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(B) Given: Frequency of source $f_A = 1800\,Hz$,Frequency received by observer $f_B = 2150\,Hz$,Speed of sound $v_s = 343\,m/s$.First,we find the term

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$2500$

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(B) Given: Frequency of source $f_A = 1800\,Hz$,Frequency received by observer $f_B = 2150\,Hz$,Speed of sound $v_s = 343\,m/s$.First,we find the terminal speed $v$ of the source using the Doppler effect formula for a moving source and stationary observer: $f_B = f_A \left( \frac{v_s}{v_s - v} ight)$.Rearranging for $v$: $\frac{v_s - v}{v_s} = \frac{f_A}{f_B} \implies 1 - \frac{v}{v_s} = \frac{1800}{2150} \implies v = v_s \left( 1 - \frac{1800}{2150} ight)$.$v = 343 	imes \left( 1 - 0.8372 ight) = 343 	imes 0.1628 \approx 55.84\,m/s$.Now,the ground acts as a stationary source reflecting the sound back to the moving source $A$. The frequency $f'$ received by the source $A$ moving towards the ground is given by: $f' = f_A \left( \frac{v_s + v}{v_s - v} ight)$.Substituting the values: $f' = 1800 	imes \left( \frac{343 + 55.84}{343 - 55.84} ight) = 1800 	imes \left( \frac{398.84}{287.16} ight) \approx 1800 	imes 1.3889 \approx 2500\,Hz$.

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