Answer
The beat frequency is $~~155~Hz$
Work Step by Step
First we can find the frequency of the sound in the reference frame of the intruder.
In this situation:
$f = 28.0~kHz$
$v_D = 0.950~m/s$
$v_S = 0$
We can find the frequency of the sound in the reference frame of the intruder:
$f' = f~\frac{v-v_D}{v}$
$f' = (28.0~kHz)~(\frac{343~m/s-0.950~m/s}{343~m/s})$
$f' = 27.92245~kHz$
Then we can find the frequency of the sound that is reflected back to the alarm.
In this situation:
$f = 27.92245~kHz$
$v_D = 0$
$v_S = 0.950~m/s$
We can find the frequency of the sound that is reflected back to the detector:
$f' = f~\frac{v}{v+v_S}$
$f' = (27.92245~kHz)~(\frac{343~m/s}{343~m/s+0.950~m/s})$
$f' = 27.845~kHz$
The frequency of the sound that is reflected back to the alarm is $~~27.845~kHz$
We can find the beat frequency:
$f = (28.0~kHz)-(27.845~kHz) = 155~Hz$
The beat frequency is $~~155~Hz$
