An observer is standing 500 m away from a vertical hill. Starting between the observer and the hill, a police van sounding a siren of frequency 1000 Hz moves towards the hill with a uniform speed. If the frequency of the sound heard directly from the siren is 970 Hz, the frequency of the sound heard after reflection from the hill (in Hz ) is about, (velocity of sound .=330 ms -1):

Question

An observer is standing 500 m away from a vertical hill. Starting between the observer and the hill, a police van sounding a siren of frequency 1000 Hz moves towards the hill with a uniform speed. If the frequency of the sound heard directly from the siren is 970 Hz, the frequency of the sound heard after reflection from the hill (in Hz ) is about, (velocity of sound .=330 ms -1): (A) 1042 (B) 1032 (C) 1022 (D) 1012

Tardigrade Admin · Accepted Answer

Sound heard directly f1 =fo((v/v+vs)) ∴ 970 =1000((330/330+vs)) or vs=10.2 m / s The frequency of reflected sound is given by f2=fo((v/v-vs)) =1000((330/330-10.2)) =(1000 × 330/319.8) ≈ 1032 Hz

1042

1032

1022

1012

Sound heard directly f1​=fo​(v+vs​v​) ∴970=1000(330+vs​330​) or vs​=10.2m/s The frequency of reflected sound is given by f2​=fo​(v−vs​v​) =1000(330−10.2330​) =319.81000×330​ ≈1032Hz

Sound heard directly
$f_{1} = f_{o} (\frac{v}{v + v _{s}})$
$∴ 970 = 1000 (\frac{330}{330 + v _{s}})$
or $v_{s} = 10.2 m / s$
The frequency of reflected sound is given by
$f_{2} = f_{o} (\frac{v}{v - v _{s}})$
$= 1000 (\frac{330}{330 - 10.2})$
$= \frac{1000 \times 330}{319.8}$
$\approx 1032 Hz$