Motion Along a Curve, Part 2

I would like to find equations for the position of a small ball rolling along a curve given by an equation, rather than a flat incline. In Part 1, I introduced the problem and set up a test case. Now let’s try to actually make some headway.

Last time, we showed that

$\displaystyle v = \sqrt{2g(y_i - y)}$

where v is the magnitude of the velocity vector v, g is the acceleration due to gravity, and y is the height. I note that if y>yi, then the quantity inside the radical is negative and v will not be real — which makes sense, since the ball should never be able to roll higher than its starting position. (Recall that I am starting with an initial velocity of zero). Since we’re assuming that the ball rolls along the function, its direction will be given by the derivative of y with respect to x. I do see a potential problem with notation arising. The value of y depends on x (by definition, since that’s how I’m defining the curve). But x and y both depend on t if we’re thinking about the motion of the ball. I want to define our variables carefully. We should think of x as a function x(t), and y as a function of x, so y(x) = y[x(t)]. I am going to reserve the prime notation y′ specifically to refer to the derivative with respect to x.

$\displaystyle y' = \frac{dy}{dx}$

We know that

$\displaystyle \mathbf{v} = \langle r, \angle\theta \rangle$

The magnitude will be v. If we take the tangent line to the curve and sketch a triangle, we can see that for a change in x of 1, y changes by y′, so tan θ = y′/1. I illustrated this below with an arbitrary curve.

$\displaystyle \mathbf{v} = \langle v, \angle \arctan y' \rangle$

Let’s now find the component vectors vx and vy, where of course vx = r cos θ and vy = r sin θ.

$\displaystyle \mathbf{v} = \langle v \cos \arctan y', v \sin \arctan y' \rangle$

There’s actually no need for these arctangents. If I go back to the earlier sketched triangle, I can see that the hypotenuse is $\sqrt{(y')^2 +1}$ and so I can read the cosine and sine directly off that triangle (1 over the hypotenuse and y’ over the hypotenuse, respectively):

$\displaystyle \mathbf{v} = \left\langle v \frac{1}{\sqrt{(y')^2 + 1}}, v \frac{y'}{\sqrt{(y')^2 +1}} \right\rangle$

$\displaystyle \mathbf{v} = \frac{v}{\sqrt{(y')^2+1}}\langle 1, y' \rangle$

Let’s substitute in our expression for the magnitude, v:

$\displaystyle \mathbf{v} = \sqrt{\frac{2g(y_i-y)}{(y')^2+1}} \langle 1, y' \rangle \blacktriangleleft$

I believe that this should be the general solution, given the constraints (ball must roll along the curve, uniform gravity, no friction, starting x and v of 0). Let’s pause to make some observations. The expression in the numerator is the magnitude of the velocity which we found in the last post. The potential/kinetic energy relationship is obvious, with the product of g and the change in height from potential energy, and the times two and the square root being inverses from kinetic energy. The <1, y′> ensures the proper direction, and the denominator adjusts for the length of that direction vector (it basically makes it a unit vector).

This equation is so far pretty straightforward. However, in order to find equations for x and y in terms of time t, we’d have to plug in the expression for y and y′, then integrate the velocity. Let’s try this with what should be a very simple example, the incline (line) from the prior entry:

$\displaystyle y = -\frac{x\sqrt{3}}{3} + 1$

$\displaystyle y' = -\frac{\sqrt{3}}{3}$

$\displaystyle y_i = y(0) = 1$

Plugging these into our equation for v gives:

$\displaystyle \mathbf{v} = \sqrt{\frac{2g(y_i-y)}{(y')^2+1}} \langle 1, y' \rangle$

$\displaystyle \mathbf{v} = \sqrt{\frac{2g[1-(-\frac{x\sqrt{3}}{3}+1)]}{(-\frac{\sqrt{3}}{3})^2+1}} \left\langle 1, -\frac{\sqrt{3}}{3} \right\rangle$

I am starting to regret setting my initial angle to π/6. There was no benefit to choosing a “nice” value for the angle; it would have been better to choose a slope with an integer value. However, it’s too late to go back and change it now. Let me see if I can simplify this.

$\displaystyle \mathbf{v} = \sqrt{\frac{2g\frac{x\sqrt{3}}{3}}{\frac{1}{3}+1}} \left\langle 1, -\frac{\sqrt{3}}{3} \right\rangle$

$\displaystyle \mathbf{v} = \sqrt{\frac{2gx\sqrt{3}}{\frac{4}{3}\cdot 3}} \left\langle 1, -\frac{\sqrt{3}}{3} \right\rangle$

$\displaystyle \mathbf{v} = \sqrt{\frac{gx\sqrt{3}}{2}} \left\langle 1, -\frac{\sqrt{3}}{3} \right\rangle$

Let’s split this into its component vectors, and switch to magnitudes:

$\displaystyle \mathbf{v} = \mathbf{v_x} + \mathbf{v_y}$

$\displaystyle \mathbf{v} = v_x \boldsymbol{\hat{\i}} + v_y \boldsymbol{\hat{\j}}$

$\displaystyle v_x = \sqrt{\frac{gx\sqrt{3}}{2}}$

and

$\displaystyle v_y = -\sqrt{\frac{gx\sqrt{3}}{2}} \cdot\frac{\sqrt{3}}{3} = -\sqrt{\frac{gx\sqrt{3}}{6}}$

Let’s now try integrating the first one:

$\displaystyle \frac{dx}{dt} = \sqrt{\frac{gx\sqrt{3}}{2}}$

$\displaystyle \frac{1}{\sqrt{x}} \, dx = \sqrt{\frac{g\sqrt{3}}{2}} \, dt$

$\displaystyle \int x^{-\frac{1}{2}} dx = \int \sqrt{\frac{g\sqrt{3}}{2}} \, dt$

$\displaystyle \frac{x^{\frac{1}{2}}}{\frac{1}{2}} + C_1 = t \sqrt{\frac{g\sqrt{3}}{2}} + C_2$

$\displaystyle 2\sqrt{x} = t \sqrt{\frac{g\sqrt{3}}{2}} + C_3$

$\displaystyle \sqrt{x} = t \sqrt{\frac{g\sqrt{3}}{8}} + C_3$

This is looking promising. Before I square both sides, let’s deal with the constant. Since we’re starting at a position x=0 at t=0,

$\displaystyle 0 = 0 + C_3$

so clearly the constant is zero. (This won’t be so simple for the y-component.)

$\displaystyle \sqrt{x} = t \sqrt{\frac{g\sqrt{3}}{8}}$

I will square both sides, realizing that I might be introducing extraneous solutions. In fact, t cannot be negative, since the square root of x is nonnegative.

$\displaystyle x(t) = \frac{g\sqrt{3}}{8} t^2, \, t\geq 0 \blacktriangleleft$

Before tackling the y-direction, let’s test this answer. Let’s plug in our value from the last post for t, for the time when it reaches the x-axis.

$\displaystyle x\left(\sqrt{\frac{8}{g}}\right) = \frac{g\sqrt{3}}{8} \left(\sqrt{\frac{8}{g}}\right)^2 = \frac{g\sqrt{3}}{8} \cdot \frac{8}{g} = \sqrt{3}$

This is indeed the value of x at the base of the triangle (recall height 1, width √3, hypotenuse 2 for a 30-60-90 triangle)! Now let’s find y. Let’s start with our equation for the y-component of the velocity:

$\displaystyle v_y = -\sqrt{\frac{gx\sqrt{3}}{6}}$

This one is going to be a little more complicated.

$\displaystyle \frac{dy}{dt} = -\sqrt{\frac{gx\sqrt{3}}{6}}$

I already see a complication. My equation has x, not y. We of course have our starting equation for y in terms of x, and it would be easy to solve that for x and substitute it in. However, if we want to try this on other functions, especially those that aren’t injective (that is, where multiple values of x might result in the same value of y — in a parabola, or sine curve for instance), then solving will be difficult. I think it would be better to try to use the chain rule.

$\displaystyle \frac{dy}{dx} \frac{dx}{dt} = -\sqrt{\frac{gx\sqrt{3}}{6}}$

$\displaystyle -\frac{\sqrt{3}}{3} \frac{dx}{dt} = -\sqrt{\frac{gx\sqrt{3}}{6}}$

I just realized something. We already have $\frac{dx}{dt}$. It was the vx we started with to find x(t) above!

Wait. That expression is in terms of x, and we have no y’s left. Substituting that in will just result in an identity.

$\displaystyle -\frac{\sqrt{3}}{3} \sqrt{\frac{gx\sqrt{3}}{2}} = -\sqrt{\frac{gx\sqrt{3}}{6}}$

$\displaystyle -\sqrt{\frac{gx\sqrt{3}}{6}} = -\sqrt{\frac{gx\sqrt{3}}{6}}$

This is bad news, because I really don’t think that solving the starter equation for y will be a practical solution for most functions. But if we don’t have a y or a dy, then we’re not going to get an equation for y.

Scratch that. There’s no need to do any of this! We just derived an equation x(t). We already know y(x). So we can find y[x(t)]!

$\displaystyle y(x) = -\frac{x\sqrt{3}}{3} + 1$

$\displaystyle x(t) = \frac{g\sqrt{3}}{8}t^2$

So, plugging in the equation for x in,

$\displaystyle y(t) = -\frac{\sqrt{3}}{3} \left(\frac{g\sqrt{3}}{8}\right) t^2 + 1$

$\displaystyle y(t) = -\frac{g}{8}t^2 + 1, \, t\geq 0 \blacktriangleleft$

This looks right. It’s of a similar form to our equation for x(t), but with a factor of -1 instead √3 (which follows from our 30-60-90 triangle, and y is decreasing. And it has a +1 to raise the initial height. Just to be sure, let’s check it with our test value for t:

$\displaystyle y\left(\sqrt{\frac{8}{g}}\right) = -\frac{g}{8}\left(\sqrt{\frac{8}{g}}\right)^2 + 1 = -1 + 1 = 0$

And that’s correct, because we specifically picked the test case for when the ball reaches the x-intercept. So our final solution is

$\displaystyle x(t) = \frac{g\sqrt{3}}{8} t^2, \,t\geq0 \blacktriangleleft$

$\displaystyle y(t) = -\frac{g}{8}t^2 + 1, \, t\geq 0 \blacktriangleleft$

The method works! In future explorations, it would be interesting to try this for a more complicated starter equation.

The restriction t≥0 is interesting. If it weren’t there, the equations for the motion of the ball would still be valid, since it would represent the ball being launched backwards up the slope, coming to rest for an instant at (0,1) at t=0, then rolling back down the slope. I’m not surprised that it is there, though, because my method explicitly assumed that the ball’s velocity vector points in the direction <1, y′>, which points in the forward x direction. Let’s trace back the restriction on t. It comes from where it times a constant equals √x, which must be positive. Why did we have √x, as opposed to ±√x? That came from the equation for the magnitude of the velocity v. That in turn came from the fact that K = ½mv2. Should we have kept the negative solution? No, because v represents the magnitude, and the magnitude is nonnegative. Still, a negative v would correspond to a vector pointing in the opposite direction, when put into our component equation. I believe that those solutions should be allowed. I could justify it by saying that v was not the magnitude, but rather ±magnitude, to allow the vector to point in either direction. Now that I think about it, this will occur for every problem we attempt with this method. Since the laws of physics and therefore the equations of motion should be symmetric over time, this means that if the ball is starting at rest at time = 0, we could “run the film in reverse” to see what path it took to get to the starting point. This is an interesting development, and I wonder if expanding the method in this manner would allow the ball to roll backwards at other times, too — say if it were to roll down a valley, then up the other side. If we allow these “negative magnitudes,” might we be able to get an equation which expresses the motion as it rolls back down into the valley and back up the original side?

Motion Along a Curve, Part 1

Earlier this week, I saw an intriguing problem suggested, one involving motion along a curve. The idea is this: in elementary mechanics, we learn about how an object will move when placed on an inclined surface. If we restrict motion to two dimensions, this surface can be represented by a line. With gravity as the propelling force, can we find a general approach to deriving its motion along an arbitrary curve? In other words, given a function f(x), can we find functions describing the position (and therefore velocity and acceleration) over time?

Here are my initial thoughts. One, I am sure that this type of problem has been analyzed before. That doesn’t matter; I am going to try figuring it out myself. Two, I think we can use conservation of energy to determine the object’s velocity at any point. However, that will depend on knowing its height, and I don’t know how this approach would handle the object flying up off a short hill, or dropping away from a cliff. So three, I am going to restrict the object’s motion to the curve — consider it a track, rather than a 1D surface. That means that four, I can use the derivative to find the direction, but that means the object can only move forward. Five, I have to ignore friction. My conservation of energy approach will require that all energy be potential or kinetic; I won’t know hot to deal with losses to friction. Six, I am using constant downward gravitation. Seven, I am starting with initial position x=0 and initial velocity of 0. I don’t believe this method requires it, but I am already concerned about the complexity of the math. And seven, I am planning to obtain a velocity equation and integrate to find position, but I don’t know if the equation will have an analytic solution (or if I will know how to integrate it).

These are some significant limitations, of course. If this approach works, I suspect that it could be expanded to deal with several of these. For instance, we could keep in terms for initial position and velocity. Gravity or whichever force could be represented by a vector field rather than a constant force; this would make the potential energy term much more complex. I also wonder if I could incorporate friction by adding in an extra term, but I think this would turn the equation into a more complex differential equation.

Let’s start with a simple case that we can solve using conventional means, so we’ll have an answer to check later. I can also use this to check my conservation of energy approach. Let’s have the object rolling down an incline of 30° (π/6), starting at a height of 1. The ball will roll down to the right (I find it more intuitive to imagine a rolling ball rather than a frictionless object sliding, especially if the surface is curved). The equation for this surface will be

$\displaystyle y(x) = -\frac{x\sqrt{3}}{3}+1 \, \bullet$

where the slope is -tan(π/6).

You can see that it would form a right triangle with height 1, length √3, and hypotenuse 2. The slope is therefore 1/√3, or √3/3. Perhaps it would have been better to use an incline of 60° (π/3) so that the slope would be -√3, but I like this one better. Let’s try the conservation of energy approach to see what the velocity would be when the ball reaches the bottom (that is, y = 0). The total energy E is the sum of the potential energy U and the kinetic energy K. This should remain constant, so

$\displaystyle E_i = E_f$

$\displaystyle U_i + K_i = U_f + K_f$

We know that potential energy is given by

$\displaystyle U = mgh$

assuming constant gravitation g (the acceleration due to gravity), with mass m and height h; and kinetic energy is given by

$\displaystyle K = \frac{1}{2}mv^2$

where v is the magnitude of the velocity vector v. I confess I am not very proficient with vectors, but clearly the ball is moving in two dimensions and we will need both components in the x and y directions. I am going to try to not be sloppy but to carefully think about what we mean by v or v every time I write it.

We assume that the initial velocity, and therefore kinetic energy, is zero.

$\displaystyle U_i + K_i = U_f + K_f$

$\displaystyle mgh_i + \frac{1}{2}m\cdot0^2 = mgh_f + \frac{1}{2}mv_f^2$

The height is y, and we can divide both sides by the nonzero mass m:

$\displaystyle gy_i = gy_f + \frac{1}{2}v_f^2$

Let’s solve for v:

$\displaystyle 2g(y_i-y_f)=v_f^2$

$\displaystyle v_f = \sqrt{2g(y_i-y_f)} \blacktriangleleft$

This should be the general case, so if this checks out, we can use this equation to develop the method. As a check, if the units of g are m/s² and y is in m, the units will be m/s, appropriate for velocity. Also note that this is simply the magnitude of the velocity vector (the speed); the actual vector could point in any direction if all we know is conservation of energy. That’s why I earlier constrained the ball to move along the curve — that will give us our direction. Plugging in our initial and final values of y, we get

$\displaystyle v_f = \sqrt{2g(1-0)} = \sqrt{2g}$

Next, to verify this solution, let’s solve using the traditional free-body approach. Gravity applies a force downwards of W=mg. This can be split into two components, one perpendicular to the surface mg cos(θ), and one along the surface mg sin(θ). The perpendicular force will be balanced out by the equal and opposite normal force, leaving a force of mg sin(θ). This is simple motion in one dimension, and let me introduce a variable s to represent its position along the surface (so I don’t generate confusion with our x’s and y’s). Think of it as placing a tape measure from the bottom of the incline up to the starting point. We know that for a given constant acceleration a,

$\displaystyle v = \int \! a \, dt = at + v_0$

$\displaystyle s = \int \! v \, dt = \int \! (at+v_0) \, dt = \frac{1}{2}at^2 + v_0 t + s_0$

By Newton’s second law, F = ma, so a = g sin(θ) and will be negative, since it’s pointing towards decreasing s (since this is motion in one dimension, we can use the sign to expression direction rather than need vectors). Initial velocity is zero, and initial position is 2 (the distance along the incline, if the bottom is at s=0).

$\displaystyle 0 = -\frac{1}{2}g \sin(\frac{\pi}{6}) \cdot t^2 + 0t + 2$

$\displaystyle \frac{1}{2} g \left( \frac{1}{2}\right)t^2 = 2$

$\displaystyle \frac{gt^2}{4} = 2$

$\displaystyle t^2 = \frac{8}{g}$

$\displaystyle t = \pm \sqrt{\frac{8}{g}}$

I can discard the negative solution, since I’m interested in the behavior starting from time = 0, not how it might have been launched before to come to a temporary stop at time = 0. Let’s plug this time to get our velocity from v = at (since the initial velocity is zero):

$\displaystyle v = at = -g\left(\frac{1}{2}\right)\sqrt{\frac{8}{g}} = - \sqrt{\frac{g^2}{4}}\sqrt{\frac{8}{g}} = -\sqrt{2g}$

This velocity is negative since it points downwards along the s direction, along the incline. Its magnitude will be the absolute value which is what we obtained earlier using the conservation of energy method. Armed with this success, and with some test values, we’ll be ready to work on the actual problem in the next post.

Simple Harmonic Motion

I came across an interesting approach to a problem involving a spring the other day, and it made me wonder if I could derive the equations for simple harmonic motion. The basic principle of a spring is Hooke’s law, that the spring exerts a restoring force proportional to its displacement from equilibrium position: F = –kx, where k is the spring constant representing the stiffness of the spring. I wondered if I could derive the formulas for its motion, and where the sinusoidal components would appear from, since clearly there are no trigonometric functions in this basic equation. To begin with, from Newton’s second law, F = ma:

$\displaystyle ma = -kx$

$\displaystyle a = -\frac{kx}{m}$

We know that acceleration is the derivative of velocity and velocity is the derivative of position:

$\displaystyle a = \frac{dv}{dt}$

and

$\displaystyle v = \frac{dx}{dt} .$

We can rewrite these as a dt = dv and v dt = dx, and integrate. However, our acceleration equation has a in terms of position x, not time t, so I would prefer to integrate with respect to x (as at this point I don’t know how x depends on t — that’s what we’re trying to derive!).

$\displaystyle a \, dt = dv$

$\displaystyle a \left( \frac{dx}{v} \right) = dv$

$\displaystyle a \, dx = v \, dv$

This equation is well-known and could have been used as a starting point, but it’s nice to derive it from the most basic principles. We can now substitute our expression in for a and integrate both sides.

$\displaystyle -\frac{kx}{m} \, dx = v \, dv$

$\displaystyle \int \! \left( -\frac{kx}{m} \right) \, dx = \int \! v \, dv$

$\displaystyle -\frac{k}{m} \int \! x \, dx = \int \! v \, dv$

$\displaystyle -\frac{k}{m} \left( \frac{x^2}{2} + C_1 \right) = \frac{v^2}{2} + C_2$

Let’s solve this for v:

$\displaystyle \frac{v^2}{2} + C_2 = -\frac{k}{m} \left( \frac{x^2}{2} + C_1 \right)$

$\displaystyle \frac{v^2}{2} + C_2 = -\frac{kx^2}{2m} + C_3$

$\displaystyle \frac{v^2}{2} = -\frac{kx^2}{2m} + C_4$

$\displaystyle v^2 = -\frac{kx^2}{m} + C_5$

$\displaystyle v = \pm \sqrt { -\frac{kx^2}{m} + C_5 }$

There’s still no sign of sine appearing anywhere (I know; I couldn’t resist). I’m also a bit uncomfortable with the plus/minus sign, but velocity can be both positive and negative, so I don’t think that discarding the negative solution is justified. I want to integrate again to remove the velocity, but we face a similar problem as before: we have velocity in terms of position x, not time t. We can rearrange the differential again:

$\displaystyle v = \frac{dx}{dt}$

to get v on the dx side:

$\displaystyle dt = \frac{dx}{v} .$

Let’s substitute in our expression for v, and integrate:

$\displaystyle dt = \frac{dx}{\pm \sqrt { -\frac{kx^2}{m} + C_5 } }$

$\displaystyle \int dt = \int \! \frac{dx}{\pm \sqrt { -\frac{kx^2}{m} + C_5 } }$

This is not pretty, and I’m not sure how to integrate the right side. I believe I can pull out the plus/minus as a constant.

$\displaystyle t + C_6 = \pm \int \! \frac{dx}{\sqrt { -\frac{k}{m} x^2 + C_5 } }$

I had to consult a table of integrals for assistance with this integration. Of course, it can be solved with trignometric substitution! All of a sudden, we have a hint of where a sine function will appear. We need to get the radical in the form $\sqrt{a^2 - b^2 x^2 }$:

$\displaystyle \sqrt{C_5 - \frac{k}{m} x^2 }$

$\displaystyle a = \sqrt{C_5} = C_7$

$\displaystyle b= \sqrt{\frac{k}{m}}$

Now, we set

$\displaystyle x = \frac{a}{b} \sin \theta :$

$\displaystyle x = C_7\sqrt{\frac{m}{k}}\sin \theta$

Let’s now take our original integral back, and start substituting.

$\displaystyle t + C_6 = \pm \int \! \frac{dx}{\sqrt { C_5 -\frac{k}{m} x^2 } }$

$\displaystyle t + C_6 = \pm \int \! \frac{dx}{\sqrt { {C_7}^2 -\frac{k}{m} x^2 } }$

$\displaystyle t + C_6 = \pm \int \! \frac{dx}{\sqrt { {C_7}^2 -\frac{k}{m} \left( C_7\sqrt{\frac{m}{k}}\sin \theta \right) ^2 } }$

This expression is complicated, but I can already see how it will simplify out.

$\displaystyle t + C_5 = \pm \int \! \frac{dx}{\sqrt{ {C_7}^2 - \frac{k}{m} \cdot {C_7}^2 \cdot \frac{m}{k} \cdot \sin^2 \theta} }$

$\displaystyle t + C_5 = \pm \int \! \frac{dx}{\sqrt{ {C_7}^2 - {C_7}^2 \sin^2 \theta } }$

$\displaystyle t + C_5 = \pm \int \! \frac{dx}{\sqrt{{C_7}^2 (1 - \sin^2 \theta)} }$

$\displaystyle t + C_5 = \pm \frac{1}{C_7} \int \! \frac{dx}{\sqrt{1 - \sin^2 \theta} }$

Let’s convert the dx to a dθ:

$\displaystyle x = C_7\sqrt{\frac{m}{k}}\sin \theta$

$\displaystyle \frac{dx}{d\theta} = C_7\sqrt{\frac{m}{k}}\cos \theta$

$\displaystyle dx = C_7\sqrt{\frac{m}{k}}\cos \theta \, d\theta$

Substituting this back in yields:

$\displaystyle t + C_5 = \pm \frac{1}{C_7} \int \! \frac{C_7\sqrt{\frac{m}{k}}\cos \theta \, d\theta }{\sqrt{1 - \sin^2 \theta} }$

$\displaystyle t + C_5 = \pm \frac{C_7}{C_7}\sqrt{\frac{m}{k}}\int \! \frac {\cos \theta}{\sqrt{1 - \sin^2 \theta}} d\theta$

$\displaystyle t + C_5 = \pm \sqrt{\frac{m}{k}}\int \! \frac{\cos \theta}{\sqrt{\cos^2 \theta}} d\theta$

$\displaystyle t + C_5 = \pm \sqrt{\frac{m}{k}}\int \! \frac{\cos \theta}{\cos \theta} d\theta$

$\displaystyle t + C_5 = \pm \sqrt{\frac{m}{k}}\int d\theta$

$\displaystyle t + C_5 = \pm \theta \sqrt{\frac{m}{k}} + C_8$

$\displaystyle t + C_9 = \pm \theta \sqrt{\frac{m}{k}}$

Remembering our equation for x in terms of θ:

$\displaystyle C_7\sqrt{\frac{m}{k}}\sin \theta = x$

$\displaystyle \sqrt{\frac{m}{k}}\sin \theta = C_{10}x$

$\displaystyle \sin \theta = C_{10}x\sqrt{\frac{k}{m}}$

$\displaystyle \theta = \arcsin \left(C_{10}x\sqrt{\frac{k}{m}} \right)$

Finally, we can substitute this back in for θ and solve for x:

$\displaystyle t + C_9 = \pm \left[ \arcsin \left(C_{10}x\sqrt{\frac{k}{m}} \right) \right] \sqrt{\frac{m}{k}}$

$\displaystyle \pm t\sqrt{\frac{k}{m}} + C_{11} = \arcsin \left(C_{10}x\sqrt{\frac{k}{m}} \right)$

The plus-minus is driving me crazy, but I can’t justify dropping it — even though I want to.

$\displaystyle \sin \left( \pm t\sqrt{\frac{k}{m}}+ C_{11} \right) = C_{10}x\sqrt{\frac{k}{m}}$

Actually, I think I can get rid of it now! Since sin(-x) = -sin(x) = sin(x+π), I can incorporate the +π into the constant!

$\displaystyle C_{10}x\sqrt{\frac{k}{m}} = \sin \left( t\sqrt{\frac{k}{m}} + C_{12} \right)$

$\displaystyle x = C_{13} \sin \left( t\sqrt{\frac{k}{m}} + C_{12} \right)$

Let’s rewrite this equation using the conventional symbols for the constants.

$\displaystyle x = A \sin \left( t\sqrt{\frac{k}{m}} - \varphi \right)$

There it is. The basic sine wave equation is readily apparent, with the constants representing the amplitude and the phase shift, and the radical representing the angular frequency. The fourth possible constant (a “+ D”) is absent, and I wondered if I had accidentally dropped a constant somewhere — since that entire family of curves should be solutions. That is, x should be able to oscillate around 1, -5, whatever. But the reason for its absence is our starting point of Hooke’s law, F=-kx, which assumes that the equilibrium point is x=0. We would have had to start with F=-k(xxeq) to have had that last constant.

It’s pretty neat to see the equation emerge from just four basic equations: Hooke’s law F = –kx, Newton’s second law F=ma, and the definitions of velocity and acceleration v=x′ and a=v′!