A stationary observer receives sound from two | vedclass

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Question

(D) Let $c$ be the speed of sound and $v$ be the speed of the tuning forks.The frequency heard from the approaching tuning fork is $
u_{1} = \left(\f

Vedclass · Accepted Answer

$\frac{1}{4} \; 	ext{m/s}$

Vedclass · Answer

(D) Let $c$ be the speed of sound and $v$ be the speed of the tuning forks.The frequency heard from the approaching tuning fork is $
u_{1} = \left(\frac{c}{c-v}ight) 
u_{0}$.The frequency heard from the receding tuning fork is $
u_{2} = \left(\frac{c}{c+v}ight) 
u_{0}$.The beat frequency is given by $\Delta 
u = 
u_{1} - 
u_{2} = 2 \; 	ext{Hz}$.Substituting the expressions:$\Delta 
u = c 
u_{0} \left(\frac{1}{c-v} - \frac{1}{c+v}ight) = c 
u_{0} \left(\frac{c+v - (c-v)}{c^{2}-v^{2}}ight) = \frac{2 c 
u_{0} v}{c^{2}-v^{2}}$.Since $v \ll c$, we can approximate $c^{2} - v^{2} \approx c^{2}$.Thus, $\Delta 
u \approx \frac{2 c 
u_{0} v}{c^{2}} = \frac{2 
u_{0} v}{c} = 2$.Substituting the given values: $\frac{2 	imes 1400 	imes v}{350} = 2$.$8v = 2 \Rightarrow v = \frac{2}{8} = \frac{1}{4} \; 	ext{m/s}$.

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