The Tautochrone

Learning Goals: students can sit back and watch the great minds of the late 17th century use the calculus they’ve been learning all year solve a very interesting problem.

When we looked at the differential equation y′′ + a2y = 0 that was a simplification of the pendulum equation, we found the general solution A cos(at) + B sin(at) where A and B are constants used to set up initial conditions (displacement and speed) of the pendulum bob. Note that when we simplified the pendulum equation, we found that so that the period of a = g / l the motion depends only on the strength of gravity and the length of the pendulum cord. But real pendulums don’t work like that. The small angle approximation goes wrong by about 1% after just a few degrees of swing. If you’re trying to navigate a ship using a pendulum clock, 1% time differences become a problem really quickly!

The problem is that real pendulums have a longer period for a larger swing. For larger θ, sin(θ) is less that θ, so a isn’t really constant, and gets smaller as θ gets larger, leader to a longer period. To make a more accurate clock, we need a pendulum that doesn’t swing in the arc of a circle. At a given height, the pendulum will need a steeper swing, so that it accelerates more and moves faster, accounting for the greater length of the path it has to follow. In other words, the tautochrone—a curve where objects that move along it under gravity all take the same amount of time to reach bottom—will be steeper than a circle. At right, the circular arc is a dashed line, while the pendulum with the same length cord that swings in a tautochrone is the solid curve.

The first person to solve for what shape the tautochrone actually is was Christian Huygens, though he did it completely geometrically. We will do it using calculus.

The trick to getting going in this problem is to think about the distance of travel along the curve. Let s be the distance along the curve, starting at the bottom and moving up to the right. The key observation is that to have a period independent of distance traveled along the curve, the motion along the curve must satisfy the simplified formula s′′ + a2s = 0 that we saw above! If this is not immediately obvious to you (and it probably shouldn’t be!) here is a very good reason why it should be true. Let there be two pendulums swinging, but start them in two different places along the curve, say at s1 and s2. Perhaps s2 is twice as far along the curve as s1. Since it has twice as far to go, it should travel twice as fast to make it to the bottom at the same time, so at any time v2(t) = 2v1(t). But to get to twice the speed, the acceleration at any point in time should be twice as much. But acceleration is d2s/dt2 = s′′. Since “twice as far” could be any multiple, we should just say that s′′ is proportional to s. (The reason we need to deal with s′′ is that it is the acceleration, and that is related to force, and we know about the forces acting on the pendulum—gravity—while we don’t know about the speed it is traveling.) So s′′ = –a2s. We have a negative because the larger s—the farther to the right—the more the acceleration should be back toward the center. And we use a2 for the constant because that’s what we did before and the a had a nice interpretation, so we’ll see what happens here.

So we have s′′ = –a2s. We know the solution to this differential equation is A cos(at) + B sin(at), but that only tells us how far along the curve the pendulum is. It doesn’t tell us what curve this is. For that, we need x’s and y’s.

Let’s say the pendulum as at some point on the curve. What actually is the acceleration experienced by the pendulum at this point? Let the tangent line to the tautochrone at this point make an angle of θ with the horizontal. When θ is zero, the pendulum is at the bottom of its swing, and the pull of the pendulum cord straight upward cancels out gravity pulling straight downward. So there is no acceleration on the pendulum. At other points, the pendulum cord pulls at angle θ to the vertical, while gravity still pulls straight down. So the acceleration along the curve is the component of gravity that is tangent to the curve—g sin(θ). So our first transformation changes our equation to g sin(θ) = –a2s.

For our next trick, we will differentiate with respect to x. So g cos(θ) dθ/dx = –a2 ds/dx (don’t forget the chain rule on the left-hand side!).

But what is ds/dx? If we change x by a little bit, how far does the pendulum actually move? 2 ds ⎛ dy ⎞ This is just arclength! = 1+ . Now dy/dx is simply the slope of the curve, which we dx ⎜ dx ⎟ ⎝ ⎠ ds 2 also know is tan(θ). So = 1+ (tan(θ )) = sec2 (θ) = sec(θ). (We can choose the positive dx square root because we will always use θ between –π/2 and π/2, otherwise the pendulum bob goes above the horizontal line where the pendulum is centered.) Substitute this inteo our equation: g cos(θ) dθ/dx = –a2 sec(θ).

This is a separable differential equation! Solve it for x: –g/a2 cos2(θ) dθ = dx. This is easy to 2 integrate and we find that x = –g/(4a ) (2θ + sin(2θ) ) + Cx.

How can we figure out y? Well, dy/dx = tan(θ) so dy = tan(θ) dx. So multiply both sides of the last differential equation by tan(θ) and obtain –g/a2 sin(θ)cos(θ) dθ = dy. We can easily integrate 2 this to find y = g/(4a ) cos(2θ) + Cy.

Finally, we set the constants so that the curve passes through the origin when θ = 0, and obtain ⎧ g x = − 2θ + sin(2θ) ⎪ 4a2 ( ) ⎨ . These happen to be the parametric equations of a cycloid! (It is ⎪ g y = 2 (cos(2θ) −1) ⎩⎪ 4a upside-down because we have cos(2θ) – 1 and not the other way around.)

Note a few more things about the final answer. The solution to the original differential equation for s is A cos(at) + B sin(at). The time required to go from the top of the curve to the bottom is then π/(2a). Since this is a tautochrone, this is the time required if we start the pendulum anywhere—it will swing less and do so slower so that it takes the same amount of time on each swing. Now the radius of the circle used to create the cycloid is g/(4a2). So in terms of the radius of the circle, the time needed to slide down the cycloid is . π r / g

Now the next question is: how can you actually get a pendulum to swing in a cycloid? The answer is to build in restraining walls that restrict the cord from swinging too far, and only lets less and less of it swing causing the effective length to shorten up as the swing becomes wider. These walls are called the “evolute” of the original cycloid.

The truly amazing thing about a cycloid is that its evolute…is another cycloid of exactly the same size, just shifted up and slid over!

Once Huygens discovered all this, he actually built a clock using a cycloidally swinging pendulum. And…it kept time worse than a regular pendulum clock! In theory it should have worked perfectly. What went wrong? The simple answer is that there was too much friction as the cord wound along the restraining walls, causing the clock to slow down. The effect turned out to be larger than the effect of approximating sin(θ) by θ.

Oh, well. Not all great ideas really work out!

Bonus—a geometric proof that the evolute of a cycloid is another cycloid.

The picture at right show an important fact about cycloids. If P is a point on the cycloid, then you consider the circle that is tracing out the cycloid, then the tangent to the cycloid at P intersects the very bottom point of the circle, and the normal to the cycloid at P touches the very top point of the circle. The proof is to consider velocities. As � the circle traces out the cycloid, the point doing the tracing has a velocity composed of two pieces: the velocity it would have if the circle were rolling in place, and the velocity it would have it the circle were not rolling but was just sliding along. So the tangent vector to the cycloid is composed of these two pieces. Now these two parts have equal magnitude (the edge of the circle’s linear speed is the speed with which the circle is rolling!) and are respectively parallel to the two vectors shown in the diagram—one tangent to the circle and one horizontal. If the vectors shown have equal length, then the end of the horizontal is tangent at the bottom of the circle—because they are two tangents from the same point. So since the components of the velocity vector are simply multiples of these two, this velocity vector—the tangle line at the point P—passes through the bottom of the circle. Then taking the normal to the cycloid, the angle inscribed in the circle will be 90°, so the arc subtended is a semicircle, and the normal passes through the exact top of the circle.

Applying this to the picture above, where we have two cycloids 180° out of phase and stacked on top of each other, the line will be tangent to one cycloid and normal to the other. Since the two circles have equal size, the circles’ point of tangency is the midpoint of the segment.

Thus, if you wrap a string around the upper cycloid, and pull it tight and use it to draw the lower curve, the only way to keep the curve being drawn perpendicular to the unwrapping cord is for the lower curve to be a cycloid!