A bus driving along at 39.6 ,km/h is approach | vedclass

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(B) The source (bus) is moving towards a stationary observer (person at the bus stop).Given:Velocity of source,$v_s = 39.6 \,km/h = 39.6 	imes \frac{

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$29 \,s$

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(B) The source (bus) is moving towards a stationary observer (person at the bus stop).Given:Velocity of source,$v_s = 39.6 \,km/h = 39.6 	imes \frac{5}{18} = 11 \,ms^{-1}$.Velocity of sound,$v = 330 \,ms^{-1}$.Time interval of source,$T = 30 \,s$.When the source moves towards a stationary observer,the apparent time interval $T^{\prime}$ between two consecutive pulses is given by the formula:$T^{\prime} = T \left( \frac{v - v_s}{v} ight)$Substituting the values:$T^{\prime} = 30 \left( \frac{330 - 11}{330} ight)$$T^{\prime} = 30 \left( \frac{319}{330} ight)$$T^{\prime} = \frac{319}{11} = 29 \,s$.Therefore,the person will hear the horn at an interval of $29 \,s$.

$A$ bus driving along at $39.6 \,km/h$ is approaching a person who is standing at the bus stop,while honking repeatedly at an interval of $30 \,s$. If the speed of sound is $330 \,ms^{-1}$,at what interval will the person hear the horn?

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