# Finding frequency of a specific mechanical oscillator -- horizontal rod on pivot

srecko97
I did the same approximation in the first part of my homework (calculating the oscillating time), when I say cos φ≈1 for small φ

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Keep terms of first order in φ. cosφ ≈ 1 - φ2/2 ≈ 1 to first order. What about sinφ?

srecko97
sinφ=φ-(φ^3)/6

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sinφ=φ-(φ^3)/6
So, to first order in φ?

srecko97
oh it is φ

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oh it is φ
That's right.

srecko97
Well, I got that M_right - M_left = mg(rφ-bφ) ...M=torque

srecko97

The frequency gets smaller, as numerator gets smaller and denominator gets bigger. I hope it is correct! I would like to say you a big THANK YOU TSny! You helped me a lot... I learned a lot from solving this task with your help. I love this forum...

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srecko97
ω2 is then negative, so there is no oscillation in real?

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ω2 is then negative, so there is no oscillation in real?
Yes, there is no oscillation. Looking back at your post#43, I think you need to add an overall negative sign to the first equation so that it's

##J \ddot{\phi} = - mg(r-b)\phi##.

If ##b > r## then you can see that ##\ddot{\phi} >0## if ##\phi## is given an initial small, positive value. So, the rod rotates away from the equilibrium position when it is released and there is no oscillation. Of course, the differential equation for ##\phi## breaks down as ##\phi## continues to increase since the differential equation was derived under the assumption that ##\phi## is always small.

Another way to investigate the general behavior of the system for various values of b is to derive the potential energy of the system as a function of ##\phi## without making a small angle approximation. Then plots of the potential energy function for various values of b will reveal a lot about the behavior. But that's probably for a rainy day.

srecko97