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Question

(B) Let $c$ be the speed of sound. $1$. Frequency heard by $B$ from $A$: Since $A$ is moving away from stationary $B$,the frequency $f_B$ heard by $B$

Vedclass · Accepted Answer

The note heard by $A$ coming from $B$ will be higher in pitch than $f_A$.

Vedclass · Answer

(B) Let $c$ be the speed of sound. $1$. Frequency heard by $B$ from $A$: Since $A$ is moving away from stationary $B$,the frequency $f_B$ heard by $B$ is $f_B = f_A \left( \frac{c}{c+u} \right)$. Since $c+u > c$,$f_B $2$. Frequency heard by $A$ from $B$: $B$ plays the note at frequency $f_B$. Now $A$ is moving away from the source $B$ at speed $u$. The frequency $f_A'$ heard by $A$ is $f_A' = f_B \left( \frac{c-u}{c} \right) = f_A \left( \frac{c}{c+u} \right) \left( \frac{c-u}{c} \right) = f_A \left( \frac{c-u}{c+u} \right)$. Since $c-u $3$. Frequency heard by $C$ from $B$: $C$ is moving toward the source $B$ at speed $u$. The frequency $f_C$ heard by $C$ is $f_C = f_B \left( \frac{c+u}{c} \right) = f_A \left( \frac{c}{c+u} \right) \left( \frac{c+u}{c} \right) = f_A$. Thus,option $A$ is correct. Therefore,the incorrect statement is $B$.

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