Consider the function $f(x) = \cos(b*\sqrt{x})$, where $b= 6.28 cm^{-0.5}$. a) Argue that this wave has crests at x=0.1 cm, 4 cm, 9cm, and so on.

b) sketch a graph of this function , and argue that it looks like a wave whose wavelength is increasing as x increases .

c) estimate the local wavelength of this function at x=4.0cm by averaging the distance to adjacent crests on either side.

d) compute the local wavelength of this function at x=4.0cm and compare with your answer for part c .

a)

the function has crests for $b*\sqrt{x} = 2n*\pi$ where n is integer

$x = (2n)^2*\pi^2/b^2 = (2n)^2 * 0.25$

n = 0, x =0

n=1 , x =1

n=2 , x =4

n=3 , x=9

b) the graph is below

The cosine function is periodic and therefore f(x) is a qusi-periodic function. Because the x is increasing the period of f(x) is increasing with x, therefore f(x) looks like a wave with wavelength increasing.

c) to the right L = 9-4 = 5 cm

to the left L = 4-1 =3 cm

the average wavelength is L = (5+3)/2 =4 cm

d) $f(x) = cos(b*\sqrt{x}) = cos((2\pi \sqrt{(x/\lambda)}$

$b = 2*\pi/\sqrt{\lambda}$

$\lambda = (4\pi^2)/b^2 = 6.28 cm$

The local wavelength at x = 4 is higher then the that found in part c)

The figure is below