Is it possible to make velocity time graph periodic?

Question

sorry i know this question is from physics but i believe it uses more of maths.

i am a highschool student but here we are not taught Fourier analysis so we can't learn those beautiful curves and therefore not being able to generate them. recently we were taught kinematics and in this topic the equation are not periodic in case of motion such as freefall.

suppose a ball falls freely and bounces of elastically to the same height and falling again and rising again..........

Now the equation of motion are unable to predict this periodic behaviour so i was curious we were also taught trigonometric graphs where we learnt how they repeat so i wondered if they are sufficient to accomplish this limitation so i googled this and came across triangular wave.

now ignoring the change in velocity during the impact from ground and assuming that its velocity becomes zero on impact can we generate a triangular wave for representing velocity in this situation like the equation so formed identifies the sign of velocity at given instant of time?

if not then can we make a triangular wave equation for this situation: suppose that the period in which balls falls from a certain height and bounces elastically to the same point, this period we call 1 cycle now the sign of velocity remain same in a cycle and changes its sign in next upcoming next cycle?

sorry again for posting here in maths community but i believe that this task relates more to maths

Answer 1

For a freely falling mass Newton's law has it that a fall or bouncing back height $h=\frac12 g\;t^2$ is proportional to the square of time but not linearly proportional to time. A triangular wave in a pure gravitational situation is violative of laws of physics unless another force is made to act to modify the acceleration. Time traces would be a series of parabolic arches. ( Multiflash photography Edgerton, MIT)

Assuming that the coefficient of restitution is $1.0$ a ball dropped from a height $h$ bounces back to this height with constant deceleration but not constant velocity.

An example of a force of resistance towards constant speed of fall is viscous resistance proportional to square of velocity. A man with parachute dropped from a hovering helicopter descends (after first few seconds of free fall ) with a constant steady -state / asymptotic or terminal velocity, so the height of fall $h-t$ parabolic relation quickly turns from free fall to linear i.e., from a stone body parabolic fall to resisting parachute ( falling steel ball in a column of oil is another example) as shown in the graph. But it is a single descent, not periodic.

In a spring mass overdamped dashpot system an approximate linear motion however it is aperiodic. A forced non-harmonic oscillation with energy input driver can be designed. We experience such bumpy rides on cobble roads.

Electrical signals can be generated by compounding individual harmonics evaluated from Fourier analysis of any triangular or saw-tooth wave.

Answer 2

In general, equations of motion in classical mechanics are based on Newton's laws, which apply to bouncing balls as well as to balls in free fall.

The equations for a ball in free fall, ignoring air resistance, happen to be especially simple. Only the force of gravity acts on the ball, so we see uniform acceleration, which can be described by a straight-line velocity graph. Note that the equations that produce this graph are simpler than reality, in which air resistance causes an additional force on the ball.

But when the ball hits the floor, additional forces occur because of the ball's contact with the floor. If the ball and the floor are both made of steel, the time period during which these forces act might be very short, so you might idealize it by assuming the ball changes velocity instantaneously. Just remember that this is an idealization you make and not what really happens.

As soon as the ball stops being in contact with the floor after the bounce, it is in free fall again (under the previous simplified assumptions).

Suppose you drop a ball from a height $0.44$ meters above a perfectly elastic slab, and suppose the acceleration of gravity is $9.8$ meters per second per second. A little calculation using the usual formulas for bodies in free fall tells us that the ball hits the slab after $0.2997$ seconds. Let's round this to $0.3$ seconds. So we have the formula for velocity as a function of time $t$ during this part of the ball's travels:

$$ v(t) = -9.8 t, \quad 0\leq t < 0.3.$$

If you suppose that the bounce is instantaneous, then the ball instantly changes velocity from $-2.94$ meters per second to $2.94$ meters per second at time $t = 0.3$ seconds after release. Does it make any sense to say what the ball's velocity is at that time? Not really, because the assumption that it changes velocity instantaneously is inconsistent with the idea of classical physics that every body has a velocity at every time. But if you want, you can define

$$ v(0.3) = 0. $$

Regardless of what you say about $v(0.3),$ you can say that (under the assumption that the ball bounces perfectly elastically) the velocity of the ball immediately after hitting the slab and for the next $0.6$ second (the time it takes to go up to a height $44$ cm above the floor and come down to the slab again) is

$$ v(t) = 2.94 - 9.8(t - 0.3), \quad 0.3 < x < 0.9. $$

The rest of the function can be defined by stipulating that

$$ v(t) = v(t - 0.6), \quad x \geq 0.9. $$

So that gives us a definition of a function "by cases" as follows:

$$ v(t) = \begin{cases} -9.8 t & 0\leq t < 0.3, \\ 0 & t = 0.3, \\ 2.94 - 9.8(t - 0.3) & 0.3 < x < 0.9, \\ v(t - 0.6) & x \geq 0.9. \end{cases} $$

The last part, where we define later values of $v(t)$ in terms of earlier values of $v$, is not only mathematically legitimate; the very definition of what it means for a function to be periodic is that we can write such an equation about the function.

The function defined in this way mostly looks like this:

This graph is just missing the disconnected dots that would show $v(0.3) = 0,$ $v(0.9) = 0,$ $v(1.5) = 0,$ and so forth.

If you are willing to let $v(0.3)$ be $2.94$ instead of $0,$ you no longer have the disconnected dots, and you can write the function as

$$ v(t) = 2.94 + 5.88\left(\left\lfloor\frac{x-0.3}{0.6}\right\rfloor-\frac{x-0.3}{0.6} \right) $$

If you really want $v(0.3)$ to be $0$, you could write

$$ v(t) = 2.94\left(\left\lceil\frac{x-0.3}{0.6}\right\rceil + \left\lfloor\frac{x-0.3}{0.6}\right\rfloor - \frac{x-0.3}{0.3} \right). $$

But I do not think either of these is as good as the first definition, because all of this jiggery-pokery with the floor and ceiling functions just gets in the way of explaining what the function actually does, which is to go from $2.94$ downward along a straight line at a slope of $-9.8$ as $t$ goes from $0.3$ (first time hitting the slab) to $0.9$ (second time hitting the slab), and then repeat the same thing every $0.6$ second.