Finding frequency of a specific mechanical oscillator -- horizontal rod on pivot
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srecko97
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But when talking about small φ that has no sense
srecko97
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sin φ ≈ 0 for small φ
srecko97
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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
srecko97
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sinφ=φ-(φ^3)/6
srecko97
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oh it is φ
srecko97
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Well, I got that M_right - M_left = mg(rφ-bφ) ...M=torque
srecko97
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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...
srecko97
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ω2 is then negative, so there is no oscillation in real?
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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'ssrecko97 said:ω2 is then negative, so there is no oscillation in real?
##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.
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