A car is driving along at $39.6 k m p h$ is approaching a person who is standing at the bus stop. The driver of the car honks the horn at an interval of $30 s$ . If the speed of sound is $330 m / s$ , at what interval (in seconds) will the person hear the horn?

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Solution

Given speed of the car $v = 39.6 k m p h = 11 m / s$ .
Distance travelled by sound so that first honk to reaches the person $= d$
Time taken for first honk to reach the person $t_{1} = \frac{d}{330}$

Next honk is blown after $30 s$ ,the car travels a distance of $11 \times 30 m$ during this interval.

Distance travelled by sound so that second honk reaches the person $= (d - 11 \times 30) m$
Time taken for second honk to reach the person $t_{2} = \frac{d - 11 \times 30}{330}$

Time interval between two consecutive honks is $Δ t = t_{1} - t_{2} = 1.$
Since $t_{1} = 30 s \Rightarrow t_{2} = 29 s;$

(OR)

Applying Doppler Effect in sound:
$f = f_{0} (\frac{v + v_{0}}{v - v_{s}})$
$v \to$ speed of sound $= 330 m / s$
$v \to$ speed of observer $= 0 m / s$
$v_{s} \to$ speed of source $= 39.6 k m p h = 39.6 \times \frac{5}{18} m / s = 11 m / s$
$f_{0} \to$ Actual sound frequency $= \frac{1}{Time period} = \frac{1}{30} H z$
$f_{0} \to$ apparent sound frequency heard by the person $= \frac{1}{T^{'}}$

So, Applying Doppler Effect in sound:
$\frac{1}{T^{'}} = \frac{1}{30} (\frac{330 + 0}{330 - 11})$
$T' = 29 s$
At interval of $29 s$ the sound will be heard.

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A car is driving along at 39.6 kmph is approaching a person who is standing at the bus stop. The driver of the car honks the horn at an interval of 30 s. If the speed of sound is 330 m/s, at what interval (in seconds) will the person hear the horn?

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