speed = v(t1) = 0\nv(t1) = 0\ncos(20*pi*t1) = 0\n20*pi*t1 = pi/2\nt1 = 1/(2*20) = 1/40
769928
T = 2pisqrt(l/g)\nl = g(T/(2pi))^2 = 9.80(1.0/(2pi))^2 = 0.25 m\n\n(The object's weight is irrelevant for small oscillations.)