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 92.5 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 50.0m x 65.0m.

Part A

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

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

Part B

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

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

Explanation / Answer

as c=3*10^8m/s
=c/f = (3*10^8)/(92.5*10^6)= 3.24 m
Constructive interference : abs(L1-L2)=n -> L1=L2+n
n=0: x= 50+0= 50[m]
n=1: x= 50+1*3.24= 53.24[m]
n=2: x= 50+2*3.24= 56.48[m]
n=3: x= 50+3*3.24= 59.72[m]
n=4: x= 50+4*3.24= 62.96[m]

Destructive interference : abs(L1-L2)=(n+1/2) -> L1=L2+(n+1/2)
n=0: x= 50+(0+1/2)*3.24= 51.62[m]
n=1: x= 50+(1+1/2)*3.24= 54.86[m]
n=2: x= 50+(2+1/2)*3.24= 58.10[m]
n=3: x= 50+(3+1/2)*3.24= 61.34[m]
n=4: x= 50+(4+1/2)*3.26= 64.52[m]

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