Question

a and V = Suppose the potential is V = 0 when x -a and when x Would you use sine or cosines for the Fourier series describing the potential? a. Which definition of k would you use for a Fourier expansion? (Choose one) b. c. 10 pts Extra Credit: Find the first three Fourier coefficients, Cn, that produce the potential V

Answer 1

Solution : (a) potential for given function graphically looks like in picture  

for region -a to a

f(x)= f(-x)  

in this region function is symmetric and even which mean fourier serious

contain only cos term because  for even functions only cos term is in fourier series

that mean only cosine term define the series this is the answer for first part

now second part

(b) fourier series formula for the cosine series is given by

n=infinite do f(z) =-+ an cosknr n=1 7l

wehere

f(x)da I - do a

$a_{0}= \frac{2}{a}(Vx)_{0}^{a} =2V \Rightarrow \frac{a_0}{2}=V$

Now for the second part we need to calculate nth coefficient for

2V akn Cl an- Ca L7t

$a_{n}$ will survive only when sin term will be non zero i.e.

sin(Ana)メ0
  

this happen only when odd multiple of  $\frac{\pi}{2a}$

so kn would be like

where n is integer

2 7t

part C now calculation for part c first three coefficient is given by

using formula  $a_n=\frac{2V}{ak_n}(\sin(k_na))$

$a_1=\frac{2V}{ak_1}(\sin(k_1a)\rightarrow \frac{4V}{\pi}$

$a_3=\frac{2V}{ak_3}(\sin(k_3a)\rightarrow- \frac{4V}{3\pi}$$a_5=\frac{2V}{ak_5}(\sin(k_5a)\rightarrow- \frac{4V}{5\pi}$

so these are the three term for the above problem so now series would be look like this$f(x)= V+ \frac{4v}{\pi}\cos(\frac{\pi}{2a}x)-\frac{4V}{3\pi}\cos(\frac{3\pi}{2a}x) +\frac{4v}{5\pi}\cos(\frac{5\pi}{2a}x)$ i also post images check out this also

プhe potential ? au CL O. akn eo ieddc 5万