Wednesday, March 2, 2016

Solutions to Solar Oscillation Problems

1)  Solution: The acoustic cutoff frequency is:

w ac   = g g/ 2c

Then:

w ac   =  (5/3) (273 ms-2)/ 2 (900 ms-1

w ac   =    0.25 s-1


2) Solution:
The spherical surface would appear as shown below:
The critical parameter is the difference  (ℓ  - m) given it yields the lines corresponding to parallels of latitude.  In this case, (ℓ  - m) =   (2 – 0) = 2, hence two parallels of latitude which are shown bounding the white region. (Since m = 0 there are no longitudinal lines)
The associated Legendre polynomial function given by:

P ℓm (q )  =  (1 – z2) m/2 / ℓ! 2   d (ℓ+m) / dz(ℓ+m)  (z2   -1)

But   =2  and  m = 0 ,  so:

P ℓm (q )  =  (1 – z2) 0 / 2! 22   d(2)  / dz(2)   (z2   -1) 2
(1 – z2) 0   = 1  so:

P ℓm (q )  =  (1)/ 2! 22   d(2)  / dz(2)   (z2   -1) 2

=   1/ 8   [d2 / dz2   (z2   -1) 2]

3) Solution:
The depth of reflection is r, and the Lamb frequency is:

L   =  2p/ T  = 2p/ 340.61 s =   0.0184 s-1

Where: L   =    Ö( ℓ(ℓ + 1) c2 / r2 )
Solving for r:     r=  Ö[(ℓ(ℓ + 1)]  c / L
r=   Ö[ (20(20 +1)] 900 m/s/ 0.0184 s-1  
r =   10 6   m

4) The sketch denotes a vibration for which m= 1 and ℓ = 2.  Then (ℓ  - m) =   (2 – 1) = 1 so a single parallel of latitude must appear in the sketch. Also since m= 1 we have at least one meriodonal or longitude line. 

Then : 

 P ℓm (q )  =  (1 – z2) m/2 / ℓ! 2   d (ℓ+m) / dz(ℓ+m)  (z2   -1)

Since   =2  and  m = 1 :

P ℓm (q )  =  (1 – z2) 1/2   / 2! 22   d(3)  / dz(3)   (z2   -1) 2

z =  cos q  =    cos  p/4  =    Ö2 / 2


But: d(3)  / dz(3)   (z2   -1) 2   =  16.97

Then:

 
ℓm (q )  =  (Ö2 / 2) / 2! 2 2   [16.97]  =  Ö2/ 16  [16.97]                                                                         

   

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