- #1
Mattara
- 347
- 1
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.
http://www.filehive.com/files/0722/image.jpg
Much of this is quite straightforward for me.
y = A sin \alpha (basic trigonometry)
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Here is the part I'm not sure I understand fully:
\alpha = \omega t
Why does that equation work?
-----
The angle alpha is replaced by \psi t so the result is:
y = A sin \omega t
The speed in the y direction is as follows:
v(t) = \frac {dy} {dt} = \omega A cos \omega t
The acceleration in the y direction is:
a(t) = \frac {d^2x} {dt^2} = -\omega^2 A sin \omega t
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):
F = -ky
If we replace y with the expression we derived earlier we get
F = ma = -m \omega^2 A sin \alpha t
F = -ky = -k A sin \alpha t
ie.
m \omega^2 = -k \Leftrightarrow \omega = \sqrt{k / m}
Combining the above expression with the commonly known
\omega = 2 \pi / T
and you get the final result
T = 2 \pi \sqrt {m / k}
------
My question is:
Why is \alpha = \omega t?
Thank you for your time. Have a nice day.
http://www.filehive.com/files/0722/image.jpg
Much of this is quite straightforward for me.
y = A sin \alpha (basic trigonometry)
----
Here is the part I'm not sure I understand fully:
\alpha = \omega t
Why does that equation work?
-----
The angle alpha is replaced by \psi t so the result is:
y = A sin \omega t
The speed in the y direction is as follows:
v(t) = \frac {dy} {dt} = \omega A cos \omega t
The acceleration in the y direction is:
a(t) = \frac {d^2x} {dt^2} = -\omega^2 A sin \omega t
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):
F = -ky
If we replace y with the expression we derived earlier we get
F = ma = -m \omega^2 A sin \alpha t
F = -ky = -k A sin \alpha t
ie.
m \omega^2 = -k \Leftrightarrow \omega = \sqrt{k / m}
Combining the above expression with the commonly known
\omega = 2 \pi / T
and you get the final result
T = 2 \pi \sqrt {m / k}
------
My question is:
Why is \alpha = \omega t?
Thank you for your time. Have a nice day.
Last edited:
