A person with a radio-wave receiver starts out equidistant from two FM radio tra

Question

A person with a radio-wave receiver starts out equidistant from two FM radio transmitters A and B that are 11.0 m apart, each one emitting in-phase radio waves at 94.5 MHz . She then walks so that she always remains 50.0 m from transmitter B. Limit your solution to the cases where 50.0m?x?65.0m.

For what values of x will she find the radio signal to be maximally enhanced?

Express your answers in increasing order to three significant figures, separated by commas.

For what values of x will she find the radio signal to be cancelled?

Express your answers in increasing order to three significant figures, separated by commas.

Explanation / Answer

for maximally enhanched,

path diff , delta(x) = n lambda

where n = 0 ,1 ,2 ,3 ......

lambda = c / f = (3 x 10^8 m) / (94.5 x 10^6 m/s)

= 3.175 m

x - 50 = 0 or 3.175 Or 6.35 Or 9.525 m

x = 59 m , 53.175 m, 56.35 m, 59.525 m .....Ans

for minimally,

delta(X) = (2n + 1) lambda /2  

n = 0 , 1 , 2 , 3....

x - 50 = 1.5875, 4.7625, 7.9375

x = 51.5875 m , 54.7625 m , 57.9375 m

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.