can someone pls help me with this question on simple harmonic motion?
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
770512
The total force on the block is mg sin (theta). The mg cos (theta) and force normal cancel out. Since there is no friction, the only force, again, is only mg sin (theta). \n\nThis is obvious if you consider that there is a force mg, and break it down into components, the solution should seem obvious.
799788
You are given P(t=0), Q(t=0); P'(t=0), Q'(t=0); and P''(t=0), Q''(t=0). Given that P'' and Q'' remain the same (that is, the motion of P and Q do not change), you need to show that there is a time T such that P(t=T) = Q(t=T).\n\nIt's late and I'm drunk, so you can do the actual working for yourself :D
24383
here are some simple mechanics formulas dealing with velocity:\n\nv=d/t\na=v/t\nvf^2=vi^2+2ad\nd=vit + 1/2at^2\n\nwhere v is velocity, vi is initial velocity, vf is final velocity, a is acceleration, and t is time.  also the i and the f after the v are just subscripts, and ^2 just means sqared.
487526
Sounds like a Math/Science type question to me. Why are you posting it here?\n\nBesides, I have the the slightest idea.
733015
a. Applying conservation of momentum\nm1 x V_i + m2 x 0 = (m1+m2) x V_f\nV_f = (m1 x V_i)/(m1+m2)\n\nb. If m1 >> m2 then m1 + m2 -> m1\nTherefore V_f = V_i\n\nc. If m1 << m2 then m1+m2 --> m2\nTherfore V_f = (m1/m2) x V_i\n\nd.m1 x V_i - m2 x V_i = (m1+m2) x V_f\n=> V_f = ((m1-m2)x(V_i))/(m1+m2)\n\ne. m2 = m1, m2 =(1/3)(m1)
716765
The potential energy at the top is mgR = 2 x10 x R = 20 R.\n\nKinetic energy at the bottom of the curve = 0.5 m U ^2 = 0.5 x 2 x U^2. = U^2.\n\nAs there is no friction  K.E = P.E\n\nU^2 = 20 R.\n\nOn the horizontal path, \n\nInitial speed is U.\n\nFinal speed is V = 0.\n\nDistance traveled = 8 m.\n\nAcceleration = Fricttional Force / mass = -8/ 2 = - 4 m/s^2. ( Negative sign since the force acts opposite to the motion.\n\nUsing V^2 - U^2 = 2as, and also noting that V=0,\n\nU^2 = 2 x 4 x 8 = 64.\n\nU = 8m/s.\n\nThe time of travel is found using V= U+ at.\n\nt = -8/ (-4) = 2 second.\n\nUsing U^2 = 20 R.\n\nR = 64/ 20 = 3.2 m.
849885
I understand your frustration in receiving no Answer yet to this question, but since I do not have any direct knowledge of how to perform the calculation(s) you need ... I hope the link below serves you well.  It appears to explain the mathematics behind all the concepts you mention.
574012
I've alread answered two of your questiions in my last answer....\n\nIn this one, the guys above are correct - however, I'd try working it out for myself if I was you. Your tutor will soon realise something is wrong if you are giving answers to a subject you don't understand. Read the chapter, ask you teacher if you need more help. Learn Newtonian mechanics for yourself.
696021
OK, you already know (a), which is the total energy. At the peak displacement (which is the amplitude), all the energy is potential energy PE, in the form of spring deformation. PE = .5*k*x^2 = .5*35*.04^2 = .028 j. At other times, the energy is divided between PE and kinetic energy KE, and PE+KE=.028. So for (b), KE = .028-.5*35*.01^2 = .02625 j. Then the velocity = sqrt(2*KE/m) = 1.025 m/s. For (c), KE = .028-.5*35*.03^2, and for (d) PE = 5*35*.03^2.
357903
When V(t)=0 the particle stops.\nIt can have a(t)=0, but still moving (constant speed)
49314
Sorry, but with the time you spent typing these questions you could of looked them up in your physics book or googled them.  Sometimes getting others to do your leg work isn't always the best answer to a problem.  Good Luck in your new class!
75525
a) 72.072 cm/sec^2\n\nb) 82.63 cm/sec^2
826674
use conservation of angular momentum in these kind of problems\nNOT energy conservation\nif you dont know how to use the conservation equation, better read up on examples
901641
Someone posted the exact same set of questions a few days back, either here or on the physics page - you ought to be able to find the methods in the responses. Have a look.\n\n\nBest of Luck - Mike
693965
If it is open on both ends, the harmonic frequencies are:\n\nf=nv/2L\n\nwhere n is the harmonic number (1st, 2nd, etc), v is the speed of sound in air (~343 m/s), and L is the length of the pipe.
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.