User Name Remember Me? Password

 Kinematics and Dynamics Kinematics and Dynamics Physics Help Forum Aug 6th 2016, 02:30 AM #1 Junior Member   Join Date: Jul 2016 Posts: 17 Kinematics of a particle A swimmer jumps using a spring board. The board bends with distance −X0. Then it accelerates him with ac = −(ω^2)x with ω = const. He loses contact at x = 0, then he flies vertically upward with af = −g. Air resistance negligible. Actually, I want to ask a simple question. I've already got v(s) = √(ω(x0)^2−x^2), then I want to search for the distance x as function of time t, but I don't know how to integrate it, as the right answer is x(t) = x0 sin (ωt−(π/2)) I don't get the idea how the π comes from. Any help would be very appreciated. Additional question: What is the distance between the unloaded board and the surface of the water x hit, if the swimmer hits the water with vhit = −2ω x0 I've calculated the v(x),x(t),v(t) but it looks like xhit doesn't depends on those 3 equations? Last edited by rndaryam; Aug 6th 2016 at 02:37 AM.   Aug 6th 2016, 06:40 AM #2 Physics Team   Join Date: Jun 2010 Location: Morristown, NJ USA Posts: 2,347 The equation for v(x) is actually: v(x) = w sqrt(x0^2-x^2) Since v = dx/dt, rearrange to get: wdt = dx/sqrt(x0^2-x^2) Integrate: wt + c= x0 arcsin(x/x0) where c is a constant of integration. Rearrange: sin((wt+c)/x0) = x/x0 x = x0 sin((wt + c)/x0 = x0 sin(wt/x0 + C) Note that I've replaced C/X0 with a new constant of intregration, C. To solve for C use the initial condition that at time t= 0 x(0) = -x0: Hence the argument of the sine function must be -pi/2. x(t) = x0 sin(wt/x0 - pi/2) For your second question, I suggest using energy principles. The KE with which he hits the water is (1/2) mv_hit^2, which must equal his initial KE when he leaves the board plus the PE he gains from the board down to the water. You know his initial velocity leaving the board is -wx0, Thus: (1/2) m v_hit^2 = (1/2)m(w x0)^2 + mgx_hit. Solve for x_hit.   Aug 6th 2016, 08:50 AM #3 Junior Member   Join Date: Jun 2016 Posts: 24 Conservation of energy is definitely the way to go to solve the last part of your problem, so ChipB has given you the best advice. However, in case you haven't yet covered conservation of energy in your studies, can you see another way to solve for final position xhit? You've got initial position, initial velocity, final velocity, and acceleration (-g). Can you find an equation from your earlier ballistic studies which includes the quantities x x0 v v0 a? Should give you the same result as the conservation of energy equation. Also note that initial position x0 in the ballistic equation for this last bit is zero according to your problem description (the point the diver leaves the board is x=0), and not the -x0 where the diving board was fully compressed. Your v(x) for the moment the diver leaves the board would contain that original x0 though. Also, I wouldn't think you'd need to find any time-dependent equations here like v(t) or x(t), unless the problem specifically asked for them. Your velocity equation is based on position x and not t. The ballistic equation I mentioned wouldn't include time either. You certainly could move off into time-dependent equations and determine the times of the different events, but it wouldn't be as efficient. Good that you know how to do it both ways though.   Aug 6th 2016, 11:08 AM   #4
Junior Member

Join Date: Jul 2016
Posts: 17
 The equation for v(x) is actually: v(x) = w sqrt(x0^2-x^2) Since v = dx/dt, rearrange to get: wdt = dx/sqrt(x0^2-x^2) Integrate: wt + c= x0 arcsin(x/x0) where c is a constant of integration. Rearrange: sin((wt+c)/x0) = x/x0 x = x0 sin((wt + c)/x0 = x0 sin(wt/x0 + C) Note that I've replaced C/X0 with a new constant of intregration, C. To solve for C use the initial condition that at time t= 0 x(0) = -x0: Hence the argument of the sine function must be -pi/2. x(t) = x0 sin(wt/x0 - pi/2) For your second question, I suggest using energy principles. The KE with which he hits the water is (1/2) mv_hit^2, which must equal his initial KE when he leaves the board plus the PE he gains from the board down to the water. You know his initial velocity leaving the board is -wx0, Thus: (1/2) m v_hit^2 = (1/2)m(w x0)^2 + mgx_hit. Solve for x_hit.
thank you so much for your help, also am I right if I replace x_hit with -x_hit since it is below x=0?

 Originally Posted by CorneliusXI Conservation of energy is definitely the way to go to solve the last part of your problem, so ChipB has given you the best advice. However, in case you haven't yet covered conservation of energy in your studies, can you see another way to solve for final position xhit? You've got initial position, initial velocity, final velocity, and acceleration (-g). Can you find an equation from your earlier ballistic studies which includes the quantities x x0 v v0 a? Should give you the same result as the conservation of energy equation. Also note that initial position x0 in the ballistic equation for this last bit is zero according to your problem description (the point the diver leaves the board is x=0), and not the -x0 where the diving board was fully compressed. Your v(x) for the moment the diver leaves the board would contain that original x0 though. Also, I wouldn't think you'd need to find any time-dependent equations here like v(t) or x(t), unless the problem specifically asked for them. Your velocity equation is based on position x and not t. The ballistic equation I mentioned wouldn't include time either. You certainly could move off into time-dependent equations and determine the times of the different events, but it wouldn't be as efficient. Good that you know how to do it both ways though.
Also, thanks again for your briefly explanation, so if Conservation of energy that we use here is EK + 0 = EK + EP (usually it's EK + EP = EK + EP), then we have another EP that equals to zero?   Aug 6th 2016, 12:04 PM #5 Junior Member   Join Date: Jun 2016 Posts: 24 Sure, you can pick one of the two x locations to be the potential energy = 0 reference point, and then use the relative displacement from there to calculate the potential energy of the other point. So you'd have only one non-zero EP quantity in your equation. Also, would be good to use some kind of sub- notation to indicate the before vs. after energy. For instance EK1 + EP1 = EK2 + EP2. Usually we go with '1' as being the 'before' case and '2' the 'after'. I've seen people use 0 and 1, and regular and prime as well. Whatever you choose, you should finish up with an equation like ChipB gave you at the end of his post.  Tags distance, kinematics, mechanics, particle, physics, velocity Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Physics Forum Discussions Thread Thread Starter Forum Replies Last Post vyomictor Special and General Relativity 0 Mar 21st 2011 12:56 AM quantum_enhanced Advanced Mechanics 2 Jan 15th 2011 05:22 AM Apprentice123 Kinematics and Dynamics 2 Apr 3rd 2009 06:13 PM Apprentice123 Kinematics and Dynamics 2 Apr 2nd 2009 11:24 AM joker1 Nuclear and Particle Physics 0 Mar 12th 2009 01:10 PM