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[tex]f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}[/tex]

I am wondering if this equation only applies for massless harmonic oscillators (or oscillators oscillating sideways)?

The proof for the equation above is

[tex]\sum {F=ma}[/tex]

[tex]-kx=ma[/tex]

[tex]-kx=-m\omega^2x[/tex]

[tex]k=m\omega^2[/tex]

Since

[tex]\omega=2\pi f[/tex]

We get the familiar equation above

[tex]f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}[/tex]

But as you can see we have neglected the force of gravity (if we assume that such a force exists). I solved an assignment today using this formula above. It gave the wrong answer when I just used the formula which neglected gravity. But when I went back and proved the formula "again", this time assuming the existense of gravity I ended up in the right answer. This would be a bummer, since I have always been taught that the frequency for a harmonic oscillator is:

[tex]f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}[/tex]

No matter what.