Linear Simple Harmonic Oscillator: period a direct linear proportion to mass?

[charlatain]

If a mass that hangs suspended vertically from a spring is increased, then won't the period increase as a direct linear proportion? (Because the larger mass has a greater inertia and will require a larger force and longer time to change the direction of motion on each oscillation?)

Some guidance would be greatly appreciated!

 

[adriank]

A simple harmonic oscillator with spring constant k and mass m has angular frequency $\omega = \sqrt{k/m}$, so the period is $T = 2\pi/\omega = 2\pi\sqrt{m/k}$.

Thus, period is proportional to the square root of the mass.

 

[sir-pinski]

Yes, you are correct. The period of a linear simple harmonic oscillator is directly proportional to the mass. This means that as the mass increases, the period will also increase in a linear manner. This can be explained by the fact that a larger mass has a greater inertia, which means it requires a larger force and longer time to change its direction of motion. This results in a longer period for each oscillation. This relationship is described by the formula T=2π√(m/k) where T is the period, m is the mass, and k is the spring constant. As the mass increases, the period will also increase in a direct linear proportion, as you suggested.