I'm trying to derive the formula for the period T as a function of the mass of the object. Here is my attempt. Note that I cheated and passed the section I had trouble with, without fully understanding it.(adsbygoogle = window.adsbygoogle || []).push({});

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Much of this is quite straightforward for me.

[tex]y = A sin \alpha[/tex] (basic trigonometry)

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Here is the part I'm not sure I understand fully:

[tex]\alpha = \omega t[/tex]

Why does that equation work?

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The angle alpha is replaced by [tex]\psi t[/tex] so the result is:

[tex]y = A sin \omega t[/tex]

The speed in the y direction is as follows:

[tex]v(t) = \frac {dy} {dt} = \omega A cos \omega t[/tex]

The acceleration in the y direction is:

[tex]a(t) = \frac {d^2x} {dt^2} = -\omega^2 A sin \omega t[/tex]

The above is simple calculus.

When a net force is acting on a body an acceleration will show.

The force that is acting on the body is (via hookes law):

[tex]F = -ky[/tex]

If we replace y with the expression we derived earlier we get

[tex]F = ma = -m \omega^2 A sin \alpha t[/tex]

[tex]F = -ky = -k A sin \alpha t[/tex]

ie.

[tex]m \omega^2 = -k \Leftrightarrow \omega = \sqrt{k / m}[/tex]

Combining the above expression with the commonly known

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

and you get the final result

[tex]T = 2 \pi \sqrt {m / k}[/tex]

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My question is:

Why is [tex]\alpha = \omega t[/tex]?

Thank you for your time. Have a nice day.

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