How many times per minute does an object pass equilibrium? (Diff Eqs)

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Homework Statement
You place a 4 pound cauliflower on a scale. This lowers scale by 9.6 inches. Gamma = 0.5. From equilibrium, you give the scale a downward velocity of 5 fps.
What kind of damping (or not) occurs?
How many times per min does cauliflower pass equilibrium?
Solve for displacement of cauliflower from equil as fn of time.
Relevant Equations
Rcos(wt - d) = 0 (general conversion of sines and cosines to a single cosine minus some phase angle)
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Essentially, my question boils down to just part b. My initial thoughts on how to solve this are illustrated above in the second image. To me, I had just interpreted omega, the frequency, as cycles per second. From there, I simply multiplied to get it into minutes. After discussing with a prof. at office hours, he said this is incorrect and is the wrong approach. He noted that what's in the blue (final image on bottom) is the correct approach to part b. This involves considering the period. The explanation was a little rushed (as he had a meeting to attend), and it involved having to do with: the fact that omega isn't just cycles per second, and that setting this equation equal to 0 is done because this is the state of equilibrium. The setting equal to zero makes sense, but I don't see how I'm supposed to know how to interpret omega and I'm actually a little bit shakey on what's in blue so if someone could please clarify this, it would be greatly appreciated.
Thanks for your time.
 
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##\omega## is the angular frequency, in rad/s. You were using it as the linear frequency ##\nu = \omega / (2\pi)##, which is expressed in cycles/s (Hz).
 
DrClaude said:
##\omega## is the angular frequency, in rad/s. You were using it as the linear frequency ##\nu = \omega / (2\pi)##, which is expressed in cycles/s (Hz).
Thanks for the reply. How would I know to interpret this frequency as angular? In the problem, it seems as though the cauliflower is moving linearly.
 
rugerts said:
How would I know to interpret this frequency as angular? In the problem, it seems as though the cauliflower is moving linearly.
It is moving linearly, but since it is harmonic motion, the displacement changes sinusoidally so there is an "angular" aspect to the motion.

Consider the situation in terms of units. When you have ##\cos(\omega t)##, what is in parenthesis is an angle, expressed in radians. For ##t## in seconds, ##\omega## has to be in radians per second.

Note that when writing ##\cos(2 \pi \nu t)##, the ##2 \pi## is assumed to have units of radians, so ##\nu## is in Hz.
 
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Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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