physics-2

We'll discuss velocity and acceleration.  I'll start with something simple - I have a motion of an object along a straight line, we call that one-dimensional motion.  And I'll tell you that this object is here at time t1; at time t2, it's here; at time t3, it's there; at time t4, it's here; and time t5 it's back where it was at t1.  And here you see the positions in X, where it is located at that moment in time.  I'll define this to be the increasing value of X, it's my free choice, but I've chosen this now.

Now, we'll introduce what we called the average of velocity.  I put a bar over it that stands for average, between time t1 and time t2 that we defined in physics that x at time t2 minus x at time t1 divided by t2 minus t1.  That is our definition.  In our case, because of the way I defined the increasing value of x, this is larger than 0.  However, if I take the average velocity between t1 and t5, that would be 0 because they're at the same position.  So the upstairs (he meant numerator) is 0.  If I had chosen t4 and t2 (writing down t2, t4 at the same time), average velocity between time t2 and t4, you would have seen that it is negative, because the upstairs (he meant numerator) is negative.  Notice that I haven't told you where I choose my 0, or my x-axis.  It’s completely unimportant for the average velocity.  It makes no differs (he meant no differences).  However, if I had chosen this to be the direction of increasing x, then, of course, these signs will flip.  Then this would have been negative, and this would have been positive.  So the direction that you are free to choose determines the signs.  Issuing where you put your 0 is not important, but signs, in Physics, do matter.  Signs are important, whether you owe me money, or I owe you money, the difference is only a minus sign, but I think it’s important for you.

Now I will give you not only the positions, as I did here, on the x-axis, at a discrete moment in time, but I’m going to tell you exactly where the object is at any moment in time.  Here you see an x-t diagram, so you see at t1, the object is at position xt1; this is the road of the object, this straight line where it’s moving.  It starts here, and it goes to this position, goes to this one, comes back to t4, and comes back here.  I will tell you now, every moment in time in between – there it goes, woala!  This is now information that is way more, you now have the information at any moment in time.

Notice that I now did choose x = 0. I chose it somewhere here, but I could have chosen it at any other point; for level floors you’ll see it makes no difference.  So I’ve chosen a zero point so I can make a graph, and now, we’ll look at the average velocity in a somewhat different way.  Say, I choose my time t2 and t3, I draw here now, this line; and this angle I call it alpha; and this, for here, I call delta x; and this here, is delta t.  And so you could write now, if you care about your sign convention, you could write down now, that the average velocity equals delta x divided by delta T.  Let’s be careful, if the angle is positive, I call this a positive angle, then the average velocity is positive.  But if I have a negative angle, then the negative velocity would be negative.  For instance, between t4 and t5, if I draw this line, then this angle here is negative, and so the average velocity here between t4 and t5 is now negative.  Again, if I had changed the zero point, you would have found the same values for the average velocity; the only difference would have been the position of the curve in that plot.  There’s a very big difference in physics between speed and velocity.  The average velocity between time t1 and t5 is zero, but the average speed is not.  The average speed is defined as the distance traveled divided by the time that it takes to travel that distance.  So what is the distance that the object traveled between time t1 and time t5?  Well, the object started at a position here, on this x-axis, and then it went up, reach the highest point here, so I’ll make a drawing for you here; reaching the highest point here, and it went down, and then it went here; it went up again, came down again, and it was back.  And in order to find the average speed, you would now have to know exactly what this distance is, add up this distance, up add this distance, and this distance.  And with all these distances all together, well, for instance, 300 meters, and if the time between t1 and t5 were 3 seconds, then the average speed would be 300 meters divided by 3 seconds, that would be 100 meters per second.  So the average speed would be 100 meters per second, yet the average velocity would be zero.

If you look at the location t3 and t2, and I bring t3 closer and closer to t2, then this angle of alpha will increase.  And I can go to the extreme that I bring t3 almost right at t2, the angle of alpha will then be tangential to this point.  This will then be my angle of alpha.  And now you’ll understand how we define the instantaneous velocity at time t, which is different from the average velocity between two time intervals.  The instantaneous velocity, v, and I pick a random time t, equals the limiting case for x measures at time t plus delta t minus x measures at time t divided by delta t, and I do that for delta t goes to zero.  So think this, think of this as being t3 and this as t2, I bring t3 closer and closer and closer to t2, and the time between them then goes to zero.  And now this is something now you undoubtly recognize, that’s the first derivative of the position versus time.  And now comes the equation, which is one of the very few that I want you to remember.  In x, in either one, v equals dx/dt.  This is one that you must remember not only in A01, but for the rest of your time at MIT.  And this would be larger than zero, this could be zero, and this could be smaller than zero.  If the angle of alpha, the tangential is positive, then it is a positive value; if it is negative, however, when you’re here, then it is a negative velocity.  And if the angle of alpha is zero, then it’s, the velocity is zero.  So if now we look at these plots, we can search for the time that the velocity is zero, so you have to look for the derivative being zero, that means the angle alpha being zero, clearly here, the velocity is zero, right here at this turning point, at means when the object is here, it would be zero; when the object is here, then it is again zero at this moment in time.  Again, the angle is zero; and it is again zero here.  So those are the times that’s the velocity is zero.  What are the times that the velocity is positive?  Well, it’s positive here, velocity is positive here, still positive, positive; it becomes negative, negative, positive, zero, negative.  So that is the definition of v, instantaneous velocity.  What is the instantaneous speed?  Well, speed is not sign-sensitive.  Suppose that the velocity here, just I call that v1, suppose that’s plus 30 meters per second, I just grabbed this number out of the blue, and suppose here somewhere, it was, I call that v2, suppose it was minus 100 meters per second.  This is negative, and it is positive; then we would have to say in physics, whether you like it or not, it’s not very pleasing, but you would have to say this velocity (v2) is lower than this one (v1), because minus 100 is lower than plus 30.  But the speed, of course, is higher here (v2) because the speed is the magnitude of the velocity, and it’s not sign-sensitive.  So this (v2) has the higher speed at 100 meters per second, and this (v1) has a lower speed; but this (v2) has the lower velocity.  It’s just an algebra game, but it is very important when you make your calculations.

I have always wondered, what the average speed, or the average velocity is of a bullet.  Now I want you to realize I’m not offended of guns, at all.  But it always interests me, how can I measure, the average speed of a bullet.  And I’ve discussed it with some people here, and we came up with an easy way to do that.  We have a wire, which goes into the blackboard, wire I; and we have another wire that goes into the blackboard, wire II; and the separation is D meters, we have to measure that.  The set up is here – so this is wire #I, and this is wire #II.  So you would see it coming in like this; so I’ll make this a one, and make this a two – that’s the way it’s set up.  And we fire a bullet, that would break this wire at the moment it starts, and then it breaks this wire, and that’s when the time will stop.  Now, I told you, a measurement is meaningless without the knowledge of uncertainty in your measurement.  So there are two uncertainties involved.  The distance, and the timing uncertainties.  This distance, I will measure for you, D, I have here a large ruler; here’s one wire, here’s the other wire, I cannot do any better, really than, maybe even half a centimeter, because the situation is not all the stable; I don’t know what will happen when the bullet hits the wire, so I would say, it is a hundred forty-eight and a half centimeters, but I cannot guarantee it better than half a centimeter.  A hundred forty-eight and half, plus or minus 0.5 centimeter.  I want you to appreciate, that is a very small percentage error.  This is only 5 points out of 1500, that is 1 out of 300, so that’s only a 1/3 percent error; that’s very small.  That’s what we call the relative error.  Then I ask myself the question, I want to measure the accuracy of the speed of the bullet to about 2 percent, that was my goal; how accurate do that due to timing?  Well, I had to make an estimate, very roughly, how fast the speed of the bullet is, and I think it would probably be lower than the speed of sound, the speed of the sound is 340 meters per second.  I don’t know if it’s 200 or 300, but it’s gotta be somewhere in that broad part, of the kind of bullets we have, 200 or 300 meters per second.  Let us assume that the speed is 300 meters per second, just a wild guess.  Then it would take 5 milliseconds for this bullet to cross from here (I) to here (II).  And if, I want to make a measurement to 2 percent accuracy, I have to know this timing, to about one tenth of a millisecond, because one tenth of a millisecond is about 2 percent of 5.  So that sets the accuracy that I need to make the time measurements.  And so we do have a timer, it is about accurate to about a tenth of a millisecond, and so now I can measure that time.  So I’m going to have here, some time that we measure, plus or minus 0.1 and we’ll do the whole thing in millisecond, but our final answer would be in meters per second.  Alright, I always have to think hard when I do this because when we deal with a bullet that is no kid’s stuff, and as I said, I have really no experience firing guns.  This is the bullet, there we go (removing the box).  Here’s the bullet, there we go – it’s in place.  Before I do that, I want to check, check the circuit; I want to make sure that the electronic circuit is properly working.  You see the timing here, right?  So I do a small test, just to see whether the circuit is working.  Yeah, should be working.  So here comes the bullet.  You’re ready, I’m ready.  3, 2, 1, 0 (firing).  What do we see?  5.8 milliseconds.  5.8, is that what you see?  Yeah! 5.8 milliseconds.  5.8, plus or minus 0.5.  So here comes the average – bullet speed, we call it velocity; it is the same thing in this case, 148.5, 5.8, and I have to convert it to meters per second; that brings it at 256 plus or minus… now you come in here with your plus and minuses.  This is a 1/3 percent error, it is neglect-able to this one – 1 out of 58 is about 1.7 percent, so this is the only one we have to worry about.  So the uncertainty in there is about 1.7 percent, that’s less than two, and that’s what I wanted.  And that gives me an error of about 4 meters per second.  So this is the result, and you see, it’s only meaningful because we have a good idea about the uncertainties in the measurement.  Just as we introduced, average velocity; now, I’m going to introduce average acceleration.  Notice that the velocity changes here throughout time.  And that brings me to the next part.  It’s a logical part, namely, we’re going to introduce average acceleration, and with a little bit of imagination, you can probably guess what that looks like.  The average acceleration between time t1 and time t2 would then be the velocity at time t2 minus the velocity at time t1, divided by t2 minus t1, and as I mentioned, it’s length per second square, so it’s meters per second square.  This is for one-dimensional situation.  This again can be larger than zero, it can be equal to zero, and it can be smaller than zero.  In our case, t1 to t2, here, notice the velocity is at 0 when it starts, and it begins to increase because the angle of alpha increases.  It’s the angle that matters!  The angle increases, so in our case, from t1 to t2, the average acceleration is larger than 0.  Look at the angle.  However, if you take the average acceleration between t1 and t5; that is smaller than 0 because here, the velocity is 0, but here, the velocity is negative.  So if you substitute that in there, you get an average acceleration which is smaller than 0.  So the signs in velocity and the signs in average acceleration depend crucially on how I have defined my increasing value of x, not where I choose my 0 point.  If I reverse the direction of my increasing x, then all my signs will change.  So you can also write down then, that average acceleration, if you like that, is delta v divided by delta t, which you have to be careful because delta v is sign-sensitive.  You must obey your sign convention.

I have here, a tennis ball, and I can bounce this tennis ball, I can throw it down; and let us assume, just for simplicity, that it hits the floor at about 5 meters per second, and that is a very very good tennis ball, and that it also bounces back of a velocity that is about 5 meters per second.  I will choose this as my increasing value of x, and so it hits the floor like this, that means, the velocity at which it hits the floor is -5 meters per second.  It bounces off, there it comes, and it goes up with +5 meters per second.  I call this v1, and I call this v2.  So, what now is the average acceleration?  Well, I’ll have to know the time that it takes for this change in direction, that we call it the “impact time”.  I would say in this case, the impact time delta t is about probably a hundreds of a second, and so my average acceleration would be v2 minus v1, that is +5 minus -5, that is 10, divided by 10^-2, and that is +10^3 meters per second square.  I have observed carefully the signs.  If now say, aha, I don’t like this – I want to go this, as the increasing value of x, no big deal, this will become a plus, this will become a minus, and this would become a minus.  So the acceleration is minus a thousand meters per second square.  I have also here, a tomato, and I have here, some eggs.  Now, imagine now that I throw the tomato down, or for that matter, the egg, and they hit the floor at 5 meters per second.  I could do that, they would not come back up, they would go “psst”.  So, therefore, the change in velocity would not be 10, apart from the sign, it would only be 5 meters per second.  The impact time would probably be much longer, maybe a quarter of a second; so therefore, the average acceleration during the impact would then be only 5 divided by one quarter, it would be something like 20 meters per second square.  Whether you call it plus or whether you call it minus 20 meters per second square, depends on your convention of what you call increasing x.  But the eggs and tomatoes don’t care what you call minus or you call plus, whether the acceleration is minus 20 meters per second square or plus 20 meters per second square, you better believe it that the egg will break.  So it is only in your convention that it matters, but of course, the physics, will not change.  The eggs couldn’t care less what you have chosen for your sign convention.  Something breaks because of the magnitude of acceleration becomes too high.  That’s why something breaks.

A few days ago, I saw a Sherlock Holmes movie; and there was a guy who fell on the floor, a marble floor, and his head was lying there motionless.  And here was Watson, Watson said to Sherlock Holmes, “What happened?”  Sherlock Holmes walked over to the guy, touched him, and he said, “He crushed his skull.”  He looked very intelligent, I must say, when he said that.  “He crushed his skull,” and I said, “Gee!  That’s really physics actions, A01 all the way!”  A modest, a really modest velocity when he hit the floor, but he hit the floor like a billiard ball, the guy was balled for one thing, and so the impact time was very short.  And when the impact time was short, even if you hit the floor with a modest speed, the acceleration is high, and that was too much, and so that’s why his skull was crushed.  So what matters is changes in velocity and the impact time.

We now want to make one last step from average acceleration; we want to go to THE acceleration at any moment in time.  Just the way we did that with velocity and that now is a natural step.  The acceleration at any moment in time would be the limit for delta t goes to 0, for t measures at t plus delta t minus vt, divided by delta t.  That is the instantaneous acceleration.  And this, you will recognize is the first derivative of velocity versus time, which is also the second derivative of position versus time.  And so here comes the second equation that I really want you to remember forever and ever, and ever – that the acceleration is dv/dt, which is also d^2x/dt^2.  We can go to our plot, and we can ask ourselves the question now, where the acceleration is 0, where it is larger than 0, and where it’s smaller than 0, because this value can be larger than 0, equal to 0, and smaller than 0.  And now, you have to be very careful, when you try to derive that from this plot.  You have to be very careful because you and I have no good feelings about second derivatives.  Velocity is easy, always have to do is looking at alpha, but when it comes to the second derivative, you have to see how alpha is changing.  Well, right here, the velocity is not changing, so the acceleration everywhere here must be 0.  Here, the velocity is increasing, so the acceleration must be larger than 0 here.  Here, the velocity is almost constant; it’s almost a straight line – what does that mean for the acceleration?  Zero!  Exactly!  Here, when it makes this rounding curve, the velocity is positive here, but on this side, it’s negative, so what does it mean for the acceleration?  Negative.  You got it.  And so you can now, roughly find where the acceleration is positive, where it is negative, and where it is zero.

Let’s do a straight-forward example, the way you could be expected on the assignment, or if you’re extraordinary lucky, you might even get something like that on the exam.  Very straight-forward, I’m going to give you, the position x as a function of time, and then ask you lots of questions about it.  So this example is a working example.  x = 8 – 6t + t^2.  So this tells you where the object is at any moment in time, and that is being in meters.  What now is the velocity at any moment in time?  Well, that’s the derivative dx/dt, and I use the following, x = t^n, then as most of you should know, dx/dt = nt^(n-1), that’s all I’m using.  So the derivative of 8 is zero, I get here a minus 6, I get here plus 2t, this would be in meters per second, and the acceleration, I have to take the derivative of the velocity, I get +2.  So notice that the acceleration is constant in time; it’s not changing, but the velocity is changing.  Well, at time t = 0, just, I’ll problem a little bit; I want a feeling of what this object is doing.  At time t = 0, x = +8, the velocity is -6 meters per second, and the acceleration equals +2.  I can also ask myself, at what time is x = 0.  What are the times that x is zero?  Well, I have to solve this second order equation, which is something you’ve all done in high school, and you’ll find that that’s the case when the time is +2 and the time is +4.  Take the +2 that makes this 4, 4+8 is 12, minus 6*2, that’s 0.  So that you see 2 works, and now you check that 4 also works.  Just for my curiosity, when is the velocity zero?  Oh, that’s easy.  That’s when this equation is zero, so that’s when t = 3.  What is, at that moment, the position?  Well I substitute t = 3 in here, and that gives me -1.  x = -1.

So now, I’m ready to plot x as a function of t.  It’s of course, a parabola, and I’ll use this information that we’ve just derived.  So here comes my plot.  Let this be increasing value of x, let this be 8, and let this be -1; this is the time-axis, I have a 0 here, and so I want to cover, let’s say, about 6 seconds.  So I have 1, 2, 3, 4, 5, 6.  Now I’m going to use this information in order to give you a curve which is similar to that one except this is a simple one, which is just a parabola, so I know that at time t = 1, the object is at position 8, I know that x is 0, that x is 0 at time 2 and at time 4; so the object is here at this time (t2), and at this time (t4).  And I know that when t = 3, it is at position -1, so the object is here, and I also know that the velocity is 0, and we can check that.  And so if I make this plot now, then we would have got a curve that sort of looks like this, and yes indeed, notice, the velocity here is 0.  The angle alpha equals 0.  The object starts out at t = 0 with a negative velocity, you can see that, the object at t = 0 is here, this is where the object is, I hope you realize that, the object is never here, this is the road, this is the one-dimensional track the object is sitting.  The object is here, and it started going this direction.  If it started going this direction, the velocity must be less than 0, for v it is, it is -6.  What’s the acceleration, which is +2 in this (up) direction.  The acceleration says “I don’t want to go down!  I want to go up!”  Velocity says, “Sure.  All I can do is slowly, slowly changing.”  And that’s what it’s doing.  It’s slowly changing the velocity, and it comes the time that the velocity is 0, so the object goes down, the velocity changes, and when it is at position -1, it has come to a grounding holes, and now it’s returning this positive value of a, is now increasing the velocity, and that’s what you see.  I therefore bet you a nickel, that if you substitute in that equation, t = 4, that the velocity better be positive.  It has changed from a minus sign to a plus sign, because of this positive acceleration.  I bet you a nickel, t = 4, what is x?  Uh, what is v?  We want to know v.  8 – 6, +2 meters per second.  You see?  Physics works!  v is now +2 meters per second.  So all the information that is in there, but I want you to be able to also digest it.  Don’t look at the curve just as some dumb parabola, some dumb curve, try to imagine what is happening.  And only then, you get some insights, and you really begin to get it in your brains.

I now would like to write down, in most general form, the equation for the position and the velocity as a function of time.  For a one-dimensional motion, whereby the acceleration is constant, so it’s going to be one-dimensional again, and we have a it is going to be a constant.  And so the equation that I write down is the most general way I can write it down.  So we already got x equals some number c1, plus some c2 times t, plus some c3 times t^2.  And notice, oh, I already erased my example; my example is gone, but you would have seen it was an 8 before, and here we had, uh, what did we have?  Minus, we have -6t, and we have +1t^2.  So you recognize these 3 c’s, and I can now take the derivative, and so I get v = c2 + 2c3*t, and then I get the acceleration equals 2c3.  And now we get some insights into these quantities.  Clearly, x1, uh, c1, is the position of x at time 0, for which we often write a x0.  When t is 0, that’s where the x is.  c2 is really the velocity at time equals 0 because when t is zero, that’s where c2 is v.  And the acceleration is now changing with time, it’s 2c3; therefore c3 is half of the acceleration. Well, these give you some insights and the meaning of these quantities, and you can see, you can read now some physics in there.  c1, c2 and c3 – they’re independently, whether they’re 0, or larger than 0, or negative.  There’s no difference, each one of these combinations is a valid possibility in physics.

When we have gravity, an object is influenced by the gravitational acceleration, and the gravitational acceleration is a constant.  And we write often for that gravitational acceleration, the letter g.  Whether I drop an object, or throw it vertically up, or I throw it vertically down, it is all one-dimensional; it becomes two-dimensional when I throw it with an angle…I keep it one-dimensional, the acceleration is always the same.  And that g, the gravitational acceleration, in Boston, is 9.80 meters per second square.  It varies a little bit for different places on earth.  This gravitational acceleration is independent of the mass of the object, that I draw, of the speed of the object, of the chemical composition of the object, of the size of the object, of the shape of the object, assuming we have no air anywhere, assuming these experiments are done in vacuum.  Is it obvious that the gravitational acceleration is independent of all these quantities?  By no means, is it true?  We think so.  But I want you to appreciate that it is not obvious, and it cannot be proven from the first principle.  Remember last time, we dropped an apple from 3 meters and we dropped another one from one and half meters.  In your second assignment, which you haven’t seen yet, I’m asking you to calculate the gravitational acceleration for me, using these both experiment, and of course, I want you to also tell me what the uncertainty is in your final answer.  And I’d like to help you a little bit, to set it up, and also, to get these equations in terms of gravity.  Whenever we deal with gravity, we get the genuine there.  So suppose here’s the object at time t = 0, it is the apple; and I call that position x0, I call that 0; I’m free to choose my 0 position, and I drop it 0 speed, I just let it go because that’s the way we did in class.  The object goes down, zzzzzz, and it hits the floor.  Well, the general equations now, which dealing with gravity, if I call this increasing value of x, you can choose it differently, it is my choice to be the following:  x = x0 + v0t + 1/2 g*t^2, and g now is 9.80 meters per second square.  The velocity at any moment in time equals v0 + gt, and acceleration is constant, is simply g.  Now in my case, I’ve chosen t = 0, x0 = 0, and I’ve chosen this 0, so these go.  And so you see, that when your object is here, which is 3 meters below this point, and you know the time, how long it took to get there, which you can now calculate g.  Because x will then be 3 meters, that’s when it’s here, we made a measure in class, how long it took, so you know the time, and so you can come up with a value for g.  You can do that for both measurements, and of course, I want you to tell me also, what the uncertainty is, in both measurements.

Remember that we derived last time, that t, that time it takes for the apple to fall is c times the square root of h/g, and we never knew what the c was.  I did a demonstration to show you that the time is proportional to the square root of h/g.  We never knew what that c was.  Now you know, because now you have the equations here, and you will see that c was simply the square root of 2.  I could not derive that from my dimensional analysis.

Now, I want you to relax, and at the same time, get a little bit alert for a change.  Look at this situation, v = gt, that means when I drop an apple, and I’m going to drop another one today, that the velocity increases with time.  So if I strobe this apple while it is falling, I would see the separation when it strobes to increase with time because the velocity goes up with time.  I have here an apple, or I’m going to put an apple up, about 3 meters from the floor, 3 meters.  So the height is 3 meters, approximately.  We know from last time, remember we did it and it was about 780 milliseconds for it to hits the floor, I would just round it off, and I think it’s about eight tenth of a second, just to get an idea.  If I flash, if I strobe it, twice per second, we call that two hertz, so my strobe is two times per second, then I should hit that ball when it’s falling twice, with my strobe lights.  I don’t know where it is though, because when we strobe it and when I let the apple go, the two of them synchronize, so maybe the first time that the light blinks may be here, and the second time it may be here.  But it is also possible that the first time is here, and the second time is there.  And so the first thing I want to do is to test your alertness – we will blink, you will tell me where you see them, but we will take a picture, we will take a picture which show us exactly where the two balls were.  That’s the first alertness test, so get ready for this, and then we’ll do a second one, which is even more intrigued.

So now I have to first, lower this pelmet, so we get a nice, dark background.  There we go.  Woah, there are my fingerprints on it; it’s not so black anymore.  There it is, that’s the background.  Oh, what am I doing?  I need this ladder again; I need to bring the apple up.  Friday is always a bad day for me.  Ok, so now I’m going to bring the apple up.  There’s a medal here, and electromagnetic… so I throw the switch here, so the electromagnetic is activated.  Very similar with what we did last time…so I put the apple up; and the apple is hanging there.  There we go.  So now I have to start the uh, the…strobe.  That’s about two hertz; that’s about 2 flashes per second, and I’m going to make it pitch black, pitch black.  I will count down 3, 2, 1, 0; and Bob there who is behind the camera will open the shutter when I say 1, and when I say 0, the ball will fall…so you may always see the ball in its highest position, that may not count there, of course, because it makes it two flashes in the time that the shutter is opened and that I drop it.  Ok, if you’re ready, I’m ready.  Make it as dark as we can.

Bob, are you ready?  Class ready?  Everyone ready?  You don’t look ready…Ok, 3, 2, 1, and 0.  So let’s look at this again – in slow motion.  There was the ball.  Oh, oh, boy, you tried that ticktack times.  You’ll never do that again.  So now we’re developing the picture, and I would like you to tell me where you saw the balls.  Where were they, roughly?  Where was the first one?  How much below the highest point?  Only this much, the first one.  And the second one was pretty low then, ok.  Sounds interesting.  We’ll take a look.  While the picture is developing, I’m now going to test your real alertness.  I’m going to strobe it, it is an unknown frequency, unknown to you – I’ll tell you a secret: it’s a higher frequency.  You’re going to see more balls on the way down.  I’m not going to ask you where they are exactly, all I want you to tell me afterwards is how many you saw.  That’s all.  To count them as it falls.  You know you only have 0.8 seconds to count.  Well, how did the picture come out?  Woah, you’re good!  Hoh, you’re good – it was very high actually, the first uh, the first flash, very high.  You see, you did very well.  We’re going to start now with the second part.  Is it always to it?  Should be.  So I activated the magnetic again – there it is.  Oh my goodness.  Working?  Ok, thank you, Bob.  Ok, Bob, if you’re ready, I’m ready.  I’m going to make it as dark as we can.  So all I want you to tell me is how many balls will you see.  Oh, oh, oh, oh, I have to change, oh my goodness.  Come on, you’re now in MIT, or you think?  Alright, ready?  Bob, you okay?  (Bob: okay.)  3, 2, 1…well, who saw 3? 4?  4! I wanna know 4.  5?  Here’s a five, there’s a five, another five.  Who saw six?  Woah!  7?  8?  9?  10?  11?  Who’s just saw blur?  Here’s the real winner I think.  Well, I’ll tell you – it’s 10 hertz.  Since it’s 0.8 seconds, depending upon where you hit it, how lucky you are; I’ll show you, you’ll either see 7 or maybe 8 balls.  But it was a good test, and for those of you who thought there was only, that they only saw 5; there you see them, and let’s count them.  Let’s count them together.  1, this is 1, 2, 3, 4, 5, 6, 7, this is a bounce.  So for those of you who saw 5, I’ll say take some rest this weekend, you’ll need it, and I’ll need it too.  See you Monday.