Why is Simple Harmoic Motion in a Spring modelable with rotation without invoking calculus?

Question

Is there a good reason why we can use uniform circular motion to get the equations for a mass on a spring, without invoking calculus?

This relationship is often used to find the equations for the spring in non-calculus physics, because obviously there is no way for those students to solve the differential equations, but no explanation is given for WHY this can be done, it is just stated it can be done and derived from there. The back argument can be made (since we can solve the Diff EQs, we see it is the same and so make the analogy), but this is an unsatisfying answer for students in the non-calc classes who are asking why we are doing this.

Without invoking "I can do the math you can't and it works", is there some reasoning we can give to why we go to a circle, look at the 1D projection, and get the equations for the motion of a mass on a spring exactly? Note that this also becomes pertinent when we get to the small angle pendulum, as the inverse question gets asked precisely because they don't know why we can go to a circle for a spring (Why CAN'T we go to a uniform circle this time when it worked before?).

Answer 1

Consider a mass $m$ in uniform circular motion with radius $r$ and angular frequency $\omega$. It experiences a force of magnitude $F = mr \omega^2$ towards the center of the circle at all times. Its position as a function of time will be something like $(x, y) = (r \cos \omega t, r \sin \omega t)$ (use trigonometry and $\theta = \omega t$ to see this.)

At any time, we can decompose the force vector into its $x$- and $y$-components. By similar triangles, when the mass is at the position $(x,y)$, we have $$ \frac{F_x}{F} = -\frac{x}{r} \quad \Rightarrow \quad F_x = - F x/r = - m \omega^2 x $$ (and similarly for the $y$-direction.) This means that the $x$-position of the mass obeys $m a_x = - m \omega^2 x$ at all times.

By comparison, the mass on a spring obeys the equation $m a_x = - k x$ at all times. By comparing these equations, we conclude that $x$-position of a mass on a spring will be exactly the same as the $x$-position of a mass in uniform circular motion, so long as we have $k = m \omega^2$.

Answer 2

why we use circular motion to explain spring force?
It is because the shadow of a body in uniform circular motion experiences a force exactly similar to that experienced by mass on a spring.

In fig, the mass $m$ is rotating at a constant angular velocity $\omega$ in a circle of radius $A$. Even though the particle is experiencing a centre seeking force, the shadow is not influenced by the perpendicular component and it (the shadow) moves as if it is under the influence of a horizontal force :

$F= - m \omega ^2 x$

where $x$ is the Displacement of the projection from mean position. Since a mass attached to a spring also experiences a similar force $F= - kx$, we should expect their motions to be similar if released with same initial conditions.

May be this does not convince you because we went from circular motion to the spring. as if we knew already that circular motion has a spring motion hidden inside. Yes it is more like - "we know to solve it... after solving we saw the analogy and now we use that to explain". But that's all someone can do. You don't know the actual way of solving. So all someone can do would be to compare it with something you know. How can I convince you that a pigeon is a bird unless I compare it with another bird. Or else I should explain the morphology of a bird which you don't know.

I can tell you that we cannot easily go from a spring to a circular motion without touching calculus. Calculus is the way of solving these problems. If we could intuitively explain it... then why calculus?

why pendulum has a small angle approximation?

In a pendulum, its not the shadow we are talking about. We wish to calculate the motion of that mass (constrained to move in a circle) itself. Also, here the mass is not in uniform circular motion. It accelerates as it falls down and decelerates on its way up. Due to these two reasons, we are approximating the motion to be in a small region. A region where the shadow of the ball is the same as the ball itself - small angles.