Lagrangian for a rod pivoting at the end of a clock hand

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  • #1
ChiralSuperfields
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Homework Statement
Please see belwo
Relevant Equations
##ml^2\ddot \phi + mRl\varphi \dot \phi\sin(\phi - \varphi t) = 0##
For (d),
1713422922316.png


I am confused how ##a(t)## oscillates harmonically. The EOM is ##ml^2\ddot \phi + mRl\varphi \dot \phi\sin(\phi - \varphi t) = 0##

Then using ##\phi - \varphi t = a(t)##
##l\ddot \phi + R\phi \varphi a(t) = 0##
Where I used the small angle approximation after substituting in ##a(t)##
Does anybody please know where to go from here to find ##\omega##?

Thanks!
 
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  • #2
Simply linearise the equation of motion around ##\phi^*##. You have attempted this but need to double check your math.

Also note: It is ##\gamma##, not ##\varphi##.
 
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  • #3
Orodruin said:
Simply linearise the equation of motion around ##\phi^*##. You have attempted this but need to double check your math.

Also note: It is ##\gamma##, not ##\varphi##.
Thank you for your reply @Orodruin!

##l^2\ddot \phi + Rl\phi \gamma a(t) = 0## is that please correct now?

Sorry I don't understand how you linearize the equation of motion around ##\phi^*##. I have not been taught that. Could you please explain a bit more?

Thanks!
 
  • #4
ChiralSuperfields said:
Thank you for your reply @Orodruin!

##l^2\ddot \phi + Rl\phi \gamma a(t) = 0## is that please correct now?

No. Carefully consider the second term. Where did the ##\phi## come from?

ChiralSuperfields said:
Sorry I don't understand how you linearize the equation of motion around ##\phi^*##. I have not been taught that. Could you please explain a bit more?
It is essentially what you are (attempting) to do. You write ##\phi = \phi^* + a(t)## and then assume that ##a(t)## is small. Perform a Taylor expansion in ##a(t)## and keep only linear terms.
 
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  • #5
Also, for the first term, you have not inserted the assumption ##\phi = \phi^* + a## ...
 

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