Morin4.23 - Morin 4.23 - Corrections to the Pendulum The...

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Morin 4.23 --- Corrections to the Pendulum The simple pendulum satisfies the nonlinear differential equation In[3]:= '' t   0^2 Sin t  Out[3]=  t  0 2 Sin t  We will agree to launch this pendulum at rest from its maximum angle a (also known as q 0), and we seek the period as a function of the input parameters w 0 and a . On dimensional grounds we know that the answer will be of the form T[ a ] = T0 f[ a ] where T0 is defined as 2 p / w 0, i.e. the period in the limit of small angles, where one can approximate Sin[ q ]= q and get the good old simple harmonic oscillator. With this definition, we know f[ a =0]=1, and we seek the rest of the terms in the series expansion for f[ a ]. By the way, since the period for - a should be exactly the same as for + a , f[ a ] is an even function, and the expansion will contain only even powers of a . Morin's problem 4.23 asks you to find the coefficient "c" in the expansion f[ a ] = 1 + c a ^2 + . .. Numerical Interlude As a first pass, we'll just find the coefficient numerically. We'll choose w 0=1, for which T0=2 p In[4]:= T0 2 Out[4]= 2 In[5]:= de  '' t   Sin t  Out[5]=  t  Sin t  In[6]:= bc1  ' 0  0 Out[6]= 0 0 Choosing a smallish angle, say a = p /10 In[7]:=  10 Out[7]= 10
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We form the numerical solution (note the use of deferred evaluation---there will be no immediate response from Mathemat- ica here) In[8]:= soln : NDSolve  de, bc1, 0  , t , t, 0, 10  And then use it by defining our my q [t] to be the expression q [t] acted up by the rule which is the first solution on the list of solutions (which happens to contain exactly one solution here) In[9]:= my t_  t  . soln  1  Out[9]= InterpolatingFunction  0., 10.  ,   t We can plot the result In[10]:= Plot my t , t, 0, 10  Out[10]= 2 4 6 8 10 0.3 0.2 0.1 0.1 0.2 0.3 and look for the period, by e.g. by finding the first zero crossing, near t=T0/4 In[11]:= quarter FindRoot my t , t, T0 4  Out[11]= t 1.58054 So we have just learned that for this a , f[ a ] is: 2 Morin4.23.nb
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In[12]:= f10 t . quarter  T0 4 Out[12]= 1.0062 This is enough to make a first guess at the constant c: In[13]:= Solve f10 1 c ^2 Out[13]=  c 0.0628554  Which is numerically close to 1/16: In[14]:= 1 c .  1  Out[14]= 15.9095 A more systematic way is to flesh out the curve with more data points. Here is a fancy way of repeating the above procedure in one line:
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This note was uploaded on 06/13/2011 for the course PHY 263 taught by Professor Kilcup during the Spring '10 term at Ohio State.

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Morin4.23 - Morin 4.23 - Corrections to the Pendulum The...

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