A- Two antennas located at points A and B are broadcasting radio waves of freque

Question

A- Two antennas located at points A and B are broadcasting radio waves of frequency 97.0 MHz, perfectly in phase with each other. The two antennas are separated by a distance d = 9.30 m. An observer, P, is located on the x axis, a distance x = 76.0 m from antenna A, so that APB forms a right triangle with PB as hypotenuse. What is the phase difference, in radians, between the waves arriving at P from antennas A and B?

B- Now observer P walks along the x axis toward antenna A. What is P's distance from A when she first observes fully destructive interference between the two waves?

Explanation / Answer

A)
wavelength of radio waves, lamda = c/f

= 3*10^8/(97*10^6)

= 3.093 m

let r1 = 76.0 m

r2 = sqrt(r1^2 + d^2)

= sqrt(76^2 + 9.3^2)

= 76.57 m

we know,

pahse diifrence = (2*pi/lamda)*path differnce

phi = (2*pi/3.093)*(r2-r1)

= (2*pi/3.093)*(76.57 - 76))

= 1.158 radians

B) let at a distance x from A the observer observes first fully distructive interefrence.

use,

r2 - r1 = lamda/2

sqrt(x^2 + d^2) - x = lamda/2

sqrt(x^2 + 9.3^2) - x = 3.093/2

==> x = 27.2 m

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