- #1

Abdullah Almosalami

- 49

- 15

There are many ways to derive that for a mass on a horizontal frictionless surface on a spring with spring constant k that the system's natural frequency is ##f = \frac {1} {2\pi} \sqrt {\frac k m}##. It makes intuitive sense. The larger the spring constant, the higher the restoring force and the faster that the mass will oscillate, and likewise the higher the mass, the more inertia it has, and the slower it responds resulting in lower frequency.

In my physics textbook, I was just doing some questions for practice and came across a question that asked "Use energy considerations to derive the oscillation frequency for a mass-on-spring oscillator ..." And ok. If I use the fact that the maximum potential energy due to the spring will be equal to the maximum kinetic energy of the mass (

*this will be the case for this system)*, I end up going through the following: $$ PE_{max} = KE_{max} \Leftrightarrow \frac 1 2 kx_{max}^2 = \frac {1} {2} mv_{max}^2 \Leftrightarrow \frac {v_{max}} {x_{max}} = \sqrt {\frac k m} $$.

Now I know that ##\omega = \sqrt {\frac k m}## and dimensionally ##\frac {v_{max}} {x_{max}}## is indeed the unit of Hertz,

**but I'm not sure that I can immediately make the jump that this is indeed the oscillation frequency if I didn't already know what ##\omega## was from other derivations; so that's my question. What would make one conclude that this is indeed the oscillation frequency without using other derivations?**I can also come to the same result by observing that if $$ x(t) = x_{max} sin(\omega t) \Leftrightarrow v(t) = \frac {dx} {dt} = x_{max}\omega cos(\omega t) = v_{max} cos(\omega t) $$ then the ratio of the amplitude of the velocity to the amplitude of the displacement is indeed ##\omega##. But this is another derivation. I guess intuitively this result makes sense and provides an interesting perspective I hadn't thought of before: the sqrt of the maximum amplitude and the sqrt of the maximum speed are proportional to each other at a given frequency, and this makes for the known non-intuitive property of oscillators that ##\omega## is independent of amplitude. But yeah. My question is highlighted in bold.