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 90.5 MHz. She then walks so that she always remains 50.0 m from transmitter B. (See (Figure 1).) Limit your solution to the cases where 50.0 m lessthanorequalto x lessthanorequalto 65.0 m. 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.

Answer 1

a)

wavelength of the wave is given by

L = c / f = 3* 10^8 / (90.5* 10^6) = 3.315 m

now path difference is given by

x - 50 = n * 3.315

for n = 0

x = 53.315 m

for n = 2

x = 56.63 m

for n = 3

x = 59.945

for n = 4

x = 63.26

for n = 5

x = 66.57

since we need to consider max x < 65 so

x = 63.26 m

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b)

for this we need to consider destructive interference

x - 50 = ( n + 0.5)* 3.315

for putting n = 0, 1, 2, 3

x1 = 51.647 m

x2 = 54.97 m

x3 = 58.3 m

x4 = 61.6 m

x5 = 64.92 m

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